0.1 Contents

• 1 Topology Spaces and Continuous Function
  • 1.1 Basic Definition of Topology
  • 1.2 Basis for a Topology
    • 1.2.1 Exercise
  • 1.3 The Order Topology
  • 1.4 The Product Topology
  • 1.5 The Subspace Topology
    • • 1.5.0.1 Exercise
  • 1.6 Closed Sets and Limit Points
    • 1.6.1 Exercise
  • 1.7 Continuous Function
    • 1.7.1 Exercise
  • 1.8 Metric Topology

1 Topology Spaces and Continuous Function

This is a note for general topology. The main reference is Topology.

The note is written in Pandoc Markdown.

My GitHub repository for this note.

1.1 Basic Definition of Topology

::: {#def:Topology .definition} Definition 1.1 (topology). A topology on a set $\mathbb{X}$ is a collection $\mathbb{T}$ of subsets of $\mathbb{X}$ having the following properties:

• $\emptyset$ and $\mathbb{X}$ are in $\mathbb{T}$

• The union of the elements of any sub collection of $\mathbb{T}$ is in $\mathbb{T}$

• The intersection of the elements of any finite sub collection of $\mathbb{T}$ is in $\mathbb{T}$ :::

::: {#def:TopologySpace .definition} Definition 1.2 (topology space). A topological space is a set $\mathbb{X}$ for which a topology $\mathbb{T}$ has been specified. :::

::: {#def:OpenSet .definition} Definition 1.3 (open set). A open set $\mathbb{U}$ is a subset of $\mathbb{X}$ that belongs to a topology $\mathbb{T}$ of $\mathbb{X}$. :::

::: {#def:OpenSets .definition} Definition 1.4 (open sets). A topology can also be called a open sets :::

::: {#def:DiscreteTopology .definition} Definition 1.5 (discrete topology). The set of all subsets of a set $\mathbb{X}$ formed a topology called discrete topology :::

::: {#def:TrivialTopology .definition} Definition 1.6 (trivial topology). The set consisting the set $\mathbb{X}$ and $\emptyset$ only formed a topology of $\mathbb{X}$ called trivial topology :::

::: {#def:FiniteComplementTopology .definition} Definition 1.7 (finite complement topology). Let $\mathbb{X}$ be a set. Let $\mathbb{T}_{\mathit{f}}$ be the collection of all subsets $\mathbb{U}$ of $\mathbb{X}$ such that $\mathbb{X} - \mathbb{U}$ either if a finite ¹ of is all of $\mathbb{X}$. Then $\mathbb{T}_{\mathit{f}}$ is a topology on $\mathbb{X}$, called the finite complement topology. :::

::: {#def:Comparable .definition} Definition 1.8 (finer, larger, strictly finer, strictly larger, coarser, smaller, strictly coarser, strictly smaller, comparable). Let $\mathbb{T}$ and $\mathbb{T'}$ be two topology on a given set $\mathbb{X}$. If $\mathbb{T}$ is a subset of $\mathbb{T'}$, we say that $\mathbb{T'}$ is finer or larger than $\mathbb{T}$. If $\mathbb{T}$ is a proper subset of $\mathbb{T'}$, we say that $\mathbb{T'}$ is strictly finer or strictly larger than $\mathbb{T}$. We also say that $\mathbb{T}$ is coarser or smaller or strictly coarser or strictly smaller than $\mathbb{T'}$. We say that $\mathbb{T}$ and $\mathbb{T'}$ is comparable if either $\mathbb{T}$ is a subset of $\mathbb{T'}$ or $\mathbb{T'}$ is a subset of $\mathbb{T}$. :::

1.2 Basis for a Topology

::: {#def:Basis .definition} Definition 1.9 (basis). If $\mathbb{X}$ is a set, a basis for a topology on $\mathbb{X}$ is a collection $\mathbb{B}$ of subsets of $\mathbb{X}$ (called basis elements) such that:

• For each $x \in \mathbb{X}$, there is at least one basis element $B$ containing $x$

• If $x$ belongs to the intersection of two basis elements $B_{1}$ and $B_{2}$, then there is another element $x \in B_{3} \in \mathbb{B}$ such that $B_{3} \subseteq B_{1} \cap B_{2}$ :::

::: {#def:TopologyGeneratedByBasis .definition} Definition 1.10 (topology generated by basis). Let $\mathbb{B}$ be a basis on $\mathbb{X}$ . Let $\mathbb{U}$ be a set containing all subsets $U$ of $\mathbb{X}$ such that for each element $x \in U$, there is $B \in \mathbb{B}$ that $x \in B \subseteq U$. Such $\mathbb{U}$ formed a topology on $\mathbb{X}$ , called topology $\mathbb{T}$ generated by $\mathbb{B}$ :::

::: lemma Lemma 1.1. Let $\mathbb{X}$ be a set. Let $\mathbb{B}$ be a basis for a topology $\mathbb{T}$ on $\mathbb{X}$ . Then $\mathbb{T}$ equals to the set of all possible unions of elements of $\mathbb{B}$ . :::

::: proof Proof. Let set $\mathbb{U}$ be the set of all possible unions of elements of $\mathbb{B}$ . For any $U \in \mathbb{U}$. $U = \cup B$ ² for some $B \in \mathbb{B}$. Thus, for every $x \in U$, there exist a $B' \in \mathbb{B}$ that $x \in B' \subseteq U$. Thus, $U \in \mathbb{T}$.

Conversely, for any $U \in \mathbb{T}$. For any $x \in U$, let $x \in B_{x} \in U$. Then, $U = \cup_{x \in U}B_{x}$. Thus, $U \in \mathbb{U}$.

Therefore, $\mathbb{U}$ equals to $\mathbb{T}$ . ◻ :::

::: lemma Lemma 1.2. ³ Let $\mathbb{X}$ be a topological space. Suppose that $\mathbb{C}$ is a collection of open sets of $\mathbb{X}$ such that for each open set $U$ of $\mathbb{X}$ and each $x \in U$, there is an element $C \in \mathbb{C}$ such that $x \in C \subseteq C$. Then $\mathbb{C}$ is a basis for the topology of $\mathbb{X}$ . :::

::: lemma Lemma 1.3. ⁴ Let $\mathbb{B}$ and $\mathbb{B'}$ be basis for the topologies $\mathbb{T}$ and $\mathbb{T'}$ , respectively, on $\mathbb{X}$ . Then the following are equivalent:

• $\mathbb{T'}$ is finer than $\mathbb{T}$

• For each $x \in \mathbb{X}$ and each basis element $B \in \mathbb{B}$ containing X, there is a basis element $B' \in \mathbb{B'}$ such that $x \in B' \subseteq B$. :::

::: {#def:StandardTopologyOnTheRealLine .definition} Definition 1.11 (standard topology on the real line). Let be $\mathbb{B} = \{ B | B = \{ x | a < x < b \}, a < b, a \in \mathbb{R}, b \in \mathbb{R} \}$. $\mathbb{B}$ formed a basis on real line. The topology generated by $\mathbb{B}$ is called the standard topology on the real line ⁵ . :::

::: {#def:LowerLimitTopologyOnTheRealLine .definition} Definition 1.12 (lower limit topology on the real line). Let be $\mathbb{B} = \{ B | B = \{ x | a \le x < b \}, a < b, a \in \mathbb{R}, b \in \mathbb{R} \}$. $\mathbb{B}$ formed a basis on real line. The topology generated by $\mathbb{B}$ is called the lower limit topology on the real line. When $\mathbb{R}$ is given this topology,we denote it by $\mathbb{R}_{l}$. :::

::: {#def:KTopologyOnTheRealLine .definition} Definition 1.13 (K-topology on the real line). Let be $\mathbb{B} = \{ B | B = \{ x | a < x < b \}, a < b, a \in \mathbb{R}, b \in \mathbb{R} \}$. Let $K = \{ x | x = \frac{1}{n}, n \in \mathbb{Z_{+}} \}$. $\mathbb{B} \cup \{ B - K | B \in \mathbb{B} \}$ formed a basis on real line. The topology generated by $\mathbb{B}$ is called the K-topology on the real line. When $\mathbb{R}$ is given this topology,we denote it by $\mathbb{R_{K}}$. :::

::: lemma Lemma 1.4. ⁶ The topologies $\mathbb{R}_{l}$ and $\mathbb{R_{K}}$ is strictly finer than the standard topology on $\mathbb{R}$ . :::

::: lemma Lemma 1.5. The topologies of $\mathbb{R}_{l}$ and $\mathbb{R_{K}}$ is not comparable. :::

::: proof Proof. Let $\mathbb{T}_{l}$ and $\mathbb{T_{K}}$ be topologies of $\mathbb{R}_{l}$ and $\mathbb{R_{K}}$ respectively. Let $K = \{ x | x = \frac{1}{n}, n \in \mathbb{Z_{+}} \}$.

We first proof that $\mathbb{T}_{l}$ is not finer than $\mathbb{T_{K}}$ . Let $U = \{ x | -1 < x < 1 \} - K, x = 0$. If there exist $B = \{ x | a \le x < b \} \in \mathbb{T}_{l}$ such that $x \in B \subseteq U$, then $0 < b < 1$. Thus, there exist $n \in \mathbb{Z_{+}}$ that $0 < \frac{1}{n} < b$. Thus $B$ is not a subset of $U$.

Then we proof that $\mathbb{T_{K}}$ is not finer than $\mathbb{T}_{l}$. Let $U' = \{ x | a' \le x < b' \}$. If there exist $B' = \{ x | a'' < x < b'' \} or \{ x | a'' < x < b'' \} - K$ such that ${a'} \in B \subseteq U$. Thus $a'' < a < b''$. Thus there exist $c$ that $a'' < x < a, x \in B ,x \notin U'$. Thus $B' \nsubseteq U'$.

Thus the topologies of $\mathbb{R}_{l}$ and $\mathbb{R_{K}}$ is not comparable. ◻ :::

::: {#def:Subbasis .definition} Definition 1.14 (subbasis). A subbasis $\mathbb{S}$ for a topology on $\mathbb{X}$ is a collection of subsets of $\mathbb{X}$ whose union equals $\mathbb{X}$ . The topology generated by the subbasis $\mathbb{S}$ is defined to be the collection $\mathbb{T}$ ⁷ of all unions of finite intersections of elements of $\mathbb{S}$ . :::

1.2.1 Exercise

1. Show that if $\mathbb{A}$ is a basis for a topology on $\mathbb{X}$ , then the topology generated by $\mathbb{A}$ equals the intersection of all topologies on $\mathbb{X}$ that contain $\mathbb{A}$ . Prove the same if $\mathbb{A}$ is a subbasis.

   ::: proof Proof. As a subbasis is also a basis, we will directly prove the case of subbasis here.

   Let $\mathbb{S} = \{ \mathbb{T}_{\alpha} \}$ be set contain all the topologies that contain $\mathbb{A}$ . Let $\mathbb{T}$ be the topology that $\mathbb{A}$ generated. Let $\displaystyle \mathbb{T}' = \cap\mathbb{T}_{\alpha}$.⁸

   First, $\mathbb{A} \subseteq \mathbb{T}_{\alpha}$. Thus, $\mathbb{T} \subseteq \mathbb{T}_{\alpha}$. Thus, $\mathbb{T} \subseteq \mathbb{T}'$.

   Also, $\mathbb{A} \subseteq \mathbb{T}$. Thus, $\mathbb{T} \in \mathbb{S}$. Thus, $\mathbb{T}' \subseteq \mathbb{T}$.

