# General Topology/Countability, density

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Definition (dense):

Let $X$ be a topological space and let $A\subseteq X$ be a subset. $A$ is called dense if and only if ${\overline {A}}=X$ .

Definition (first-countable):

Let $X$ be a topological space. $X$ is called first-countable iff for all $x\in X$ the neighbourhood filter $N(x)$ has a countable basis.

Definition (second-countable):

Let $X$ be a topological space. $X$ is called second-countable iff the topology of $X$ has a countable basis.

Since subsets of countable sets are countable and the open neighbourhoods generate $N(x)$ , second-countability implies first-countability.

Definition (separable space):

Let $X$ be a topological space. $X$ is called separable if and only if there exists a countable set $S\subseteq X$ which is dense in $X$ .

Proposition (second-countable spaces are separable):

Let $X$ be a second-countable space. Then $X$ is separable.

(On the condition of the axiom of countable choice.)

Proof: Let $(U_{n})_{n\in \mathbb {N} }$ be a basis of the topology of $X$ and choose $x_{n}\in U_{n}$ . Then $S:=\{x_{n}|n\in \mathbb {N} \}$ is countable and dense. $\Box$ Proposition (subspace of second-countable space is second-countable):

Let $X$ be a second-countable space, and $S\subseteq X$ a subset. Turn $S$ into a topological space using the subspace topology. $S$ is then a second-countable space.

Proof: Any countable basis $(U_{n})_{n\in \mathbb {N} }$ of the topology of $X$ induces a countable basis $(S\cap U_{n})_{n\in \mathbb {N} }$ of the subspace topology on $S$ . $\Box$ Proposition (continuous function into Hausdorff space is uniquely determined by dense subspace):

Let $X,Y$ be topological spaces, where $Y$ is Hausdorff. Let $A\subseteq X$ be a dense subspace, and suppose $f:A\to Y$ is continuous. Whenever $F,G:X\to Y$ are continuous functions such that $F|_{A}=G|_{A}$ , then $F=G$ .

Proof: Let $x\in X$ be arbitrary, and let $V\subseteq Y$ be any neighbourhood of $G(x)$ . By continuity of $G/math>wemayfindaneighbourhood[itex]U$ of $x$ that is mapped completely into $V$ . Analogously, whenever $V'$ is a neighbourhood of $F(x)$ , we find a neighbourhood $U'$ mapping completely into $V'$ . Then $U\cap U'$ is mapped completely into $V\cap V'$ , so that $F(x),G(x)\subseteq V\cap V'$ for any open neighbourhoods $V$ of $G(x)$ and $V'$ of $F(x)$ . If $F(x)\neq G(x)$ , then $V\cap V'=\emptyset$ for suitable $V,V'$ as above by the Hausdorff condition, a contradiction to $G(x)\in V\cap V'$ . Hence, $F(x)=G(x)$ . Since $x$ was arbitrary, we conclude. $\Box$ 