<Home>Teaching><MATH 131B (Fall 2024)>Proof of Proposition 1.2.15

Proof of Proposition 1.2.15

Note that I used (i)-(viii) instead of (a)-(h) to label the statements. Everything else should be exactly the same as in the textbook.

Proposition (Basic properties of open and closed sets)

Let $(X, d)$ be a metric space.

Let $E$ be a subset of $X$ . Then $E$ is open if and only if $E = \int\p{E}$ . In other words, $E$ is open if and only if for every $x \in E$ , there exists an $r > 0$ such that $B\p{x, r} \subseteq E$ .
Let $E$ be a subset of $X$ . Then $E$ is closed if and only if $E$ contains all its adherent points. In other words, $E$ is closed if and only if for every convergent sequence $\p{x_n}_{n=m}^\infty$ in $E$ , the limit $\lim_{n\to\infty} x_n$ of that sequence also lies in $E$ .
For any $x_0 \in X$ and $r > 0$ , then the ball $B\p{x_0, r}$ is an open set. The set $\set{x \in X \st d\p{x, x_0} \leq r}$ is a closed set. (This set is sometimes called the closed ball of radius $r$ centered at $x_0$ .)
Any singleton set $\set{x_0}$ , where $x_0 \in X$ , is automatically closed.
If $E$ is a subset of $X$ , then $E$ is open if and only if the complement $X \setminus E \coloneqq \set{x \notin X \st x \notin E}$ is closed.
If $E_1, \ldots, E_n$ is a finite collection of open sets in $X$ , then $E_1 \cap E_2 \cap \cdots \cap E_n$ is also open. If $F_1, \ldots, F_n$ is a finite collection of closed sets in $X$ , then $F_1 \cup F_2 \cup \cdots \cup F_n$ is also closed.
If $\set{E_\alpha}_{\alpha \in I}$ is a collection of open sets in $X$ (where the index set $I$ could be finite, countable, or uncountable), then the union $\bigcup_{\alpha \in I} E_\alpha \coloneqq \set{x \in X \st x \in E_\alpha \text{ for some } \alpha \in I}$ is also open. If $\set{F_\alpha}_{\alpha \in I}$ is a collection of closed sets in $X$ , then the intersection $\bigcap_{\alpha \in I} F_\alpha \coloneqq \set{x \in X \mid x \in F_\alpha \text{ for all } \alpha \in I}$ is also closed.
If $E$ is any subset of $X$ , then $\int\p{E}$ is the largest open set which is contained in $E$ ; in other words, $\int\p{E}$ is open, and given any other open set $V \subseteq E$ , we have $V \subseteq\int\p{E}$ . Similarly $\overline{E}$ is the smallest closed set which contains $E$ ; in other words, $\overline{E}$ is closed, and given any other closed set $K \supseteq E$ , $K \supseteq \cl{E}$ .

Proof.

" $\implies$ "

Assume $E$ is open. By definition, this means that $\bd{E} \cap E = \emptyset$ .

First, notice that $E \supseteq \int\p{E}$ by definition of $\int\p{E}$ (check this!). For the reverse inclusion, let $x \in E$ . Recall that $\int\p{E}, \ext\p{E}, \bd{E}$ partition $X$ ; that is, the three sets are disjoint and their union is all of $X$ :
$X = \int\p{E} \cup \ext\p{E} \cup \bd{E}.$
Since $x \in E$ , we know that $x \notin \ext\p{E}$ (check this!). Because $\bd{E} \cap E = \emptyset$ and $x \in E$ , this means that $x \notin \bd{E}$ . Thus, $x \in \int\p{E}$ .

This shows that $E \subseteq \int\p{E}$ , so $E = \int\p{E}$ by double-inclusion.

" $\impliedby$ "

Assume $E = \int\p{E}$ . By definition, we need to check that $\bd{E} \cap E = \emptyset$ . We have
$\bd\p{E} \cap E = \bd\p{E} \cap \int\p{E} = \emptyset,$
where in the last step, we used the fact that $\int\p{E}, \ext\p{E}, \bd{E}$ partition $X$ .
" $\implies$ "

Assume $E$ is closed. This means that $\bd{E} \subseteq E$ . We know that the set of adherent points of $E$ satisfies $\cl{E} = E \cup \bd{E}$ (this is Proposition 1.2.10 in the textbook), so because both $E \subseteq E$ and $\bd{E} \subseteq E$ , their union $\cl{E} \subseteq E$ also.