   Thus, $\mathbb{T} = \mathbb{T}'$ ◻ :::

1.3 The Order Topology

::: {#def:Interval .definition} Definition 1.15 (interval). Let $\mathbb{X}$ is a set having a simple order relation $<$. Given elements $a$ and $b$ of $\mathbb{X}$ such that $a < b$, there are four subsets of $\mathbb{X}$ that are called intervals determined by $a$ and $b$:

• $(a,b) = \{ x | a < x < b \}$

• $(a,b] = \{ x | a < x \le b \}$

• $[a,b) = \{ x | a \le x < b \}$

• $[a,b] = \{ x | a \le x \le b \}$

$(a,b)$ is called an open interval on $\mathbb{X}$ . $[a,b]$ is called an closed interval on $\mathbb{X}$ . $(a,b]$ and $[a,b)$ is called half-open intervals. :::

::: {#def:OrderTopology .definition} Definition 1.16 (order topology). ⁹ Let $\mathbb{X}$ be a set with a simple order relation; assume $\mathbb{X}$ has more than one element. Let $\mathbb{B}$ be the collection of all sets of the following types:

• All open intervals $(a,b)$ in $\mathbb{X}$ .

• All intervals of the form $[a_{0},b)$, where $a_0$ is the smallest element(if exist) of $\mathbb{X}$ .

• All intervals of the form $(a,b_{0}]$, where $b_0$ is the largest element(if exist) of $\mathbb{X}$ .

The collection $\mathbb{B}$ formed a basis for a topology on $\mathbb{X}$ , which is called the order topology. :::

::: {#def:Ray .definition} Definition 1.17 (ray). ¹⁰¹¹ If $\mathbb{X}$ is an ordered set, and $a$ is an element of $\mathbb{X}$ , there are four subsets of $\mathbb{X}$ that are called rays determined by $a$:

• $(a,+\infty) = \{ x | x > a \}$

• $(-\infty,a) = \{ x | x < a \}$

• $[a,+\infty) = \{ x | x \ge a \}$

• $(-\infty,a] = \{ x | x \le a \}$

$(a,+\infty)$ and $(-\infty,a)$ are called open rays. $[a,+\infty)$ and $(-\infty,a]$ are called closed rays. :::

1.4 The Product Topology

::: {#def:ProductTopology .definition} Definition 1.18 (product topology). Let $\mathbb{X}$ and $\mathbb{Y}$ be topological spaces. The product topology on $\mathbb{X} \times \mathbb{Y}$ having a basis $\mathbb{B}$ containing all sets of the form $U \times V$, where $U$ and $V$ is open sets of $\mathbb{X}$ and $\mathbb{Y}$ respectively. :::

::: theorem Theorem 1.1. ¹² If $\mathbb{B}$ and $\mathbb{C}$ is basis for the topology of $\mathbb{X}$ and $\mathbb{Y}$ respectively, then the collection $\mathbb{D} = \{ B \times C | B \in \mathbb{B} and C \in \mathbb{C} \}$ is a basis for the topology of $\mathbb{X} \times \mathbb{Y}$ :::

::: {#def:Projection .definition} Definition 1.19 (projection). Let $\pi_{1}: \mathbb{X} \times \mathbb{Y} \rightarrow \mathbb{X}$ be defined by the equation: $\pi_{1}(x,y) = x$

Let $\pi_{2}: \mathbb{X} \times \mathbb{Y} \rightarrow \mathbb{Y}$ be defined by the equation: $\pi_{1}(x,y) = y$

The maps $\pi_{1}$ and $\pi_{2}$ are called the projections of $\mathbb{X} \times \mathbb{Y}$ onto its first and second factors, respectively. :::

::: theorem Theorem 1.2. ¹³ The collection $\mathbb{S} = \{ \pi_{1}^{-1}(U) | U open in \mathbb{X} \} \cup \{ \pi_{2}^{-1}(V) | V open in \mathbb{Y} \}$ is a subbasis for the product topology on $\mathbb{X} \times \mathbb{Y}$. :::

::: {#def:BoxTopology .definition} Definition 1.20 (box topology). Let, $\mathbb{X} = \mathbb{X}_{1} \times \mathbb{X}_{2} \times \dots \times \mathbb{X}_{n} \text{or} \mathbb{X}_{1} \times \mathbb{X}_{2} \times \dots$

In the first case, all the sets of the form $U_{1} \times \dots \times U_{n}$ where $U_{i}$ is a open set of $\mathbb{X}_{i}$ form a basis.

In the second case, all the sets of the form $U_{1} \times U_{2} \times \dots$ where $U_{i}$ is a open set of $\mathbb{X}_{i}$ also form a basis.

Topology defined in this way was called a box topology. :::

::: {#def:ProjectionMapping .definition} Definition 1.21 (product topology). []{#def:ProductTopologyInfinite label="def:ProductTopologyInfinite"}¹⁴ Let, $\mathbb{X} = \mathbb{X}_{1} \times \mathbb{X}_{2} \times \dots \times \mathbb{X}_{n} \text{or} \mathbb{X}_{1} \times \mathbb{X}_{2} \times \dots$

Let $\pi_{i}$ be the projection function¹⁵ that $\pi_{i}: \mathbb{X} \rightarrow \mathbb{X}_{i}$

And if $x \in \mathbb{X}$ $\pi_{i}(x) = x_{i}$

All the set of the form $\pi_{i}^{-1}(U_{i})$ where $i$ is arbitrary and $U_{i}$ is an open set of $\mathbb{X}_{i}$ , form a subbasis of $\mathbb{X}$ . The topology generated by this subbasis is called product topology. And $\mathbb{X}$ is called a product space. :::

::: {#def:JTuple .definition} Definition 1.22 (J-tuple). Let $J$ be an index set¹⁶. Give a set $\mathbb{X}$ , a J-tuple is defined as a function $x: J \rightarrow \mathbb{X}$ . If $\alpha$ is an element of $J$ , $x(\alpha)$ is often denoted by $x_{\alpha}$ and is called the $\alpha\text{th}$ coordinate of $x$ . And the function $x$ itself is often denoted by the symbol $(x_{\alpha})_{\alpha \in J}$

The set of all J-tuples of elements of $\mathbb{X}$ is often denoted by $\mathbb{X}^{J}$ . :::

::: {#def:CartesianProduct .definition} Definition 1.23 (cartesian product). Let $\{A_{\alpha}\}_{\alpha \in J}$ be an indexed family of sets; let $\mathbb{X} = \bigcup_{\alpha \in J} A_{\alpha}$ . The cartesian product of this indexed family is denoted by $\displaystyle \prod_{\alpha \in J} A_{\alpha}$

And is defined to be the set of all J-tuples $(x_{\alpha})_{\alpha \in J}$ of elements of $\mathbb{X}$ such that $x_{\alpha} \in A_{\alpha}$ for each $\alpha \in J$ . That is, it is the set of all functions $x: J \rightarrow \bigcup_{\alpha \in J} A_{\alpha}$ such that $x(\alpha) \in A_{\alpha}$ for each $\alpha \in J$ . :::

::: {#theorem:ComparisonOfBoxProductTopology .theorem} Theorem 1.3 (Comparison of the box and product topologies). ¹⁷ The box topology on $\prod \mathbb{X}_{\alpha}$ has a basis all sets of the form $\prod U_{\alpha}$ where $U_{\alpha}$ is open in $X_{\alpha}$ for each $\alpha$ . The product topology on $\prod \mathbb{X}_{\alpha}$ has a basis all sets of the form $\prod U_{\alpha}$ where $U_{\alpha}$ is open in $X_{\alpha}$ for each $\alpha$ and $U_{\alpha}$ equals $\mathbb{X}_{\alpha}$ except for finitely many values of $\alpha$ . :::

::: theorem Theorem 1.4. ¹⁸ Suppose the topology on each space $\mathbb{X}_{\alpha}$ is given by a basis $\mathbb{X}_{\alpha}$ . The collection of all sets of the form $\prod_{\alpha \in J} B_{\alpha}$ where $B_{\alpha} \in \mathbb{B}_{\alpha}$ form a basis for the box topology on $\prod _{\alpha \in J} \mathbb{X}_{\alpha}$ .

The collection of all sets of the same form, where $B_{\alpha} \in \mathbb{B}_{\alpha}$ for finitely many indices $\alpha$ and $B_{\alpha} = \mathbb{X}_{\alpha}$ for all the remaining indices, will form a basis for the product topology $\prod_{\alpha \in J}\mathbb{X}_{\alpha}$ . :::

::: theorem Theorem 1.5. ¹⁹ Let $A_{\alpha}$ be a subspace of $\mathbb{X}_{\alpha}$ , for each $\alpha \in J$ . Then $\prod A_{\alpha}$ is a subspace of $\prod \mathbb{X}_{\alpha}$ if both products are given the box topology, or if both products are given the product topology. :::

::: theorem Theorem 1.6. ²⁰ If each space $\mathbb{X}_{\alpha}$ is a Hausdorff space, then $\prod \mathbb{X}_{\alpha}$ is a Hausdorff space in both the box and product topologies. :::

::: theorem Theorem 1.7. Let $\{ \mathbb{X}_{\alpha} \}$ be an indexed family of spaces; let $A_{\alpha} \subseteq \mathbb{X}_{\alpha}$ for each $\alpha$ . If $\prod \mathbb{X}_{\alpha}$ is given either the product or the box topology, then $\prod \overline{A_{\alpha}} = \overline{\prod A_{\alpha}}$ :::

::: proof Proof. Let $\pi_{\alpha}$ represent the projection mapping.

Let $x$ be an element of $\prod \mathbb{X}_{\alpha}$ . Let $V$ be an open set in $\prod \mathbb{X}_{\alpha}$ that containing $x$ .

If $x \in \prod \overline{A_{\alpha}}$ , then $\pi_{\alpha}(V)$ is a open set in $\mathbb{X}_{\alpha}$ that containing $x_{\alpha}$ . Thus $\pi_{\alpha}(V)$ intersect with $A_{\alpha}$ . Thus $V$ intersect with $\prod A_{\alpha}$ . Thus $x \in \overline{\prod A_{\alpha}}$ .

If $x \in \overline{\prod A_{\alpha}}$ . Let $U_{\alpha}$ be an open set of $A_{\alpha}$ that contain $x_{\alpha}$ . Let $V = \prod U_{\beta}$ such that $U_{\beta} = \begin{cases} \mathbb{X}_{\beta}, & \beta \neq \alpha \\ U_{\alpha}, & \beta = \alpha \end{cases}$ . It is obvious that $V$ is an open set that contain $x$ . Thus $V$ intersect with $\prod A_{\alpha}$ . Thus $U_{\alpha}$ intersect with $A_{\alpha}$ . Thus $x \in \prod \overline{A_{\alpha}}$ . ◻ :::

::: theorem Theorem 1.8. Let $f: A \rightarrow \prod_{\alpha \in J} \mathbb{X}_{\alpha}$ be given by the equation $f(a) = ( f_{\alpha}(a) )_{\alpha \in J}$ where $f_{\alpha} : A \rightarrow \mathbb{X}_{\alpha}$ for each $\alpha$ . Let $\prod \mathbb{X}_{\alpha}$ have the product topology. Then the function $f$ is continuous if and only if each function $f_{\alpha}$ is continuous. :::

::: proof Proof. Let $\pi_{\alpha}$ be the projection mapping

It is obvious that $f^{-1}(U) = \bigcap_{\alpha \in J} f_{\alpha}^{-1}(\pi_{\alpha}(U))$

If $f_{\alpha}$ is continuous. Let $V$ be a closed set of $\prod_{\alpha \in J} \mathbb{X}_{\alpha}$ . Then $\pi_{\alpha}(V)$ is closed. Then $f^{-1}(V)$ is intersect of closed set. Thus $\pi_{\alpha}(V)$ is closed. So $f$ is continuous.

