# General Topology/Countability, density

Definition (dense):

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

Definition (first-countable):

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

Definition (second-countable):

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

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

Definition (separable space):

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

Proposition (second-countable spaces are separable):

Let ${\displaystyle X}$ be a second-countable space. Then ${\displaystyle X}$ is separable.

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

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

Proposition (subspace of second-countable space is second-countable):

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

Proof: Any countable basis ${\displaystyle (U_{n})_{n\in \mathbb {N} }}$ of the topology of ${\displaystyle X}$ induces a countable basis ${\displaystyle (S\cap U_{n})_{n\in \mathbb {N} }}$ of the subspace topology on ${\displaystyle S}$. ${\displaystyle \Box }$

Proposition (continuous function into Hausdorff space is uniquely determined by dense subspace):

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

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