properties of first countability

Proposition 1.

Let $X$ be a first countable topological space and $x\mathrm{\in }X$. Then $x\mathrm{\in }\overline{A}$ iff there is a sequence $\mathrm{(}{x}_i\mathrm{)}$ in $A$ that converges to $x$.

Proof.

One side is true for all topological spaces: if $(x_i)$ is in $A$ converging $x$, then for any open set $U$ of $x$, there is some $i$ such that $x_i\in U$, whence $U\cap A\ne \mathrm{\varnothing }$. As a result, $x\in \overline{A}$.

Conversely, suppose $x\in \overline{A}$. Let $\{B_i\mid i=1,2,\mathrm{\dots }\}$ be a neighborhood base around $x$. We may as well assume each $B_i$ open. Next, let

$$N_n:=B_1\cap B_2\cap \mathrm{\cdots }\cap B_n,$$

then we obtain a set of nested open sets containing $x$:

$$N_1\supseteq N_2\supseteq \mathrm{\cdots }.$$

Since each $N_i$ is open, its intersection with $A$ is non-empty. So we may choose $x_i\in N_i\cap A$. We want to show that $(x_i)$ converges to $x$. First notice tat for any fixed $j$, $x_i\in N_j$ for all $i\ge j$. Pick any open set $U$ containing $x$. Then $N_j\subseteq B_j\subseteq U$. Hence $x_i\in U$ for all $i\ge j$.∎

From this, we can prove the following corollaries (assuming all spaces involved are first countable):

Corollary 1.

$C$ is closed iff every sequence $\mathrm{(}{x}_i\mathrm{)}$ in $C$ that converges to $x$ implies that $x\mathrm{\in }C$.

Proof.

First, assume $(x_i)$ is in a closed set $C$ converging to $x$. Then $x\in \overline{C}$ by the proposition above. As $C$ is closed, we have $x\in \overline{C}=C$.

Conversely, pick any $x\in \overline{C}$. By the proposition above, there is a sequence $(x_i)$ in $C$ converging to $x$. By assumption $x\in C$. So $\overline{C}\subseteq C$, which means that $C$ is closed.∎

Corollary 2.

$U$ is open iff every sequence $\mathrm{(}{x}_i\mathrm{)}$ that converges to $x\mathrm{\in }U$ is eventually in $U$.

Proof.

First, suppose $U$ is open and $(x_i)$ converges to $x\in U$. If none of $x_i$ is in $U$, then all of $x_i$ is in its complement $X-U$, which is closed. Then by the proposition, $x$ must be in the closure of $X-U$, which is just $X-U$, contradicting the assumption that $x\in U$. Hence $x_i\in U$ for some $i$.

Conversely, assume the right hand side statement. Suppose $x\notin U^{\circ }=X-\overline{X-U}$. Then $x\in \overline{X-U}$. By the proposition, there is a sequence $(x_i)$ in $X-U$ converging to $x$. If $x\in U$, then by assumption, $(x_i)$ is eventually in $U$, which means $x_i\in U$ for some $i$, contradicting the earlier statement that $(x_i)$ is in $X-U$. Therefore, $x\notin U$, which implies that $U\subseteq U^{\circ }$, or $U$ is open.∎

Corollary 3.

A function $f\mathrm{:}X\mathrm{\to }Y$ is continuous iff it preserves converging sequences.

Proof.

Suppose first that $f$ is continuous, and $(x_i)$ in $X$ converging to $x$. We want to show that $(f(x_i))$ converges to $f(x)$. Let $V$ be an open set containing $f(x)$. So $f^{-1}(V)$ is open containing $x$, which implies that there is some $j$ such that $x_i\in f^{-1}(V)$ for all $i\ge j$, or $f(x_i)\in V$ for all $i\ge j$, which means that $(f(x_i))\to f(x)$.

Conversely, suppose $f$ preserves converging sequences and $C$ a closed set in $Y$. We want to show that $D:=f^{-1}(C)$ is closed. Suppose $(x_i)$ is a sequence in $D$ converging to $x$. Then $(f(x_i))$ converges to $f(x)$. Since $(f(x_i))$ is in $C$ and $C$ is closed, $f(x)\in C$ by the first corollary above. So $x\in f^{-1}(C)=D$ too. Hence $D$ is closed, again by the same corollary.∎