If $f$ is continuous. Let $U_{\alpha}$ be an open set of $\mathbb{X}_{\alpha}$ . Let $U_{\beta} = \mathbb{X}_{\beta}$ if $\beta \neq \alpha$ . Let $V = \prod_{\beta \in J} U_{\beta}$ . It is obvious that $V$ is an open set of $\prod \mathbb{X}_{\alpha}$ . And

$$\begin{aligned} f^{-1}{V} &=& \bigcap_{\alpha \in J} f_{\alpha}^{-1}(\pi_{\alpha}(U)) \\ &=& f_{\alpha}^{-1}(U_{\alpha}) \end{aligned}$$

which is an open set in $A$ . Thus $f_{\alpha}$ is continuous. ◻ :::

1.5 The Subspace Topology

::: {#def:SubspaceTopology .definition} Definition 1.24 (subspace topology). Let $\mathbb{X}$ be a topological space with topology $\mathbb{T}$ . If $Y$ is a subset of $\mathbb{X}$ , the collection $\mathbb{T}_{Y} = \{ Y \cap U | U \in \mathbb{T} \}$ is a topology on $Y$ , called the subspace topology.

$Y$ is also called a subspace of $\mathbb{X}$ :::

::: lemma Lemma 1.6. ²¹ If $\mathbb{B}$ is basis for the topology of $\mathbb{X}$ , $Y$ is a subset of $\mathbb{X}$ then the collection $\mathbb{B}_{Y} = \{ B \cap Y | B \in \mathbb{B} \}$ is a basis for the subspace topology on $Y$ :::

::: lemma Lemma 1.7. ²² Let $Y$ be a subspace of $\mathbb{X}$ . If $U$ is open in $Y$ and $Y$ is open in $\mathbb{X}$ , then $U$ is open in $\mathbb{X}$ . :::

::: theorem Theorem 1.9. ²³ If $A$ is a subspace of $\mathbb{X}$ and $B$ is a subspace of $\mathbb{Y}$ , then the product topology on $A \times B$ is the same as the topology $A \times B$ inherits as a subspace of $\mathbb{X} \times \mathbb{Y}$ :::

::: proof Proof. Let $\mathbb{B}_{\mathbb{X}}$ and $\mathbb{B}_{\mathbb{Y}}$ and $\mathbb{B}_{\mathbb{XY}}$ be basis of topology of $\mathbb{X}$ and $\mathbb{Y}$ and $\mathbb{X} \times \mathbb{Y}$ respectively. Let $\mathbb{B}_{\mathbb{X}}'$ and $\mathbb{B}_{\mathbb{Y}}'$ and $\mathbb{B}_{\mathbb{XY}}'$ be basis of topology of $A$ and $A$ and $A \times B$ respectively. We will show that $\mathbb{B}_{\mathbb{X}}' \times \mathbb{B}_{\mathbb{Y}}' = \mathbb{B}_{\mathbb{XY}}'$. Thus, the product topology on $A \times B$ is the same as the topology $A \times B$ inherits as a subspace of $\mathbb{X} \times \mathbb{Y}$ .

First, every element in $\mathbb{B}_{\mathbb{X}}' \times \mathbb{B}_{\mathbb{Y}}'$ can be represented by $B_{A} \cap A \times B_{B} \cap B = B_{A} \times B_{B} \cap A \times B \in \mathbb{B}_{\mathbb{XY}}'$ where $B_{A} \in \mathbb{B}_{\mathbb{X}}', B_{B} \in \mathbb{B}_{\mathbb{Y}}'$. Thus $\mathbb{B}_{\mathbb{X}}' \times \mathbb{B}_{\mathbb{Y}}' \subseteq \mathbb{B}_{\mathbb{XY}}'$.

Next, we show that $\mathbb{B}_{\mathbb{X}}' \times \mathbb{B}_{\mathbb{Y}}'$ generate the topology $A \times B$ inherits as a subspace of $\mathbb{X} \times \mathbb{Y}$ . For any open set $U$ in $\mathbb{X} \times \mathbb{Y}$ , and $\forall x \in U \cap A \times B, \exists B_{\mathbb{X}} \times B_{\mathbb{Y}} \in \mathbb{B}_{\mathbb{XY}}, x \in B_{\mathbb{X}} \times B_{\mathbb{Y}} \subseteq \mathbb{X} \times \mathbb{Y}$. Thus $x \in B_{\mathbb{X}} \times B_{\mathbb{Y}} \cap A \times B \subseteq A \times B, B_{\mathbb{X}} \times B_{\mathbb{Y}} \cap A \times B \in \mathbb{B}_{\mathbb{X}}' \times \mathbb{B}_{\mathbb{Y}}'$. Thus $\mathbb{B}_{\mathbb{X}}' \times \mathbb{B}_{\mathbb{Y}}'$ generate the topology $A \times B$ inherits as a subspace of $\mathbb{X} \times \mathbb{Y}$ .gi ◻ :::

::: {#def:OrderedSquare .definition} Definition 1.25 (ordered square). Let $I = [0,1]$. The set $I \times I$ in the dictionary order ²⁴ topology will be called ordered square, and denoted by $I_{o}^{2}$ :::

::: {#def:Convex .definition} Definition 1.26 (convex). Given an ordered set $\mathbb{X}$ , let us say that a subset $\mathbb{Y}$ of $\mathbb{X}$ is convex in $\mathbb{X}$ if for each pair of points $a < b$ of $\mathbb{Y}$ , the entire interval $(a,b)$ of points of $\mathbb{X}$ lies in $\mathbb{Y}$ :::

::: theorem Theorem 1.10. ²⁵ Let $\mathbb{X}$ be an ordered set in the order topology. Let $\mathbb{Y}$ be a subset of $\mathbb{X}$ that is convex in $\mathbb{X}$ . Then the order topology on $\mathbb{Y}$ is the same as the topology $\mathbb{Y}$ inherits as a subspace of $\mathbb{X}$ . :::

::: proof Proof. Consider the ray $(a,+\infty)$ in $\mathbb{X}$ . If $a \in \mathbb{Y}$, then $(a,+\infty) \cap \mathbb{Y} = \{ x | x \in \mathbb{Y} and x > a \}$

This is an open ray of the ordered set of $\mathbb{Y}$ . if $a \notin Y$, then $a$ is either a lower bound on $\mathbb{Y}$ or an upper bound on $\mathbb{Y}$ , since $\mathbb{Y}$ is convex. In the former case, the set $(a,+\infty) \cap \mathbb{Y}$ equals all of $\mathbb{Y}$ , in the latter case, it is empty.

A similar remark shows that the intersection of the rat $(-\infty,a)$ with $\mathbb{Y}$ is either an open ray of $\mathbb{Y}$ , or $\mathbb{Y}$ itself, or empty. Since the sets $(a,+\infty) \cal \mathbb{Y}$ and $(-\infty,a) \cap \mathbb{Y}$ form a subbasis for the subspace topology on $\mathbb{Y}$ , and since each is open in the order topology, the order topology contains the subspace topology.

To prove the reverse, note that any open ray of $\mathbb{Y}$ equals the intersection of an open ray of $\mathbb{X}$ with $\mathbb{Y}$ , so it is open in the subspace topology on $\mathbb{Y}$ . Since the open rays of $\mathbb{Y}$ are a subbasis for the order topology on $\mathbb{Y}$ , this topology is contained in the subspace topology. ◻ :::

1.5.0.1 Exercise

1. A map $\mathit{f} : \mathbb{X} \rightarrow \mathbb{Y}$ is said to be a open map[]{#def:OpenMap label="def:OpenMap"} if for every open set $U \subseteq \mathbb{X}$, the set $\mathit{f}(U)$ is open in $\mathbb{Y}$ . Show that $\pi : \mathbb{X} \times \mathbb{Y} \rightarrow \mathbb{X}$ is open map.

   ::: proof Proof. An open set in $\mathbb{X} \times \mathbb{Y}$ can be represented by $\cup( U_{i} \times U_{i}' )$ where $U_{i}, U_{i}'$ are open sets in $\mathbb{X}$ , $\mathbb{Y}$ , respectively.

   Also, $\cup( U_{i} \times U_{i}' ) = \cup( U_{i} ) \times \cup( U_{i}' )$

   Thus, $\pi(\cup( U_{i} \times U_{i}' )) = \cup( U_{i} )$

   Thus, $\pi(U)$ is open in $\mathbb{X}$ . ◻ :::

2. Let $\mathbb{X}$ and $\mathbb{X'}$ denote a single set in the topologies $\mathbb{T}$ and $\mathbb{T'}$ , respectively; let $\mathbb{Y}$ and $\mathbb{Y'}$ denote a single set in the topologies $\mathbb{U}$ and $\mathbb{U'}$ , respectively. ²⁶ Assume these sets are nonempty.

   1. Show that if $\mathbb{T}' \supseteq \mathbb{T}$ and $\mathbb{U}' \supseteq \mathbb{U}$ , then the product topologies $\mathbb{X'} \times \mathbb{Y'}$ is finer than the product topology on $\mathbb{X} \times \mathbb{Y}$ .

   2. Does the converse of the previous statement hold?

3. Show that the countable collection²⁷ $\{ (a,b) \times (c,d) | a < b, c < d, a \in \mathbb{Q}, b \in \mathbb{Q}, c \in \mathbb{Q}, d \in \mathbb{Q} \}$ is a basis for $\mathbb{R}^{2}$

   ::: proof Proof. This is obvious if you prove that $(a,b) \times (c,d)$ is a rectangle in the $\mathbb{R}^{2}$ plane. ◻ :::

4. Let $\mathbb{X}$ be an ordered set. If $\mathbb{Y}$ is a proper subset of $\mathbb{X}$ that is convex in $\mathbb{X}$ prove that $\mathbb{Y}$ may not be an interval or a ray in $\mathbb{X}$ .

   ::: proof Proof. Let $\mathbb{X} = \mathbb{R}^{2}$ with dictionary order. Then $Y = \{ (x,y) | -1 \le x \le 1 \}$ is convex in $\mathbb{X}$ , however it is not an interval or a ray. ◻ :::

   There is a false prove given by myself.

   ::: proof Proof. Let $\mathbb{S}$ be a set that contain all intervals and rays of $\mathbb{Y}$ . We define a partial order on $\mathbb{S}$ by inclusion. So if there is a chain in $\mathbb{S}$ : $S_{1} \subseteq S_{2} \subseteq S_{3} \dots$

   Let $S = S_{1} \cup S_{2} \cup S_{3} \cup \dots$

   Thus, $S$ is an upper bound of the chain.

   Thus, by Zorn's Lemma, there is a maximal element of $\mathbb{S}$ , say $U$ , then we prove that $U = \mathbb{Y}$ .

   If $U \neq \mathbb{Y}$ , then $\exists x, x \in \mathbb{Y} - U$ .

   If $U$ is a ray say $(a,+\infty)$ . Then $x < a$ , thus $U \subseteq (x,+\infty) \subseteq \mathbb{B}$ , then there is contradiction with the maximal element.

   If $U$ is an interval, the circumstance is similar with the proof of $U$ is a ray.

   Thus $\mathbb{Y}$ is a ray or an interval. ◻ :::

   However, there is issue with this proof, the set $S$ does exists. However, it may not be an interval or ray, so it may not be contained in $\mathbb{S}$

1.6 Closed Sets and Limit Points

::: {#def:Closed .definition} Definition 1.27 (closed). ²⁸ A subset $A$ of a topological space is said to be closed if the set $\mathbb{X}-A$ is open. :::

::: theorem Theorem 1.11. ²⁹ Let $\mathbb{X}$ be a topological space. Then the following conditions hold

1. $\emptyset$ and $\mathbb{X}$ are closed.

