Hausdorff space that's not first countable?

Question

What's an example of a Hausdorff space that's not first countable?

Answer 1

Let $X=\mathbb{R}^\mathbb{R}$ denote the set of real functions.

Given a countable set $C\subseteq\mathbb{R}$ and a function $f\in X$, let $B(f,C)=\{g\in X:f_C=g_C\}$ where $f_C$ denotes the restriction of $f$ to $C$. (This set might be taken as a ball containing all the real functions that "agree" with $f$ at least on $C$; the ball "narrows down" as $C$ grows larger.) Now let $\mathcal{B}=\{B(f,C):f\in X, C\subseteq\mathbb{R}, C \text{ finite}\}$. Then $\mathcal{B}$ has the following two properties.

• $\bigcup\mathcal{B}=X$. (Proof: simply $B(f,\emptyset)=X$ for all $f\in X$.)
• For any $B_1,B_2\in\mathcal{B}$ and for any $h\in B_1\cap B_2$ there exists a $B_3\in\mathcal{B}$ such that $$h\in B_3\subseteq B_1\cap B_2.$$ (Proof: let $B_1=B(f,C)$ and $B_2=B(g,D)$ for some $f,g\in X$ and some countable $C,D\subseteq\mathbb{R}$. Let $B_3=B(h,C\cup D)$. Obviously $h\in B_3$. Let's show that $B_3\subseteq B_1\cap B_2$. Let $s\in B_3$. Then $$s_{C\cup D}=h_{C\cup D}\implies s_{C}=h_{C}=f_C\implies s\in B_1$$ and similarly $$s_{C\cup D}=h_{C\cup D}\implies s_{D}=h_{D}=g_D\implies s\in B_2$$ and we're done.)

These properties imply that the set $$\mathcal{T}=\{A\subseteq X:\forall f\in A\,\exists B\in\mathcal{B}:f\in B\subseteq A\}\cup\{\emptyset\}$$ is a topology on $X$ of which $\mathcal{B}$ is a base (in particular, $\mathcal{B}\subseteq\mathcal{T}$).

$(X,\mathcal{T})$ is Hausdorff

Let $f,g\in X$ with $f\neq g$. Let $x\in\mathbb{R}$ such that $f(x)\neq g(x)$. Let $B_1=B(f,\{x\})$ and $B_2=B(g,\{x\})$. Then $B_1$ and $B_2$ are neighbourhoods of $f$ and $g$ respectively (since $B_1,B_2\in\mathcal{T}$) which don't intersect. To show the latter, observe that $$h\in B_1\implies h_{\{x\}}=f_{\{x\}}\implies h(x)=f(x)\neq g(x)\implies h_{\{x\}}\neq g_{\{x\}}\implies h\nsubseteq B_2$$ implying that $B_1\cap B_2=\emptyset$. We conclude that $(X,\mathcal{T})$ is a Hausdorff space.

$(X,\mathcal{T})$ is not first countable

Let $\mathcal{A}=\{A_n\}_{n\in\mathbb{N}}$ be a family of open sets all containing the zero function $0\in X$. Let's see that $\mathcal{A}$ can't be a neighbourhood basis for $0$. For each $n\in\mathbb{N}$, let $C_n\subseteq\mathbb{R}$ a finite set such that $B(0,C_n)\subseteq A_n$. Let $y\in\mathbb{R}$ such that $y\notin C_1\cup C_2 \cup C_3\ldots$ (which is countable hence not $\mathbb{R}$). Now let $B=B(0,\{y\})$ and let $\chi\in X$ be the function defined as $$\chi(x)=\begin{cases}1\quad\text{if $x=y$}\\ 0\quad\text{if $x\neq y$}\end{cases}.$$

Then note that $\chi\notin B$ because $\chi(y)\neq 0(y)$ but, for any $n\in\mathbb{N}$ and any $x\in C_n$, $\chi(x)=0$; therefore $\chi_{C_n}=0_{C_n}$ and thus $\chi\in B(0,C_n)\subseteq A_n$.

But then the aforementioned $\mathcal{A}$ cannot be a neighbourhood basis for $0$ since no $A_n$ could ever be contained in $B$, as we would have $\chi\in A_n\subseteq B\implies \chi\in B$ which we've seen to be false.

Hence $(X,\mathcal{T})$ is not first countable.