Basic Topology

Theorem 2.23

A set $E$ is open if and only if its complement is closed.

Proof

1. $E^c$ is closed implies $E$ is open.

Suppose $E^c$ is closed. This means every limit point of $E^c$ is a point of $E^c$. Equivalently, if $x \notin E^c$, then $x$ is not a limit point of $E^c$.

Choose $x \in E$. Then $x \notin E^c$, and $x$ is not a limit point of $E^c$.

Hence, there must exist a neighborhood $N$ of $x$ such that $E^c \cap N$ is empty.

This implies that $N \subset E$. Hence, $x$ is an interior point of $E$.

Since every point of $E$ is an interior point of $E$, it is open.

2. $E$ is open implies $E^c$ is closed.

Suppose $E$ is open. This means every point of $E$ is an interior point of $E$. Equivalently, if $x$ is not an interior point of $E$, then $x \notin E$.

Let $x$ be a limit point of $E^c$. Then every neighborhood of $x$ contains a point of $E^c$.

Hence, there cannot exist a neighborhood $N$ such that $N \subset E$. Therefore, $x$ is not an interior point of $E$. As $E$ is open, it follows that $x \notin E$ and therefore $x \in E^c$.

Since every limit point of $E^c$ is a point of $E^c$, it is closed.

Theorem 2.24a

For any collection $\{ G_{\alpha} \}$ of open sets, $\cup_{\alpha} G_{\alpha}$ is open.

Proof

Let $G = \cup_{\alpha} G_{\alpha}$. If $x \in G$, then $x \in G_{\alpha}$ for some $\alpha$.

Since $G_{\alpha}$ is open, $x$ is an interior point of $G_{\alpha}$.

Clearly, $x$ must also be an interior point of $G$ and therefore, $G$ is open.

Theorem 2.24b

For any collection $\{ F_{\alpha} \}$ of closed sets, $\cap_{\alpha} F_{\alpha}$ is closed.

Proof

By Theorem 2.22,

$$\left ( \cap_{\alpha} F_{\alpha} \right )^c = \cup_{\alpha} F_{\alpha}^c.$$

Since $F_{\alpha}$ is closed, $F_{\alpha}^c$ is open, by Theorem 2.23.

Since $F_{\alpha}^c$ is open, $\cup_{\alpha} F_{\alpha}^c$ is open, by Theorem 2.24a.

This means $\left ( \cap_{\alpha} F_{\alpha} \right )^c$ is open, and implies $\cap_{\alpha} F_{\alpha}$ is closed, again by Theorem 2.23.

Theorem 2.24c

For any finite collection $G_1, \dotsc, G_n$ of open sets, $\cap_i^n G_i$ is open.

Proof

Let $H = \cap_i^n G_i$. Then, $x \in H$ implies $x \in G_i$ for all $i = 1, \dotsc, n$.

Since $G_i$ is open, there exists a neighborhood $N_i$ of $x$ with radius $r_i$, such that $N_i \subset G_i$ for each $i = 1, \dotsc, n$.

Let $$r = \min (r_1, \dotsc, r_n),$$ and let $N$ be the neighborhood of $x$ with radius $r$.

Clearly, $N \subset G_i$ for all $i = 1, \dotsc, n$, so that $N \subset H$.

Since any point in $H$ is an interior point of $H$, it is open.

Theorem 2.27a

If $X$ is a metric space and $E \subset X$, then $\bar{E}$ is closed.

Proof

If $p \in X$ and $p \notin \bar{E}$ (i.e. $p \in \bar{E^c}$) then $p$ is neither a point of $E$ nor a limit point of $E$.

Hence, there must exist a neighborhood $N$ of $p$ such that $\bar{E} \cap N$ is empty.

This implies that $N \subset \bar{E^c}$. Hence, $p$ is an interior point of $\bar{E^c}$, and $\bar{E^c}$ is open.

Since $\bar{E^c}$ is open, $\bar{E}$ is closed, by Theorem 2.23.

Theorem 2.27b

If $X$ is a metric space and $E \subset X$, then $E = \bar{E}$ if and only if $E$ is closed.

Proof

1. $E = \bar{E}$ implies $E$ is closed, by Theorem 2.27a.

2. $E$ is closed implies $E = \bar{E}$.

Assume $E$ is closed. Then, $x \in E’$ implies $x \in E’$, or equivalently, $E’ \subset E$.

Hence, $\bar{E} = E’ \cup E = E$.

Theorem 2.27c

If $X$ is a metric space and $E \subset X$, then $\bar{E} \subset F$ for every closed set $F \subset X$ such that $E \subset F$.

Proof

If $F$ is closed, then $F’ \subset F$.

If $F$ is closed and $E \subset F$, then $E’ \subset F$.

Since $E \subset F$ and $E’ \subset F$, we have $\bar{E} = E \cup E’ \subset F$.