2. Arbitrary intersections of closed sets are closed

3. Finite unions of closed sets are closed :::

::: {#def:ClosedIn .definition} Definition 1.28 (closed in). Let $\mathbb{X}$ be a topological space; let $\mathbb{Y}$ be a subspace of $\mathbb{X}$ . We say that a set $A$ is closed in $\mathbb{Y}$ if $A$ is a subset of $\mathbb{Y}$ and $A$ is closed in the subspace topology of $\mathbb{Y}$ :::

::: theorem Theorem 1.12. Let $\mathbb{Y}$ be a subspace of $\mathbb{X}$ . Then a set $A$ is closed in $\mathbb{Y}$ if and only if it equals the intersection of a closed set of $\mathbb{X}$ with $\mathbb{Y}$ :::

::: proof Proof. First we proof that if $A$ is closed in $\mathbb{Y}$ , then $\exists B \subseteq \mathbb{X}, B \cap \mathbb{Y} = A$ . As the origin topology form a surjective map to its subspace topology, there exists a $B$ closed in $\mathbb{X}$ that $\mathbb{Y} - A = ( \mathbb{X} - B ) \cap \mathbb{Y}$ . Then $B \cap \mathbb{Y} = A$

Conversely, if $\exists B \subseteq \mathbb{X}, B \cap \mathbb{Y} = A$ . Then, $\mathbb{Y} - A = ( \mathbb{X} - B ) \cap \mathbb{Y}$ . Then $\mathbb{X}-B$ is open in $\mathbb{Y}$ , $\mathbb{Y}-A$ is open in $\mathbb{Y}$ . Then $A$ is closed in $\mathbb{Y}$ ◻ :::

::: theorem Theorem 1.13. ³⁰ Let $\mathbb{Y}$ be a subspace of $\mathbb{X}$ . If $A$ is closed in $\mathbb{Y}$ and $\mathbb{Y}$ is closed in $\mathbb{X}$ , then $A$ is closed in $\mathbb{X}$ . :::

::: {#def:Interior .definition} Definition 1.29 (interior). Given a subset $A$ of a topological space $\mathbb{X}$ , the interior of $A$ is defined as the union of all open sets contained in $A$ . Denoted by $Int(A)$. :::

::: {#def:Closure .definition} Definition 1.30 (closure). Given a subset $A$ of a topological space $\mathbb{X}$ , the closure of $A$ is defined as the intersection of all closed sets containing $A$ . Denoted by $Cl(A)$ or $\overline{A}$ :::

::: theorem Theorem 1.14. ³¹³² Let $\mathbb{Y}$ be a subspace of a topological space $\mathbb{X}$ ; let $A$ be a subset of $\mathbb{X}$ . Let $\overline{A}$ denote the closure of $A$ in $\mathbb{X}$ . Then the closure of $A$ in $\mathbb{Y}$ equals $\overline{A} \cap \mathbb{Y}$ :::

::: {#def:Intersect .definition} Definition 1.31 (intersect). We say that a set $A$ intersects $B$ if $A \cap B$ is not empty. :::

::: theorem Theorem 1.15. Let $A$ be a subset of the topological space $\mathbb{X}$

1. The $x \in \overline{A}$ if and only if every open set $U$ containing $x$ intersect $A$ .

2. Supposing the topology of $\mathbb{X}$ is given by a basis, then $x \in \overline{A}$ if and only if every basis element $B$ containing $x$ intersects $A$ :::

::: proof Proof. There are only two types of closed set $U$ in $\mathbb{X}$ :

1. $U \supseteq \overline{A}$

2. $U \cap A \neq A$

Thus, there are only two types of open set $U$ in $\mathbb{X}$ respectively.

1. $U$ does not intersects $A$ .

2. $U \cap \overline{A} \neq \emptyset$

3. If $x \in \overline{A}$ , then every open set containing $x$ is the open set of second type, thus every open set containing $x$ intersects $A$

   If every open set containing $x$ intersect $\mathbb{A}$ , suppose $x \notin \overline{A}$ . Then $\mathbb{X} - \overline{A}$ is a open set containing x, however, it does not intersects $A$ . Thus, $x \in \overline{A}$ .

4. If $x \in \overline{A}$ , as every basis element of $\mathbb{X}$ is a open set, thus every basis element containing $x$ intersects $\mathbb{A}$

   If every open set containing $x$ intersect $\mathbb{A}$ , suppose $x \notin \overline{A}$ .

   As every open sets can be represented by union of basis. Let $\mathbb{X} - \overline{A} = B_{1} \cup B_{2} \cup B_{3} \cup \dots \cup B_{1}' \cup B_{2}' \cup B_{3}' \cup \dots$ where $B$ are bases containing $x$ , and $B'$ are bases that does not contain $x$ .

   Thus, $x \in B_{1} \cup B_{2} \cup B_{3} \cup \dots \subseteq \mathbb{X} - \overline{A}$

   Then $B_{1} \cup B_{2} \cup B_{3} \cup \dots$ that is a open set can be generated by all the bases containing $x$ , however, that does not intersects $A$ . So, $x \in \overline{A}$ .

 ◻ :::

::: {#def:Neighbourhood .definition} Definition 1.32 (neighbourhood). ³³ If we say $U$ is a neighbourhood of $x$ in $\mathbb{X}$ , then $U$ is an open set in $\mathbb{X}$ containing $x$ :::

::: {#def:LimitPoint .definition} Definition 1.33 (limit point, point of accumulation, cluster point). ³⁴ If $A$ is a subset of topological space $\mathbb{X}$ .We say that $x$ is a limit point of $A$ if and only if every open sets containing $x$ intersects A with some points other than $x$ .

This condition is also equivalent to the condition that if $x$ is a limit point of $A$ if and only if $x \in \overline{A-\{x\}}$ :::

::: theorem Theorem 1.16. ³⁵ Let $A$ be a subset of topological space $\mathbb{X}$ ; let $A'$ be the set of all limit points of $A$ . Then $\overline{A} = A \cup A'$ :::

::: corollary Corollary 1.1. ³⁶ A subset of a topological space is closed if and only if it contains all its limit point. :::

::: {#def:Converge .definition} Definition 1.34 (converge). ³⁷ We say that a sequence of $x_{1}, x_{2}, x_{3} \dots$ converge to $x$ . When for every neighbourhood $U$ of $x$ , there exists a positive integer $N$ , such that for all $n > N$ , $x_{n} \in U$ . :::

::: {#def:HausdorffSpace .definition} Definition 1.35 (Hausdorff space). A topological space is called a Hausdorff space, if for every distinct $x_{1}$ , $x_{2}$ in $\mathbb{X}$ , there exists disjoint neighbourhood of $U_{1}$ , $U_{2}$ of $x_{1}$ , $x_{2}$ in $\mathbb{X}$ . :::

::: theorem Theorem 1.17. ³⁸³⁹ Every finite point set in a Hausdorff space $\mathbb{X}$ is closed. :::

::: proof Proof. Let $A$ be a finite point set in a Hausdorff space $\mathbb{X}$ .

Suppose $A$ only have one element. Then for every $x\in\mathbb{X}-A$ , there exists a neighbourhood of $x$ that does not intersect with $A$ . So $A$ is closed.

Suppose $A$ is a closed finite point set. We take $x_{0}\in\mathbb{X}-A$ . As finite union of closed set is closed, $A\cup\{x_{0}\}$ is closed.

Then, from induction, all finite point set in a Hausdorff space is closed. ◻ :::

::: theorem Theorem 1.18. If $\mathbb{X}$ is a Hausdorff space, then a sequence of points in $\mathbb{X}$ converges to at most one point. :::

::: proof Proof. Suppose that the following sequence $x_{1}, x_{2}, x_{3}\dots$

Converge to more than one points say $y_{1}, y_{2}, y_{3}\dots$

Then there exists $n_{1}, n_{2}, n_{3}\dots, U_{1}, U_{2}, U_{3}\dots$

Such that for $n > n_{i}$ $x_{n} \in U_{i}, y_{i} \in U_{i}$

If we take disjoint $U_{1}, U_{2}$ which is possible as this is a Hausdorff space.

Then the previews condition does not stand. So, every sequence of points in a Hausdorff space can only converge to at most one point. ◻ :::

::: {#def:Limit .definition} Definition 1.36 (limit). If a sequence $x_{n}$ of points in Hausdorff space converge to the point $x$ , we denote this by $x_{n} \rightarrow x$ and we say the limit of $x_{n}$ is $x$ . :::

::: {#def:T1Axiom .definition} Definition 1.37 ( $T_{1}$ axiom). The condition that all finite point set of a topological space is closed is called $T_{1}$ axiom. :::

::: theorem Theorem 1.19. Let $\mathbb{X}$ be a space satisfying the $T_{1}$ axiom; let $A$ be a subset of $\mathbb{X}$ . Then the point $x$ is a limit point of $A$ if and only if every neighbourhood of $x$ contains infinitely many points of $A$ . :::

::: proof Proof. If every neighbourhood of $x$ contains infinitely many point of $A$ . Than every neighbourhood of $x$ intersect with $A$ with infinite element other than $x$ , then $x$ is a limit point of $A$ .

If $x$ is a limit point of $A$ . Suppose that there exists a open set $U$ containing $x$ and intersect with $A$ for finite many points. Let $U' = U \cap ( A - x )$

Then, $x \notin U'$ . Let $U'' = U - U'$

Then $U''$ is open as $U'$ is a finite point set and $U'' = U - U' = U \cap (\mathbb{X} - U')$

Also, $x \in U''$ . Thus, $U''$ is a open set containing $x$ that only intersect $A$ with $x$ or do not intersect $A$ . This is a contradiction of x is a limit point. Thus there does not exists a open set $U$ containing $x$ and intersect with $A$ for finite many points. ◻ :::

::: theorem Theorem 1.20. ⁴⁰ Every simply ordered set is a Hausdorff space in order topology. :::

::: theorem Theorem 1.21. ⁴¹ The product of two Hausdorff space is a Hausdorff space. :::

::: theorem Theorem 1.22. ⁴² A subspace of a Hausdorff space is a Hausdorff space. :::

1.6.1 Exercise

1. Give an counter example why $\overline{\cup A_{\alpha}} = \cup \overline{A_{\alpha}}$ dose not hold.

   ::: proof Proof. Consider the X be the K-topology on the real line.

   Let

   $$\begin{aligned} A_{n} &=& (\frac{1}{n+1},\frac{1}{n}), n \in \mathbb{Z}_{+} \\ A &=& \cup A_{n} \end{aligned}$$

   Then

   $$\begin{aligned} \overline{A_{n}} &=& [\frac{1}{n+1},\frac{1}{n}] \\ \cup \overline{A_{n}} &=& (0,1] \end{aligned}$$

   However, as every neighbourhood of $0$ intersect $\cup A_{\alpha}$ . $0 \in \overline{\cup A_{\alpha}}$ .

   Thus, $\overline{\cup A_{\alpha}} \neq \cup \overline{A_{\alpha}}$ ◻ :::

2. Prove that $\overline{A-B} \supseteq \overline{A} - \overline{B}$

   ::: proof Proof. If $x \in \overline{A} - \overline{B}$ . Then

   $$\begin{aligned} x \in \overline{A}, x \notin \overline{B} \end{aligned}$$

   .

   Thus for open set $U$ containing $x$

   $$\begin{aligned} &\exists& U_{1} \cap B = \emptyset \\ &\forall& U \cap A \neq \emptyset \end{aligned}$$

   Suppose that $x \notin \overline{A-B}$ . Then\

   $$\begin{aligned} \exists U_{0} \cap (A-B) = \emptyset \end{aligned}$$

   Thus,

   $$\begin{aligned} U_{0} \cap A \subseteq B \end{aligned}$$

   Thus,

   $$\begin{aligned} U_{1} \cap B &=& \emptyset \\ U_{1} \cap U_{0} \cap A &=& \emptyset \end{aligned}$$

   As $U_{1} \cap U_{0}$ is an open set containing $x$ , so there is contradiction with $x \in \overline{A}$ . Thus $x \in \overline{A-B}$ . ◻ :::

3. A diagonal[]{#def:Diagonal label="def:Diagonal"} is a subset $\Delta = \{ x \times x | x \in \mathbb{X} \}$ of the product topology $\mathbb{X \times X}$ where $\mathbb{X}$ is a topological space. Show that the diagonal is closed in $\mathbb{X \times X}$ if and only if $\mathbb{X}$ is a Hausdorff space.

