# 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

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

Similarly,

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

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

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

$\begin{split} S &= \bigcup_{x \in S} B_Y(x, r_x) \\ &= \bigcup_{x \in S}(B_X(x, r_x) \cap Y) \\ &= (\bigcup_{x \in S} B_X(x, r_x)) \cap Y \\ &= T \cap Y. \end{split}$
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 $$\RR_+$$.

## 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. $$\EmptySet$$ is open in $$(X, d)$$.

2. Hence $$\EmptySet \cap Y = \EmptySet$$ 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 $$\{ A_i \ST i \in I \}$$ 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,

$\begin{split} S &= Y \setminus (Y \setminus S) \\ &= Y \setminus (Y \cap O) \\ &= Y \setminus O \\ &= Y \cap (X \setminus O). \end{split}$
4. But $$C = X \setminus O$$ is closed in $$X$$.