Recently me and some friends were discussing first countability, and noticed that Wikipedia gave an example of space that was not first countable, but with no proof. So here’s my attempt at a proof.

Let $X=\mathbb R / \mathbb N$, where we identify $1 \simeq 2 \simeq \ldots$. Geometrically, this space is an “infinite wedge product” of circles. It looks like a flower with a countable infinite number of “petals”.

The open sets in $X$ are of two types: those containing $[1]$ (the equivalence class of $1 \in X$), and those not containing $[1]$. The latter ones are just shorts intervals not containing any natural numbers. Let $\pi: \mathbb R \to X$ be the projection map. Note that if $[1] \in V$, then $\mathbb N \subset \pi^{-1}(V)$ (since $\pi^{-1}([1])=\mathbb N$, by definition of $X$). This implies that any open set in $X$ containing $[1]$ must intersect all petals (and so must be a union of intervals around each natural number).

# Proposition: $X$ is not first countable.

Recall that a space is first countable if every point $x \in X$ has a countable neigbourhood basis. A neigbourhood basis is a collection of open subsets that is “arbitrarily small”.

Note that we can consider a neigbourhood basis as indexed by the natural numbers, i.e. a function $f: \mathbb N \to \mathscr P(X)$. We will denote $f(i)$ by $N_i$, and its preimage in $\mathbb R$ by $N_i’$.

We will construct an open set $U \subset X$ such that $N_i \not \subset U$ for all $i$ given any proposed neigbourhood basis ${ N_i }$ around $[1]$.

For $N_1$, since $\mathbb N \subset N_1’$, we can find a small interval $I_1’$ in $\mathbb R$ around $1 \in \mathbb R$ strictly contained in $N_1’$ (make it small enough so that it doesn’t intersect any other natural numbers).

We continue: choose a small interval $I_2’$ around $2 \in \mathbb R$ strictly contained in $N_2’$.

And so on… We define $U’$ by taking the union of all the $I_n’$’s. We denote its image in $X$ by $U$. Note that $U=\pi(U’)$ is also open, since $\pi^{-1}(\pi(U))=U’$. Geometrically, $U$ looks like a small interval around every petal on $X$, made in such a way that $N_i \not \subset U$ for every $i$.

We conclude that $X$ cannot be first countable, since given any indexed collection of open subsets containing $[1] \in X$, we can find an open $U$ that is “too big”.