   ::: proof Proof. If $\mathbb{X}$ is a Hausdorff space. For every element $x \times y$ of $\mathbb{X} \times \mathbb{X}$ that not in $\Delta$ . We take disjoint set $U_{x},U_{y}$ where $x \in U_{x}, y \in U_{y}$ . Then $\mathbb{X} \times \mathbb{X} - \Delta = \cup_{x \neq y} U_{x} \times U_{y}$ . Where $\cup_{x \neq y} U_{x} \times U_{y}$ is an open set. Thus $\Delta$ is a closed set.

   Conversely, if $\Delta$ is a closed set, suppose that $\mathbb{X}$ is not a Hausdorff space. Then there exists distinct $x, y$ such that every neighbourhood of $x$ and $y$ intersect. Let $\mathbb{B}$ be a basis of topology of $\mathbb{X}$ . Then $x \times y \in \mathbb{X} \times \mathbb{X} - \Delta$ . However we cannot find $B_{1}, B_{2} \in \mathbb{B}, x \times y \in B_{1} \times B_{2} \subset \mathbb{X} \times \mathbb{X} - \Delta$ . Then $\Delta$ is not a closed set. So there is a contradiction, then $\mathbb{X}$ must be a Hausdorff space. ◻ :::

4. Prove that $T_{1}$ axiom is equivalent to the condition such that for every distinct pair $x,y$ of $\mathbb{X}$ , there exists neighbourhood of $x$ does not contain $y$ .

   ::: proof Proof. First if $T_{1}$ axiom hold, then for every pair $x,y$ , the neighbourhood $\mathbb{X}-\{y\}$ of $x$ does not contain $y$ , so the second condition hold.

   Conversely, if the second condition hold. Suppose that we can find a finite points set say $\{x_{1}, x_{2}, x_{3} \dots$ }, then there must exists $x \in \{x_{1}, x_{2}, x_{3} \dots$ } such that the set $\{x\}$ is not closed. Then $\overline{\{x\}} - \{x\} \neq \emptyset$ . Let $y \in \overline{\{x\}} - \{x\}$ , then every neighbourhood of y must contain $x$ , this is a contradiction to the second condition, so the $T_{1}$ axiom must hold. ◻ :::

5. If $A \subseteq \mathbb{X}$ , we define the boundary[]{#def:Boundary label="def:Boundary"} of $A$ by the equation $\text{Bd} A = \overline{A} \cap \overline{\mathbb{X}-A}$

   1. Show that $\text{Int}A$ and $\text{Bd} A$ are disjoint and $\overline{A} = \text{Int} A \cup \text{Bd} A$ .

      ::: proof Proof. For every $x \in \text{Bd}A$ , every open set contain $x$ must intersect $A$ and $\mathbb{X}-A$ so, there is no open set $U$ contain $x$ , $U \subseteq A$ .

      For every $x' \in \text{Int}A$ , there exists $U' \subseteq A$ , so $\text{Bd}A$ and $\text{Int}A$ are disjoint sets.

      For every $x \in \overline{A}$ , $x \in \text{Bd}A$ or $x \notin \text{Bd}A$ . We discuss the condition that $x \notin \text{Bd}A$ .

      Then $x \notin \overline{\mathbb{X}-A}$ , then there exists a open set $U$ containing $x$ , that does not intersect with $\mathbb{X}-A$ . Thus $U \subseteq A$ , thus $x \in \text{Int}A$ . So $\overline{A} \subseteq \text{Int} A \cup \text{Bd} A$ .

      Then, $\text{Bd}A \subseteq \overline{A}$ , $\text{Int}A \subseteq A \subseteq \overline{A}$ . Thus, $\overline{A} \supseteq \text{Int} A \cup \text{Bd} A$

      So, $\overline{A} = \text{Int} A \cup \text{Bd} A$ ◻ :::

   2. Show that $\text{Bd}A = \emptyset$ if and only if $A$ is both open and closed.

      ::: proof Proof. So, $\text{Int}A = \overline{A}$ , then $\text{Bd}A = \emptyset$ follows directly from $\overline{A} = \text{Int} A \cup \text{Bd} A$ . ◻ :::

   3. Show that $U$ is open if and only if $\text{Bd}U = \overline{U}-U$ .

      ::: proof Proof. Suppose U is open. Then $\overline{\mathbb{X}-U} = \mathbb{X} - U$ . Then for every $x \in U$ , $x \notin \mathbb{X} - U, x \notin \overline{\mathbb{X}-U}$ . Thus $\overline{U} \cap \overline{\mathbb{X}-U}=\overline{U}-U$ .

      Conversely, suppose $\text{Bd}U = \overline{U}-U$ . Then for every $x \in U$ , $x \notin \text{Bd}U$ . Then as $\overline{U} = \text{Int}U\cup \text{Bd}U$ , $x \in \text{Int}U$ . So $\text{Int}U \supseteq U$ . Thus $U = \text{Int}U$ . Thus, $U$ is open. ◻ :::

1.7 Continuous Function

::: {#def:ContinuousRelativeTo .definition} Definition 1.38 (continuous). []{#def:Continuous label="def:Continuous"}⁴³ Let $\mathbb{X}$ and $\mathbb{Y}$ be topological spaces. A function $f: \mathbb{X}\rightarrow \mathbb{Y}$ is said to be continuous if for each open subset $V$ of $\mathbb{Y}$ , the set $f^{-1}(V)$ is an open subset of $\mathbb{X}$ . :::

::: theorem Theorem 1.23. Let $\mathbb{X}$ and $\mathbb{Y}$ be topological spaces; let $f: \mathbb{X}\rightarrow\mathbb{Y}$ . Then the following are equivalent.

1. $f$ is continuous.

2. For every subset $A$ of $X$ , one has $f(\overline{A})\subseteq\overline{f(A)}$ .

3. For every closed set $B$ of $\mathbb{Y}$ , the set $f^{-1}(B)$ is closed in $\mathbb{X}$ .

4. For each $x\in\mathbb{X}$ and each neighbourhood of $V$ of $f(x)$ , there is a neighbourhood $U$ of $x$ such that $f(U) \subseteq V$ . :::

::: proof Proof. 1 $\Rightarrow$ 3:

Let $A$ be a open set in $\mathbb{Y}$ . $f^{-1}(\mathbb{Y}-A) = \mathbb{X} - f^{-1}(A)$ .

3 $\Rightarrow$ 1:

Let $A$ be a closed set in $\mathbb{Y}$ . $f^{-1}(\mathbb{Y}-A) = \mathbb{X} - f^{-1}(A)$ .

1 $\Rightarrow$ 2:

For $x \in \overline{A}$ , we take a open set $f(x) \in U \subseteq \mathbb{Y}$ . Thus $x \in f^{-1}(U) \cap A \neq \emptyset$ . Thus $U \cap f(A) \neq \emptyset$ . So $f(x) \in \overline{f(A)}$ . Thus $f(\overline{A})\subseteq\overline{f(A)}$ .

2 $\Rightarrow$ 3:

Suppose $f$ is not continuous. Then there must exists $V$ , such that $f^{-1}(V) = U$ is not closed. Thus $\overline{U} \supset B = f^{-1}(A)$ . Thus $f{\overline{B}} \supset A$ . However $f(\overline{B}) \subseteq \overline{f(B)} = A$ . There is a contradiction. So $f$ must be continuous.

1 $\Rightarrow$ 4:

For every neighbourhood $V$ of $f(x)$ , $f^{-1}(V)$ is a neighbourhood of $x$ that $f(f^{-1}(V)) \subseteq V$ .

4 $\Rightarrow$ 1:

We take a open set $V$ of $\mathbb{Y}$ . Let $S$ be the collection of all open set $U$ in $\mathbb{X}$ such that $f(U) \subseteq V$ . The set cannot be empty unless $f^{-1}(V) = \emptyset$ . Let $U_{0}$ denote the union of all the element in $S$ . We prove that $U_{0} = f^{-1}(V)$ .

For all element $x \in U_{0}$ , $f(x) \in V$ . Thus $U_{0} \subseteq f^{-1}(V)$ .

For all element $x \in f^{-1}(V)$ . There is a $U'$ such that $x \in U', f(U')\subseteq V$ . This follows from the condition 4. Thus $U' \in S$ . Thus $x \in U_{0}$ . Thus $U_{0} \subseteq f^{-1}(V)$ . As $U_{0}$ is union of open set, $U_{0}$ is also open. Thus, $f^{-1}(V)$ is also open.

Thus $f$ is continuous. ◻ :::

::: {#def:Homeomorphism .definition} Definition 1.39 (homeomorphism). ⁴⁴ Let $\mathbb{X}$ and $\mathbb{Y}$ be topological space; let $f: \mathbb{X} \rightarrow \mathbb{Y}$ be a bijection. If both the function $f$ and the inverse function $f^{-1}: \mathbb{Y} \rightarrow \mathbb{X}$ are continuous, then f is called a homeomorphism :::

::: {#def:TopologicalImbedding .definition} Definition 1.40 (topological imbedding). Suppose that $f: \mathbb{X} \rightarrow \mathbb{Y}$ is an injective continuous map, where $\mathbb{X}$ and $\mathbb{Y}$ are topological spaces. Let $\mathbb{Z}$ be the image set $f(\mathbb{X})$ , considered as a subspace of $\mathbb{Y}$ ; then the function $f': \mathbb{X} \rightarrow \mathbb{Z}$ obtained by restricting the range of $f$ is bijective. If $f'$ happens to be a homeomorphism of $\mathbb{X}$ with $\mathbb{Z}$ , we say that the map $f: \mathbb{X} \rightarrow \mathbb{Y}$ is a topological imbedding, or simply an imbedding, of $\mathbb{X}$ in $\mathbb{Y}$ . :::

::: {#theorem:RulesForConstructingContinuousFunctions .theorem} Theorem 1.24 (Rules for constructing continuous functions). Let $\mathbb{X}$ , $\mathbb{Y}$ , and $\mathbb{Z}$ be topological spaces.

1. (Constant function) If $f: \mathbb{X} \rightarrow \mathbb{Y}$ maps all of $\mathbb{X}$ into the single point $y_{0}$ of $\mathbb{Y}$ , then $f$ is continuous.

2. (Inclusion) If $A$ is a subspace of $\mathbb{X}$ , the inclusion function $j: A \rightarrow \mathbb{X}$ is continuous.

3. (Composites) If $f: \mathbb{X}\rightarrow \mathbb{Y}$ and $g:\mathbb{Y}\rightarrow\mathbb{Z}$ are continuous, then the map $g \circ f: \mathbb{X} \rightarrow \mathbb{Z}$ is continuous.

4. (Restricting the domain) If $f: \mathbb{X} \rightarrow \mathbb{Y}$ is continuous, and if $A$ is a subspace of $\mathbb{X}$ , then the restriction function $f|A : A \rightarrow \mathbb{Y}$ is continuous.

5. (Restricting or expanding the range) Let $f:\mathbb{X}\rightarrow\mathbb{Y}$ is continuous. Let $\mathbb{Z}$ be a subspace of $\mathbb{Y}$ containing the image $f(\mathbb{X})$ , the function $h: \mathbb{X}\rightarrow \mathbb{Z}$ obtained by restricting the range of $f$ is continuous. If $\mathbb{Z}$ is a space having $\mathbb{Y}$ as a subspace, then the function $h: \mathbb{X}\rightarrow\mathbb{Y}$ obtained by expanding the range of $f$ is continuous.

6. (Local formulation of continuity) The map $f: \mathbb{X}\rightarrow\mathbb{Y}$ is continuous if $\mathbb{X}$ can be written as the union of open sets $U_{\alpha}$ such set $f|U_{\alpha}$ is continuous for each $\alpha$ :::

::: proof Proof.

1. $f^{-1}(U)$ of any open set $U$ is $\mathbb{X}$ , thus $f$ is continuous.

2. For every open subset $U$ of $\mathbb{X}$ , $j^{-1}(U) = U\cap A$ is continuous in $A$ . Thus $j$ is a continuous function.

