3.5. Subspace Topology

Let $(X,d)$ be a metric space. Let $Y\subseteq X$. Recall from Definition 3.3 that $(Y,d)$ is a metric subspace with the distance function $d$ restricted to $Y\times Y$.

Remark 3.6

Let $Y$ be a metric subspace of $X$ and $S\subseteq Y$. The interior, closure and boundary of $S$ w.r.t. $X$ and w.r.t. $Y$ may be different.

If a subspace hasn’t been specified, by default, we shall assume that we are computing the interior, closure and boundary w.r.t. the metric space $X$.

3.5.1. Open Balls

Theorem 3.38 (Open balls in the metric subspace)

Let $(Y,d)$ be a metric subspace of $(X,d)$. Let $B_X(p,r)$ denote an open ball of radius $r$ at $p\in X$. Let $B_Y(p,r)$ denote an open ball of radius $r$ at $p\in Y$. Then

$$B_Y(p,r)=B_X(p,r)\cap Y.$$

Proof. We recall that

$$B_X(p,r)=\{y\in X|d(p,y)<r\}.$$

Similarly,

$$B_Y(p,r)=\{y\in Y|d(p,y)<r\}.$$

We first show that $B_Y(p,r)\subseteq B_X(p,r)\cap Y$

1. Let $y\in B_Y(p,r)$.

2. Then $y\in Y\subseteq X$ and $d(p,y)<r$.

3. Hence $y\in B_X(p,r)\cap Y$.

We now show that $B_X(p,r)\cap Y\subseteq B_Y(p,r)$.

1. Let $y\in B_X(p,r)\cap Y$.

2. Then $y\in Y$ and $y\in X$ and $d(p,y)<r$.

3. Hence $y\in Y$ and $d(p,y)<r$.

4. Hence $y\in B_Y(p,r)$.

3.5.2. Open Sets

Theorem 3.39 (Open sets in the metric subspace)

Let $(Y,d)$ be a metric subspace of $(X,d)$. Let $S\subseteq Y$. Then $S$ is open in $(Y,d)$ if and only if $S=O\cap Y$ where $O$ is an open subset of $(X,d)$.

Proof. A subset $S$ of $Y$ is open in $(Y,d)$ if for every $x\in S$, there exists an open ball $B_Y(x,r)\subseteq S\subseteq Y$.

Assume that $S=O\cap Y$ where $O$ is open in $X$.

1. Let $x\in S$.

2. Then $x\in O$ and $x\in Y$.

3. Since $O$ is open in $X$, hence there is an open ball $B_X(x,r)\subseteq O$.

4. Then $B_X(x,r)\cap Y\subseteq O\cap Y=S$.

5. By Theorem 3.38,

   $$B_X(x,r)\cap Y=B_Y(x,r)$$

   is an open ball in the metric subspace $(Y,d)$ of radius $r>0$ around $x\in Y$.

6. Hence $S$ is open in $(Y,d)$.

For the converse, assume that $S$ is open in $Y$.

1. For every $x\in S$, there is an open ball $B_Y(x,r_x)\subseteq S$.

2. Hence $\bigcup _{x\in S}{B}_Y(x,r_x)\subseteq S$.

3. Also, $x\in B_Y(x,r_x)$ implies that $S\subseteq \bigcup _{x\in S}{B}_Y(x,r_x)$.

4. Thus, $S=\bigcup _{x\in S}{B}_Y(x,r_x)$.

5. By Theorem 3.38

   $$B_Y(x,r_x)=B_X(x,r_x)\cap Y$$

   for every $x\in S$.

6. Define $T=\bigcup _{x\in S}{B}_X(x,r_x)$.

7. Then $T$ is a union of open balls of $X$.

8. Hence $T$ is open in $X$.

9. Also

   $$\begin{array}{rl}S& =\bigcup _{x\in S}{B}_Y(x,r_x)\\ & =\bigcup _{x\in S}(B_X(x,r_x)\cap Y)\\ & =(\bigcup _{x\in S}{B}_X(x,r_x))\cap Y\\ & =T\cap Y.\end{array}$$

10. Hence $S=T\cap Y$ where $T$ is an open set of $(X,d)$.

Example 3.13

• $[0,1)$ is open in the metric space ${\mathbb{R}}_{+}$.

3.5.3. Subspace Topology

Recall from Definition 3.7 that a topology on a set $S$ is a collection of sets $T$ such that

1. Empty set and the whole set are elements of $T$.

2. $T$ is closed under arbitrary union.

3. $T$ is closed under finite intersection.

Theorem 3.40 (The subspace topology)

Let $(Y,d)$ be a metric subspace of $(X,d)$. The open sets of the $(Y,d)$ form a topology.

1. $\varnothing$ is open in $(X,d)$.

2. Hence $\varnothing \cap Y=\varnothing$ is open in $(Y,d)$.

3. $X$ is open in $(X,d)$.

4. Hence $X\cap Y=Y$ is open in $(Y,d)$.

5. Let $\{A_i|i\in I\}$ be a family of open sets of $(Y,d)$.

6. Then $A_i=B_i\cap Y$ for every $i\in I$ with $B_i$ open in $X$.

7. Let $B=\bigcup B_i$. Then $B$ is open in $(X,d)$.

8. But $\bigcup A_i=\bigcup (B_i\cap Y)=(\bigcup B_i)\cap Y=B\cap Y$.

9. Since $B$ is open in $(X,d)$, hence $B\cap Y$ is open in $(Y,d)$.

10. Let $\{A_1,\dots ,A_n\}$ be a finite collection of open sets of $(Y,d)$.

11. Then $A_i=B_i\cap Y$ for every $i\in [1,\dots ,n]$ such that $B_i$ is open in $(X,d)$.

12. Then $\bigcap A_i=\bigcap (B_i\cap Y)=(\bigcap B_i)\cap Y$.

13. Since $\bigcap B_i$ is open in $(X,d)$ (a finite intersection), hence $\bigcap A_i$ is open in $(Y,d)$.

3.5.4. Closed Sets

Theorem 3.41 (Closed sets in subspace topology)

Let $Y$ be a metric subspace of $X$. Let $S\subseteq Y$.

$S$ is closed in $Y$ if and only if $S=C\cap Y$ where $C$ is a closed subset of $X$.

Proof. Let $S$ be closed in $Y$.

1. Then, $Y\setminus S$ is open in $Y$.

2. By definition of subspace topology,

   $$Y\setminus S=Y\cap O$$

   where $O$ is some open subset of $X$.

3. Then,

   $$\begin{array}{rl}S& =Y\setminus (Y\setminus S)\\ & =Y\setminus (Y\cap O)\\ & =Y\setminus O\\ & =Y\cap (X\setminus O).\end{array}$$

4. But $C=X\setminus O$ is closed in $X$.