(Proof pattern) Diagonal argument

Examples §

(Topology Homework 8.2) Let $X=\mathbb{Z}\times [0,1]$ and let $Y=X/(\mathbb{Z}\times \{0\})$, where $\mathbb{Z}$ has the discrete topology. Show that $X$ is second-countable (and hence first-countable). Show that $Y$ is not first-countable (and hence not second-countable).

Proof.

We know $\mathbb{Z}$ with the discrete topology is second-countable since it has countably many singletons as basis elements, and $[0,1]$ is second countable since it is a subset of the second-countable space $\mathbb{R}$, so their product $X=\mathbb{Z}\times [0,1]$ is second-countable as well.

We use a diagonal argument to show that $Y$ is not first-countable. Suppose, towards a contradiction, that every point of $Y$ has a countable local basis. In particular, let $\mathcal{B}=\{B_n{\}}_{n\in \mathbb{N}}$ be a countable local basis for $[0]\in Y$, and let $p:X\to Y$ be the canonical projection. If $I_k$ is the $k$-th copy of $[0,1]$ in $X$, there exists some $B_k\in \mathcal{B}$ such that $p^{-1}(B_k)\cap I_k$ is a non-empty open subset. In particular, there is some $x_k\in [0,1]$ such that $p^{-1}(B_k)\cap I_k=[0,x_k)\subset I_k$. Then the union of all such intersections $U={\bigcup }_{n\in \mathbb{N}}[0,x_n/2)$ is open in $X$, so $p(U)$ is open in $Y$. But then $p(U)$ is an open neighborhood of $[0]$ that does not contain any element of $\mathcal{B}$, so $Y$ does not have a countable local basis at $[0]$. $\square$