3. For every open subset $U$ of $\mathbb{Z}$ , $f^{-1}(U)$ is open in $\mathbb{Y}$ , and $g^{-1}(f^{-1}(U))$ is open in $\mathbb{X}$ . Thus, $g \circ f$ is continuous

4. For every open subset $U$ of $\mathbb{Y}$ , $f^{-1}(U)$ is open in $\mathbb{X}$ , thus $f^{-1}(U)\cap A$ is open in $A$ . Thus the function $f|A$ is continuous.

5. If $\mathbb{Z}$ is a subspace of $\mathbb{Y}$ , then every open subset of $\mathbb{Z}$ can be represented as $U\cap\mathbb{Z}$ , where $U$ is a open subset of $\mathbb{Y}$ . Thus $h^{-1}(U\cap\mathbb{Z})=g^{-1}(\mathbb{Z})\cap g^{-1}(U) = \mathbb{X}\cap g^{-1}(U)$ which is a open subset of $X$ , thus $h$ is continuous.

   If $\mathbb{Y}$ is a subspace of $\mathbb{Z}$ . Then we take a open subset $U$ of $\mathbb{Z}$ . $h^{-1}(U) = g^(-1)(U\cap \mathbb{Y})$ which is open in $\mathbb{X}$ , thus $h$ is continuous.

6. if $f|U_{\alpha}$ is continuous for each $\alpha$ . For every open subset $U$ of $\mathbb{Y}$ . $U = \cup_{\alpha} (U_{\alpha}\cap U)$ where $U_{\alpha}\cap U$ is open both in $U_{\alpha}$ and in $\mathbb{Y}$ .

   Thus,

   $$\begin{aligned} f^{-1}(U) &=& f^{-1}(\cup_{\alpha} (U_{\alpha}\cap U)) \\ &=& \cup_{\alpha} ((f|U_{\alpha})^{-1}(U_{\alpha}\cap U)) \end{aligned}$$

   and each $(f|U_{\alpha})^{-1}(U_{\alpha}\cap U)$ is open, thus $f^{-1}(U)$ is open.

 ◻ :::

::: {#theorem:ThePastingLemma .theorem} Theorem 1.25 (The pasting lemma). ⁴⁵ Let $\mathbb{X} = A \cup B$ , where $A,B$ are closed in $\mathbb{X}$ . Let $f: A\rightarrow \mathbb{Y}$ and $g: B \rightarrow \mathbb{Y}$ be continuous. If $f(x)=g(x)$ for every $x\in A\cap B$ , then $f,g$ combine to give a continuous function $h: \mathbb{X}\rightarrow\mathbb{Y}$ , defined by setting $h(x)=f(x),x\in A$ and $h(x)=g(x),x\in B$ . :::

::: {#def:CoordinateFunctions .theorem} Theorem 1.26 (Maps into products). []{#theorem:MapsIntoProducts label="theorem:MapsIntoProducts"} ⁴⁶ Let $f: A \rightarrow \mathbb{X}\times\mathbb{Y}$ be given by the equation $f(a) = (f_{1}(a),f_{2}(a))$

Then, the function $f$ is continuous if and only if the functions $f_{1}: A \rightarrow \mathbb{X}, f_{2}: A \rightarrow \mathbb{Y}$ are continuous. :::

::: proof Proof. Let $\pi_{1},\pi_{2}$ be the projection function

$$\begin{aligned} \pi_{1}&:& \mathbb{X}\times\mathbb{Y} \rightarrow \mathbb{X} \\ \pi_{2}&:& \mathbb{X}\times\mathbb{Y} \rightarrow \mathbb{Y} \end{aligned}$$

We first proof that if $U$ is an open subset of $\mathbb{X} \times \mathbb{Y}$ , $f^{-1}(U) = f_{1}^{-1}(\pi_{1}(U)) \cap f_{2}^{-1}(\pi_{2}(U))$

Let $x \times y \in U$ , $f^{-1}(x \times y)$ contains all $a$ such that $f(a) = x \times y$ .

Then for any $a \in f^{-1}(x \times y)$ , $a \in f_{1}^{-1}(\pi_{1}(x \times y)), a \in f_{2}^{-1}(\pi_{2}(x \times y))$ .

Thus, $f^{-1}(x \times y) \subseteq f_{1}^{-1}(\pi_{1}(x \times y)) \cap f_{2}^{-1}(\pi_{2}(x \times y))$ .

Thus $f^{-1}(U) \subseteq f_{1}^{-1}(\pi_{1}(U)) \cap f_{2}^{-1}(\pi_{2}(U))$ .

Also, if $a \in f_{1}^{-1}(\pi_{1}(x \times y)), a \in f_{2}^{-1}(\pi_{2}(x \times y))$ , $f_{1}(a) = x, f_{2}(a) = y$ .

Thus $f(a) = x \times y$ . Thus $a \in f^{-1}(x \times y)$ .

Thus $f^{-1}(U) = f_{1}^{-1}(\pi_{1}(U)) \cap f_{2}^{-1}(\pi_{2}(U))$

Let $U$ be any open subset of $\mathbb{X} \times \mathbb{Y}$

$f^{-1}(U) = f_{1}^{-1}(\pi_{1}(U)) \cap f_{2}^{-1}(\pi_{2}(U))$

Where $f_{1}^{-1}(\pi_{1}(U))$ and $f_{2}^{-1}(\pi_{2}(U))$ are both open set. Thus $f^{-1}(U)$ is open. ◻ :::

1.7.1 Exercise

1. Let $\mathbb{Y}$ be an ordered set in the order topology. Let $f,g: \mathbb{X} \rightarrow \mathbb{Y}$ be continuous, show that the set $A$ $\{ x | f(x) \le g(x) \}$ is closed in $\mathbb{X}$ .

   ::: proof Proof. We only need to proof $\mathbb{X}-A$ is open in $\mathbb{X}$ . We take $x \in \mathbb{X}-A$ . Thus $f(x) > g(x)$ .

   Let $U_{1},U{2}$ be the open set in $\mathbb{Y}$ that met the following demand

   $$\begin{aligned} &&\forall y_{1} \in U_{1}, y_{2} \in U_{2}, y_{1} > y_{2} \\ &&f(x) \in U_{1}, g_{x} \in U_{2} \end{aligned}$$

   As $\mathbb{Y}$ is an ordered set, $U_{1},U{2}$ must exist.

   Let $U = f^{-1}(U_{1}) \cap g^{-1}(U_{2})$ . It is obvious that $U$ is a open set, and $x \in U$ .

   Also, for any $y \in U$ . $f(y)>g(y)$ . Thus $U \subseteq A$ . Thus $A$ is an open set. ◻ :::

2. Let $\{A_{\alpha}\}$ be a collection of subsets of $\mathbb{X}$ ; let $\mathbb{X} = \cup_{\alpha}A_{\alpha}$ . Lef $f: \mathbb{X}\rightarrow \mathbb{Y}$ ; suppose that $f|A_{\alpha}$ is continuous for each $\alpha$ . An indexed family of sets $\{A_{\alpha}\}$ is said to be locally finite[]{#def:LocallyFinite label="def:LocallyFinite"} if each point $x$ of $\mathbb{X}$ has a neighbourhood that intersect $A_{\alpha}$ for only finitely main values of $\alpha$ . Show that if the family $\{A_{\alpha}\}$ is locally finite and each $A_{\alpha}$ is closed, then $f$ is continuous.

   ::: proof Proof. For any closed subset $U$ of $\mathbb{Y}$ . Let $V = \cup f|A_{\alpha}(U)$

   We prove that $V$ is closed, so, $f$ is continuous.

   To prove that $V$ is closed, we prove that $\overline{V} = V$ . That is for any $x \in \overline{V}$ , we prove $x \in V$ . For any neighbourhood $B$ if $x$ , let $C_{B}$ denote the set that contain all $\alpha$ , such that $f|A_{\alpha(U)}$ intersect with $B$ . As $B$ intersect with $V$ , $C_{B}$ can not be empty.

   Let $\mathbb{C} = \{ C_{B} | B \text{ be a neighbourhood of } x \}$

   As $\{A_{\alpha}\}$ is locally definite, $\mathbb{C}$ contain at least one element with finite elements.

   Also $C_{B_{1} \cap B_{2}} \subseteq C_{B_{1}} \cap C_{B_{2}}$

   Let $\le$ be a partial order on the $\mathbb{C}$ . If $C_{B_{1}} \subseteq C_{B_{2}}$ , we say that $C_{B_{1}} \ge C_{B_{2}}$ .

   If there is chain in $\mathbb{C}$ $C_{B_{1}} \le C_{B_{2}} \dots$

   Let $C_{B_{0}}$ be a element of $\mathbb{C}$ with finite element. If $C_{B_{0}} \subseteq C_{B_{1}}, C_{B_{0}} \subseteq C_{B_{2}} \dots$ . Then $C_{B_{0}}$ is a upper bound of the chain.

   If $C$ is not a subset of all element of the chain. Then we construct a new set say $D = \{ C_{B_{0} \cap B_{1}}, C_{B_{0} \cap B_{2}} \dots \}$

   Let $\mathbb{D} = \{ C_{D_{1} \cap D_{2} \cap \dots } | C_{D_1}, C_{D_2}\dots \in D \}$

   As $C_{B_{0}}$ is a finite set, $D$ is a finite set, $\mathbb{D}$ is also a finite set. Thus there must be a maximal element $E \in \mathbb{D}$ that is the subset of all element of $\mathbb{D}$ . Then $E$ is a subset of all element of the chain. Thus $E$ is a upper bound of the chain.

   Thus, there must be a maximal element $C_{F}$ of $\mathbb{C}$ , that is a subset of all element of $\mathbb{C}$ .

   Let $G$ be the set be the union of all element of $C_{F}$ .

   As $C_{F}$ is finite, $G$ is closed. And all neighbourhood of $x$ intersect with $G$ . Thus $x \in G$

   As $G$ is a subset of $V$ , $x \in V$ . So $V$ is closed. And $f$ is a continuous function on $\mathbb{X}$ . ◻ :::

3. Let $A$ be a subset of topological space $\mathbb{X}$ , let $\mathbb{Y}$ be a Hausdorff space. Let $f: A \rightarrow \mathbb{Y}$ be a continuous function. Let $g: \overline{A} \rightarrow \mathbb{Y}$ also be a continuous function where $g(x) = f(x), x \in A$ . Prove that $g$ us uniquely determined by $f$ .⁴⁷

   ::: proof Proof. Say $g$ and $h$ are two distinct function that met the demand.

   So there exist $x_{0}$ such that $g(x_{0}) \neq h(x_{0})$ .

   As $\mathbb{Y}$ is a Hausdorff space, so there exist adjoint open subset $g(x_{0}) \in U$ and $h(x_{0}) \in V$ .

   Then $g^{-1}(U)$ and $h^{-1}(V)$ are both open subset of $\mathbb{X}$ that contain $x_{0}$ .

   If $g^{-1}(U) \cap h^{-1}(V) \cap A \neq \emptyset$ . Then there exist $x_{1} \in g^{-1}(U) \cap h^{-1}(V) \cap A$ such that $g(x_{1}) \in U$ and $h(x_{1}) \in V$ and $g(x_{1})=h(x_{1})$ . However $U$ and $V$ are disjoint. So there is a contradiction.

   As $^{-1}(U) \cap h^{-1}(V)$ is a open subset contain $x_{0}$ . So $^{-1}(U) \cap h^{-1}(V)$ must intersect with $A$ . So it is impossible that $g^{-1}(U) \cap h^{-1}(V) \cap A = \emptyset$ .

