General Topology/Countability, density

Definition (dense):

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

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

Proposition ():

second-countable spaces are 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. $◻$

Proposition ():

subspace of second-countable space is second-countable

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 $S$. $◻$

Proposition ():

continuous function into Hausdorff space is uniquely determined by dense subspace

Proof: Let $x\in X$ be arbitrary, and let $V\subseteq Y$ be any neighbourhood of $G(x)$. By continuity of $G$ we may find a neighbourhood $U$ of $x$ that is mapped completely into $V$. Analogously, whenever $V^{\prime }$ is a neighbourhood of $F(x)$, we find a neighbourhood $U^{\prime }$ mapping completely into $V^{\prime }$. Then $U\cap U^{\prime }$ is mapped completely into $V\cap V^{\prime }$, so that $F(x),G(x)\subseteq V\cap V^{\prime }$ for any open neighbourhoods $V$ of $G(x)$ and $V^{\prime }$ of $F(x)$. If $F(x)\ne G(x)$, then $V\cap V^{\prime }=\mathrm{\emptyset }$ for suitable $V,V^{\prime }$ as above by the Hausdorff condition, a contradiction to $G(x)\in V\cap V^{\prime }$. Hence, $F(x)=G(x)$. Since $x$ was arbitrary, we conclude. $◻$