" $\impliedby$ "

Assume that $\cl{E} \subseteq E$ . By Proposition 1.2.10 again, we know $E \cup \bd{E} \subseteq E$ . In particular, $\bd{E} \subseteq E$ , so $E$ is closed by definition,
By part (i), we need to show that for all $y \in B\p{x_0, r}$ , there exists $s > 0$ such that $B(y, s) \subseteq B\p{x_0, r}$ . (It's a little confusing, but in (i), I replaced $E$ with $B\p{x_0, r}$ , $x_0$ with $y$ , and $r$ with $s$ .)

Let $s = r - d\p{x_0, y}$ (draw a picture to see where I got this from). Then if $z \in B\p{y, s}$ ,
$d\p{z, x} \leq d\p{z, y} + d\p{y, x_0} < s + d\p{x_0, y} = r.$
In other words, $d\p{z, x_0} < r$ , which means $z \in B\p{x_0, r}$ . Thus, $B\p{y, s} \subseteq B\p{x_0, r}$ , so $B\p{x_0, r}$ is open.

For brevity, I will denote $K\p{x_0, r} \coloneqq \set{x \in X \st d\p{x, x_0} \leq r}$ . We'll use (ii): let $\p{x_n}_{n=m}^\infty$ be a sequence of elements in $K\p{x_0, r}$ which converges to $x' \in X$ . We need to show that $x' \in K\p{x_0, r}$ . Since $x_n \in K\p{x_0, r}$ , we have
$d\p{x', x_0} \leq d\p{x', x_n} + d\p{x_n, x_0} \leq d\p{x', x_n} + r.$
Lastly, because $x_n \xrightarrow{n\to\infty} x'$ , we can send $n \to \infty$ to get $d\p{x', x_0} \leq r$ .

(I used the following fact from 131A: if $a_n \xrightarrow{n\to\infty} a$ and $a_n \leq L$ for all $n$ , then $a \leq L$ .)
We'll use (ii). Let $\p{x_n}_{n=m}^\infty$ be a sequence of elements in $\set{x_0}$ which converges to $x' \in X$ . Since $x_n \in \set{x_0}$ , this means that $x_n = x_0$ for all $n$ , i.e., $\p{x_n}_{n=m}^\infty$ is a constant sequence. Thus, $x_n \xrightarrow{n\to\infty} x_0 \in \set{x_0}$ , so $\set{x_0}$ is closed.
" $\implies$ "

Assume $E$ is open. Suppose $\p{x_n}_{n=m}^\infty$ is a sequence of elements in $X \setminus E$ which converges to $x_0 \in X$ . We need to show that $x_0 \in X \setminus E$ . If $x_0 \in E$ , then because $E$ is open, there exists $r > 0$ such that $B\p{x_0, r} \subseteq E$ . But because $x_n \xrightarrow{n\to\infty} x_0$ , there exists $n \geq m$ such that $x_n \in B\p{x_0, r}$ . This means $x_n \in E \cap \p{X \setminus E} = \emptyset$ , which is a contradiction.

" $\impliedby$ "

Assume $X \setminus E$ is closed. We will show that $\bd{E} \cap E = \emptyset$ . Let $x_0 \in \bd{E}$ . By definition, this means that for all $r > 0$ , we have $B\p{x_0, r} \cap \p{X \setminus E} \neq \emptyset$ . But this is exactly what it means for $x_0 \in \cl{X \setminus E} = X \setminus E$ .
Let $x_0 \in E_1 \cap \cdots \cap E_n$ . For each $1 \leq i \leq n$ , $x_0 \in E_i$ , so because $E_i$ is open, there exists $r_i > 0$ such that $B\p{x_0, r_i} \subseteq E_i$ . Let $r = \min\set{r_1, \ldots, r_n} > 0$ (this is where we use the fact that we have a finite intersection). Then for each $1 \leq i \leq n$ ,
$B\p{x_0, r} \subseteq B\p{x_0, r_i} \subseteq E_i.$
Thus, $B\p{x_0, r} \subseteq E_1 \cap \cdots \cap E_n$ , so $E_1 \cap \cdots \cap E_n$ is open.

For the second statement, we can use (v) by taking complements. Note that
$X \setminus \p{F_1 \cup \cdots \cup F_n} = \p{X \setminus F_1} \cap \cdots \cap \p{X \setminus F_n}.$
By (v), $X \setminus F_1, \ldots, X \setminus F_n$ are all open sets, so the first statement (where we plug in $E_i = X \setminus F_i$ ) implies that $X \setminus \p{F_1 \cup \cdots \cup F_n}$ is open. By (v) again, $F_1 \cup \cdots \cup F_n$ is closed.
Let $x_0 \in \bigcup_{\alpha \in I} E_\alpha$ . By definition, this means there exists $\alpha \in I$ such that $x_0 \in E_\alpha$ . Since this $E_\alpha$ is open, there exists $r > 0$ such that
$B\p{x_0, r} \subseteq E_\alpha \subseteq \bigcup_{\alpha \in I} E_\alpha,$
so $\bigcup_{\alpha \in I} E_\alpha$ is open.