   So $g=h$ . ◻ :::

1.8 Metric Topology

::: {#def:Distance .definition} Definition 1.41 (metric). []{#def:Metric label="def:Metric"} A metric on a set $\mathbb{X}$ is a function $d: \mathbb{X} \times \mathbb{X} \rightarrow \mathbb{R}$ having the following properties:

1. $d(x,y) \geq 0$ for all $x,y \in \mathbb{X}$ ; equality hold if and only if $x = y$

2. $d(x,y) = d(y,x), \forall x,y \in \mathbb{X}$

3. (Triangle Inequality) $d(x,y) + d(y,z) \geq d(x,z), \forall x,y,z \in \mathbb{X}$

Given a metric $d$ on $\mathbb{X}$ , the number $d(x,y)$ is often called the distancebetween $x$ and $y$ in the metric $d$ . :::

::: {#def:EpsilonBallCenteredAtX .definition} Definition 1.42 ( $\epsilon$ -ball centered at $x$ ). *⁴⁸ Given metric $d$ on a set $\mathbb{X}$ and $\epsilon > 0$ . The set

$$B_{d}(x,\epsilon) = \{ y | d(x,y) < \epsilon \}$$

is called $\epsilon$ -ball centered at $x$ .* :::

::: {#def:MetricTopology .definition} Definition 1.43 (metric topology). If $d$ is a metric on the set $\mathbb{X}$ , then the collection of all $\epsilon$ -balls $B_{d}(x,\epsilon)$ , such that $x \in \mathbb{X}$ and $\epsilon > 0$ , is a basis for a topology on $\mathbb{X}$ , called the metric topology induced by $d$ . :::

::: {#def:MetricSpace .definition} Definition 1.44 (metrizable). []{#def:Metrizable label="def:Metrizable"} If $\mathbb{X}$ is topological space, $\mathbb{X}$ is said to be metrizable if there exists a metric $d$ on the set $\mathbb{X}$ that induces the topology of $\mathbb{X}$ . A metric spaceis a metrizable space $\mathbb{X}$ together with a specific metric $d$ that gives the topology of $\mathbb{X}$ . :::

::: {#def:Bounded .definition} Definition 1.45 (bounded). Let $\mathbb{X}$ be a metric space with metric $d$ . A subset $A$ of $\mathbb{X}$ is said to be bounded if there is some number $M$ such that $d(a_{1},a_{2}) \leq M$ for every pair $a_{1}$ and $a_{2}$ if points of $A$ . :::

::: {#def:Diameter .definition} Definition 1.46 (diameter). Let $\mathbb{X}$ be a metric space with metric $d$ . Let $A$ be a bounded subset of $\mathbb{X}$ . Then diameter is defined to be $\mathrm{diam}A = \sup\{ d(a_{1},a_{2}) | a_{1}, a_{2} \in A \}$ :::

::: {#def:StandardBoundedMetric .theorem} Theorem 1.27. Let $\mathbb{X}$ be a metric space with metric $d$ . Define $\overline{d}: \mathbb{X} \times \mathbb{X} \rightarrow \mathbb{R}$ by the equation $\overline{d}(x,y) = min\{ d(x,y),1 \}$

Then $\overline{d}$ is a metric that induces the same topology as $d$ .

The metric $\overline{d}$ is called the standard bounded metriccorresponding to $d$ :::

::: proof Proof. It is obvious that $\overline{d}$ is a metric.

To prove that $d$ and $\overline{d}$ induces the same topology, it is suffice to prove that for all $a \in X$ and $\epsilon > 0$ there exists $\{a_{\alpha}\}$ and $\{\epsilon_{\alpha}\}$ where $\epsilon_{\alpha} \leq 1$ such that $B_{d}(a,\epsilon) = \bigcup B_{\overline{d}}(a_{\alpha},\epsilon_{\alpha})$

For every $x \in B_{d}(a,\epsilon)$ take $a_{x} = x$ and $\epsilon_{x} < min(\epsilon - d(a,x),1)$ . Then $B_{d}(a,\epsilon) \supseteq B_{\overline{d}}(a_{x},\epsilon_{x})$ as for all $y \in B_{\overline{d}}(a_{x},\epsilon_{x})$

$$\begin{aligned} d(a,y) &\leq& d(a,a_{x}) + d(a_{x},y) \\ &<& min(\epsilon - d(a,x),1) + d(a,a_{x}) \\ &\leq& \epsilon \end{aligned}$$

Thus $B_{d}(a,\epsilon) = \bigcup_{ x \in B_{d}(a,\epsilon) } B_{\overline{d}}(a_{x},\epsilon_{x})$ ◻ :::

::: {#def:Norm .definition} Definition 1.47 (norm). *Given $x = (x_{1},\dots,x_{n})$ in $\mathbb{R}^{n}$ . The norm of $x$ is defined by the equation

$$||x|| = ( x_{1}^{2}+\dots+x_{n}^{2} )^{\frac{1}{2}}$$

:::

::: {#def:EuclideanMetric .definition} Definition 1.48 (euclidean metric). The euclidean metric $d$ on $\mathbb{R}^{n}$ is defined by $d(x,y) = ||x-y||$ :::

::: {#def:SquareMetric .definition} Definition 1.49 (square metric). The square metric $\rho$ on $\mathbb{R}^{n}$ is defined by $\rho(x,y) = max\{ |x_{1}-y_{1}|,\dots,|x_{n}-y_{n}| \}$ :::

::: lemma Lemma 1.8. Let $d$ and $d'$ be two metrics on the set $\mathbb{X}$ ; let $\mathbb{T}$ and $\mathbb{T'}$ be the topology induced by $d$ and $d'$ respectively. Then $\mathbb{T'}$ is finer than $T$ if and only if for all $x \in \mathbb{X}$ and $\epsilon > 0$ , there exists a $\delta > 0$ such that $B_{d'}(x,\delta) \subseteq B_{d}(x,\epsilon)$ :::

::: proof Proof. If $\mathbb{T'}$ is finer than $\mathbb{T}$ . Then for all $B_{d}(x,\epsilon)$ there exists a open set $U$ that containing $x$ such that $U \subseteq B_{d}(x,\epsilon)$ . As $\{{B_{d'}(x,\delta)}\}$ is a basis of $T'$ , then there exists $B_{d'}(x,\delta) \subseteq U$ that containing $x$ .

If for all $B_{d}(x,\epsilon)$ , there exists $B_{d'}(x,\delta) \subseteq B_{d}(x,\epsilon)$ . Then as $\{{B_{d'}(x,\epsilon)}\}$ and $\{{B_{d}(x,\epsilon)}\}$ are both basis, then $\mathbb{T'}$ is finer than $T$ . ◻ :::

::: theorem Theorem 1.28. ⁴⁹ The topologies on $\mathbb{R}^{n}$ induced by the euclidean metric $d$ and the square metric $\rho$ are the same as the product topology on $\mathbb{R}^{n}$ . :::

::: {#def:UniformMetric .definition} Definition 1.50 (uniform metric, uniform topology). *Given an index set $J$ , and given points $x = (x_{\alpha})_{\alpha \in J}$ and $y = (y_{\alpha})_{\alpha \in J}$ of $\mathbb{R}^{J}$ , let us define a metric $\overline{\rho}$ on $\mathbb{R}^{J}$ by the equation

$$\overline{\rho}(x,y) = \sup\{ \overline{d}(x_{\alpha},y_{\alpha}) | \alpha \in J \}$$

where $\overline{d}$ is the standard bounded metric on $\mathbb{R}$ . $\overline{\rho}$ is called the uniform metric on $\mathbb{R}^{J}$ , and the topology it induces is called the uniform topology* :::

::: theorem Theorem 1.29. ⁵⁰ The uniform topology on $\mathbb{R}^{J}$ is finer than the product topology and is coarser than the box topology. :::

::: theorem Theorem 1.30. *Let $\overline{d}(a,b) = \min\{ |a-b|,1 \}$ be the standard bounded metric on $\mathbb{R}$ . If $x$ nad $y$ are two points of $\mathbb{R}^{\omega}$ , define

$$D(x,y) = \sup\left\{ \frac{\overline{d}(x_{i},y_{i})}{i} \right\}$$

Then $D$ is a metric that induces the product topology on $\mathbb{R}^{\omega}$ :::

::: proof Proof. The properties of a metric are satisfied trivially except for the triangle inequality, which is proved by noting that for all $i$ ,

$$\begin{aligned} \frac{\overline{d}(x_{i},z_{i})}{i} &\le& \frac{\overline{d}(x_{i},y_{i})}{i} + \frac{\overline{d}(y_{i},z_{i})}{i} \\ &\le& D(x,y) + D(y,z) \end{aligned}$$

so that

$$\sup\left\{ \frac{\overline{d}(x_{i},z_{i})}{i} \right\} \le D(x,y) + D(y,z)$$

The fact that $D$ gives the product topology requires a little more work. First, let $U$ be open in the metric topology and let $x \in U$ ; we find an open set $V$ in the product topology such that $x \in V \supseteq U$ . Choose an $\epsilon-ball$ $B_{D}(x,\epsilon)$ lying in $U$ . Then choose $N$ large enough that $\frac{1}{N} < \epsilon$ . Finally, let $V$ be the basis element for the product topology $V = (x_{1}-\epsilon,x_{1}+\epsilon) \times \dots \times (x_{N}-\epsilon,x_{N}+\epsilon) \times R \times R \times \dots$

We assert that $V \in B_{D}(x,\epsilon)$ : Given any $y$ in $\mathbb{R}^{\omega}$ $\frac{\overline{d}(x_{i},y_{i})}{i} \le \frac{1}{N}, \forall i \ge N$

Therefore, $D(x,y) \le \max\left\{ \frac{\overline{d}(x_{1},y_{1})}{1},\dots,\frac{\overline{d}(x_{N},y_{N})}{N},\frac{1}{N} \right\}$

If $y$ is in $V$ , this expression is less than $\epsilon$ , so that $V \subseteq B_{D}(x,\epsilon)$ , as desired.

Conversely, consider a basis element $U = \prod_{i \in \mathbb{Z}_{+}} U_{i}$ for the product topology, where $U_{i}$ is open in $\mathbb{R}$ for $i = \alpha_{1},\dots,\alpha_{n}$ and $U_{i} = \mathbb{R}$ for all other indices $i$ . Given $x \in U$ , we find an open set $V$ of the metric topology such that $x \in V \supseteq U$ . Choose an interval $(x_{i}-\epsilon_{i},x_{i}+\epsilon_{i})$ in $\mathbb{R}$ centered about $x_{i}$ and lying in $U_{i}$ for $i = \alpha_{1},\dots,\alpha_{n}$ ; choose each $\epsilon_{i} \le 1$ . Then define $\epsilon = \min \left\{ \frac{\epsilon_{i}}{i} | i = \alpha_{1},\dots,\alpha_{n} \right\}$

We assert that $x \in B_{D}(x,\epsilon) \subseteq U$

Let $y$ be a point of $B_{D}(x,\epsilon)$ . Then for all $i$ $\frac{\overline{d}(x_{i},y_{i})}{i} \le D(x,y) < \epsilon$

Now if $i = \alpha_{1},\dots,\alpha_{n}$ , then $\epsilon \le \frac{\epsilon_{i}}{i}$ , so that $\overline{d}(x_{i},y_{i}) < \epsilon_{i} \le 1$ ; it follows that $|x_{i}-y_{i}| < \epsilon_{i}$ . Therefore $y \in \prod U_{i}$ , as desired. ◻ :::

::: {#def:HilbertCube .definition} Definition 1.51 (Hilbert Cube). The set $H = \prod_{ n \in \mathbb{Z}_{+} } [0,\frac{1}{n}]$ is called Hilbert cube :::

::: {#def:L2Topology .definition} Definition 1.52 ($l^{2}$-topology). Let $\mathbb{X}$ be the subset of $\mathbb{R}^{\omega}$ consisting of all sequences $x$ such that $\sum x_{i}^{2}$ converges.

