properties of first countability

Proposition 1.

Let $X$ be a first countable topological space  and $x\in X$. Then $x\in\overline{A}$ iff there is a sequence $(x_{i})$ 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 $U\cap A\neq\varnothing$. As a result, $x\in\overline{A}$.

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

 $N_{n}:=B_{1}\cap B_{2}\cap\cdots\cap B_{n},$

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

 $N_{1}\supseteq N_{2}\supseteq\cdots.$

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 $i\geq j$. Pick any open set $U$ containing $x$. Then $N_{j}\subseteq B_{j}\subseteq U$. Hence $x_{i}\in U$ for all $i\geq j$. ∎

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

Corollary 1.

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

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 $(x_{i})$ that converges to $x\in U$ 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. ∎

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 $i\geq j$, or $f(x_{i})\in V$ for all $i\geq j$, 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. ∎

Title properties of first countability PropertiesOfFirstCountability 2013-03-22 19:09:26 2013-03-22 19:09:26 CWoo (3771) CWoo (3771) 11 CWoo (3771) Result msc 54D99