As in the proof of (vi), we can prove the second statement by using (v) and taking complements.
$X \setminus \bigcap_{\alpha \in I} F_\alpha = \bigcup_{\alpha \in I} \p{X \setminus F_\alpha}$
is a union of open sets, hence open by the first statement. By (v), this means $\bigcap_{\alpha \in I} F_\alpha$ is closed.
First, we show that $\int\p{E}$ is open. Let $x_0 \in \int\p{E}$ so that there exists $r > 0$ such that $B\p{x_0, r} \subseteq E$ . By (iii), $B\p{x_0, r}$ is an open set, so given any $y \in B\p{x_0, r}$ , there exists $s > 0$ so that
$B\p{y, s} \subseteq B\p{x_0, r} \subseteq E,$
so $y \in \int\p{E}$ . Thus, $B\p{x_0, r} \subseteq \int\p{E}$ , so $\int\p{E}$ is open.

Now, assume $V \subseteq E$ , where $V$ is an open set. Then for any $x_0 \in V$ , there exists $r > 0$ so that
$B\p{x_0, r} \subseteq V \subseteq E,$
so $x_0 \in \int\p{E}$ . Thus, $V \subseteq \int\p{E}$ .

Next, we show that $\cl{E}$ is a closed set. Let $\p{\cl{x}_n}_{n=m}^\infty$ be a sequence of elements in $\cl{E}$ which converges to $x_0 \in E$ . We need to show that $x_0 \in \cl{E}$ , which means we need to find a sequence of elements in $E$ which converges to $x_0$ . Since $\cl{x}_n \in \cl{E}$ , we know that for any $r > 0$ , $B\p{\cl{x}, r} \cap E \neq \emptyset$ . In particular, if $r = \frac{1}{n}$ , then there exists $x_n \in E$ such that $x_n \in B\p{\cl{x}, \frac{1}{n}}$ . The sequence $\p{x_n}_{n=m}^\infty$ also converges to $x_0$ :
$d\p{x_n, x_0} \leq d\p{x_n, \cl{x}_n} + d\p{\cl{x}_n, x_0} \leq \frac{1}{n} + d\p{\cl{x}_n, x_0}.$
Sending $n \to \infty$ (as we did in the proof of (iii)), the right-hand side tends to $0$ , so $x_n \xrightarrow{n\to\infty} x_0$ . This shows that $x_0 \in \cl{E}$ , so $\cl{E}$ is closed.

Lastly, if $K$ is a closed set such that $K \supseteq E$ . Let $x_0 \in \cl{E}$ . Then there exists a sequence $\p{x_n}_{n=m}^\infty$ of elements in $E$ such that $x_n \xrightarrow{n\to\infty} x_0$ . But because $E \subseteq K$ , this means $\p{x_n}_{n=m}^\infty$ is also a sequence of elements in $K$ which converges to $x_0$ , so $x_0 \in \cl{K} = K$ . Thus, $\cl{E} \subseteq K$ . $\square$

Remark.

It's important in (vi) (but not in (vii)) that the number of open/closed sets is finite. Here are two counter-examples from $\p{\R, \abs{\,\cdot\,}}$ .

Let $E_n = \p{-\frac{1}{n}, \frac{1}{n}}$ . Then
$\bigcap_{n=1}^\infty E_n = \set{0},$
which is not an open set. Indeed, these sets are equal since $0 \in E_n$ for all $n \geq 1$ , and if $x_0 \in \bigcap_{n=1}^\infty E_n$ , then $x_0$ satisfies
$-\frac{1}{n} < x_0 < \frac{1}{n} \iff \abs{x_0} < \frac{1}{n}.$
If $x_0 \neq 0$ , then if $n_0$ is large enough, we have $\abs{x_0} < \frac{1}{n_0} < \abs{x_0}$ , which is not possible. Thus, it must be the case that $x_0 = 0$ , which proves that the two sets above are equal.

(Notice how I didn't mention that $\set{0}$ is closed. In general, a set can be open, closed, both open and closed, or neither open nor closed. In other words, $\set{0}$ being closed doesn't tell you whether or not it's open.)
$F_n = \br{-1 + \frac{1}{n}, 1 - \frac{1}{n}}$ gives another counter-example (exercise below).

Exercise 1.

Let $F_n = \br{-1 + \frac{1}{n}, 1 - \frac{1}{n}}$ . Compute $\bigcup_{n=1}^\infty F_n$ and show that it is not closed (which is not the same as showing that it's open).