Then the formula $d(x,y) = \left[ \sum_{i=1}^{\infty}(x_{i}-y_{i})^{2} \right]^{\frac{1}{2}}$ defines a metric on $\mathbb{X}$ . The topology induced by $d$ is called the $l^{2}$-topology. :::

::: {#def:FirstCountabilityAxiom .definition} Definition 1.53 (countable basis at point $x$ ). *[]{#def:CountableBasisAtPointX label="def:CountableBasisAtPointX"} A space is said to be have countable basis at point $x$ if there is a countable collection $\{ U_{n} \}_{n \in \mathbb{Z}_{+}}$ of neighbourhoods of $x$ such that any neighbourhood $U$ of $x$ contains at least on of the sets $U_{n}$ . A space $\mathbb{X}$ that has a countable basis at each of its point is said to satisfy the first countability axiom * :::

::: theorem Theorem 1.31. Let $f: \mathbf{X} \rightarrow \mathbf{Y}$ be metrizable with metric $d_{\mathbf{X}}$ and $d_{\mathbf{Y}}$ , respectively. Then continuity of $f$ is equivalent to the requirement that given $x \in \mathbb{X}$ and given $\epsilon > 0$ , there exists $\delta > 0$ such that $d_{\mathbf{X}}(x,y) < \delta \implies d_{\mathbb{Y}}(f(x),f(y)) < \epsilon$ :::

::: proof Proof. Suppose $f$ is continuous. Given $x$ and $\epsilon$ , consider the set $f^{-1}(B(f(x),\epsilon))$ which is open in $\mathbb{X}$ and contains the point $x$ . It contains some $\delta$ -ball $B(x,\delta)$ centered at $x$ . If $y$ is in this $\delta$ -ball, then $f(y)$ is in this $\delta$ -ball as desired.

Conversely, suppose that the $\epsilon-\delta$ condition is satisfied. Let $V$ be open in $\mathbb{Y}$ ; we show that $f^{-1}(V)$ is open in $\mathbb{X}$ . Let $x$ be a point of the set $f^{-1}(V)$ . Since $f(x) \in V$ there is an $\epsilon$ -ball $B(f(x),\epsilon)$ centered at $f(x)$ and contained in $V$ . By the $\epsilon-\delta$ condition, there exists a $\delta$ -ball centered at $x$ such that $f(B(x,\delta)) \subseteq B(f(x),\epsilon)$ . Then $B(x,\delta)$ is a neighbourhood of $x$ contained in $f^{-1}(V)$ , so that $f^{-1}(V)$ is open, as desired. ◻ :::

::: {#def:TheSequenceLemma .lemma} Lemma 1.9 (The sequence lemma). ⁵¹ Let $\mathbb{X}$ be a topological space; let $A \subseteq \mathbb{X}$ If there is a sequence of points of $A$ converging to $x$ , then $x \in \overline{A}$ , the converse holds if $\mathbb{X}$ is metrizable. :::

::: theorem Theorem 1.32. ⁵² Let $f: \mathbb{X} \rightarrow \mathbb{Y}$ . If the function $f$ is continuous, then for every convergent sequence $x_{n} \rightarrow x$ , the sequence $f(x_{n})$ converges to $f(x)$ . The converse holds if $\mathbb{X}$ is metrizable. :::

::: lemma Lemma 1.10. ⁵³ The addition, subtraction, and multiplication operations are continuous functions from $\mathbb{R} \times \mathbb{R}$ into $\mathbb{R}$ ; and the quotient operation is continuous function from $\mathbb{R} \times (\mathbb{R} - \{0\})$ into $\mathbb{R}$ . :::

::: theorem Theorem 1.33. ⁵⁴ If $\mathbb{X}$ is a topological space, and if $f,g: \mathbb{X} \rightarrow \mathbb{R}$ are continuous functions, then $f + g$ , $f - g$ and $f \cdot g$ are continuous. If $g(x) \neq 0$ for all $x$ , then $\frac{f}{g}$ is continuous. :::

::: {#def:ConvergeUniformly .definition} Definition 1.54 (converge uniformly). Let $f_{n}: \mathbb{X} \rightarrow \mathbb{Y}$ be a sequence of functions from the set $\mathbb{X}$ to the metric space $\mathbb{Y}$ . Let $d$ be the metric for $\mathbb{Y}$ . We say that the sequence $(f_{n})$ converges uniformly to the function $f : \mathbb{X} \rightarrow \mathbb{Y}$ if given $\epsilon > 0$ , there exists an integer $N$ such that $d(f_{n}(x),f(x)) < \epsilon$ for all $n > N$ and all $x \in \mathbb{X}$ :::

::: {#def:UniformLimitTheorem .theorem} Theorem 1.34 (Uniform limit theorem). Let $f_{n} : \mathbb{X} \rightarrow \mathbb{Y}$ be a sequence of continuous functions from the topological space $\mathbb{X}$ to the metric space $\mathbb{Y}$ . If $(f_{n})$ converges uniformly to $f$ , then $f$ is continuous. :::

::: {#def:IsometricImbedding .definition} Definition 1.55 (isometric imbedding). Let $\mathbb{X}$ and $\mathbb{Y}$ be metric spaces with metric $d_{\mathbb{X}}$ and $d_{\mathbb{Y}}$ , respectively. Let $f: \mathbb{X} \rightarrow \mathbb{Y}$ have the property that for every pair of points $x_{1}$ , $x_{2}$ of $\mathbb{X}$ , and $d_{\mathbb{Y}}(f(x_{1}),f(x_{2})) = d_{\mathbb{X}}(x_{1},x_{2})$

$f$ is an topological imbedding and is called an isometric imbedding of $\mathbb{X}$ in $\mathbb{Y}$ :::

Footnotes

1. The set $\mathbb{U}$ can form a topology because of the definition of topology is intersection of finite sub collection. If this can be intersection of infinite sub collection, $\mathbb{U}$ will not be a topology. ↩

2. Note that this expression may not be unique. ↩

3. We omit the proof of this lemma as it is obvious. ↩

4. We omit the proof of this lemma as it is obvious. ↩

5. Whenever we consider $\mathbb{R}$ , we shall suppose it is given this topology unless we specifically state otherwise. ↩

6. We omit the proof of this change@code@ as it is obvious. ↩

7. It is obvious that $\mathbb{T}$ is a topology, we just omit the proof here. ↩

8. It is obvious that $\mathbb{T'}$ is also a topology, we just omit the proof here. ↩

9. The standard topology on $\mathbb{R}$ is an order topology derived from the usual order on $\mathbb{R}$ . ↩

10. open rays are always open sets in the order topology ↩

11. the open rays also formed a subbasis of the order topology ↩

12. We omit the proof of this change@code@ as it is obvious. ↩

13. We omit the proof of this change@code@ as it is obvious. ↩

14. In the finite case, the product topology and box topology are the same, however they differ when $\mathbb{X}$ is a infinite cartesian product. ↩

15. This is also called a projection mappingin a cartesian product. ↩

16. A index set was the set $\{1,\dots,n\}$ or the set $\mathbb{Z}_{+}$ . ↩

17. It is assumed that it is given product topology when considering $\prod X_{\alpha}$ unless it state specifically. ↩

18. We omit the proof of this change@code@ as it is obvious. ↩

19. We omit the proof of this change@code@ as it is obvious. ↩

20. We omit the proof of this change@code@ as it is obvious. ↩

21. We omit the proof of this change@code@ as it is obvious. ↩

22. We omit the proof of this change@code@ as it is obvious. ↩

23. If $\mathbb{X}$ is an ordered set in the order topology, and $\mathbb{Y}$ is a subset of $\mathbb{X}$ . The order relation, when restricted to $\mathbb{Y}$ , makes $\mathbb{Y}$ into and ordered set. However, the resulting order topology on $\mathbb{Y}$ need not be the same as the topology that $\mathbb{Y}$ inherits as a subspace of $\mathbb{X}$ . ↩

24. *the dictionary means for $X_{1}, X_{2} \in \mathbb{Y} = \mathbb{X}_{1} \times \mathbb{X}_{2} \times \mathbb{X}_{3} \dots$ which:

    $$\begin{aligned} X_{1} &=& (x_{1},x_{2},x_{3}\dots)\\ X_{2} &=& (x_{1}',x_{2}',x_{3}'\dots) \end{aligned}$$

    $X_1 > X_2$ only when

    $$\begin{aligned} && \exists k \in \mathbb{Z}_{+}, \forall i \in \mathbb{Z}_{+}, 0 < i < k \\ && x_{i} = x_{i}'\\ && x_{k} > x_{k}' \end{aligned}$$
    ↩

25. Given $\mathbb{X}$ is an ordered set in the order topology and $\mathbb{Y}$ is a subset of $\mathbb{X}$ , we shall assume that $\mathbb{Y}$ is given the subspace topology unless we specifically state otherwise. ↩

26. what does $\mathbb{X}$ , $\mathbb{X'}$ , $\mathbb{Y}$ , $\mathbb{Y'}$ really mean here?? I do not know, so I just put the exercise here without a proof. ↩

27. The prove of this set is countable is typically similar to Cantor's enumeration of a countable collection of countable sets. ↩

28. A set can be open, or closed, or both, or neither ↩

29. We omit the proof of this change@code@ as it is obvious. ↩

30. As the proof is similar to the case in the open set, so we omit the proof here. ↩

31. We omit the proof of this change@code@ as it is obvious. ↩

32. As the closure of $A$ in $\mathbb{X}$ and the closure $A$ in $\mathbb{Y}$ will sometimes be different. We always use $\overline{A}$ to denote the closure of $A$ in $\mathbb{X}$ ↩

33. Some other mathematicians use neighbourhood to say that $U$ merely contains an open set containing $x$ . The book does not give a formal definition for the word merely, and I am not sure either. ↩

34. Note that, $x$ may belong to $A$ or not, this does not matter. ↩

35. We omit the proof of this change@code@ as it is obvious. ↩

36. We omit the proof of this change@code@ as it is obvious. ↩

37. In real line, a sequence can not converge to multiple points, but for an arbitrary topological space, this is possible. ↩

38. This implies that a sequence in a Hausdorff space cannot converge to multiple points. The following theorem prove this. ↩

39. The condition every finite point set is closed is weaker than the Hausdorff space condition. For instance, the finite complement topology of $\mathbb{R}$ met the condition of finite point set. However it is not a Hausdorff space. ↩

40. We omit the proof of this change@code@ as it is obvious. ↩

41. We omit the proof of this change@code@ as it is obvious. ↩

42. We omit the proof of this change@code@ as it is obvious. ↩

43. As the continuity of a function is different as the topological spaces are different. So if we want to emphasis this fact, we say that $f$ is continuous relativeto specific topologies on $\mathbb{X}$ and $\mathbb{Y}$ . ↩

44. A equivalent way to define homeomorphism, is that for any open subset $U$ of $\mathbb{X}$ , $f(U)$ is open if and only if $U$ is open. ↩

45. The proof of this theorem is similar to the "Local formulation of continuity" condition of "Rules for constructing continuous functions", so we omit the proof here. ↩

46. The map $f_{1},f_{2}$ are called the coordinate functionsof $f$ ↩

47. It is possible that $g$ does not exist.

    Let $\mathbb{X}$ be the real line with order topology. Let $\mathbb{Y}$ be $\{0,1\}$ .

    Let $A = \mathbb{X} - \{0\}$ .

    Let, $f(x) = \begin{cases} 1, & x > 0 \\ 0, & x < 0 \end{cases}$

    So, it is obvious that $f$ is a continuous function on $\mathbb{X}$ . However $g$ does not exist in this case. ↩

48. When no confusion will arise, the metric $d$ may be omit in $B_{d}(x,\epsilon)$ ↩

49. We omit the proof of this change@code@ as it is obvious. ↩

50. We omit the proof of this change@code@ as it is obvious. ↩

51. We omit the proof of this change@code@ as it is obvious. ↩

52. We omit the proof of this change@code@ as it is obvious. ↩

53. We omit the proof of this change@code@ as it is obvious. ↩

54. We omit the proof of this change@code@ as it is obvious. ↩