Quotient Space Gluing the Integers Together is Sequential

Question

I have been attempting to prove this statement, namely that when one takes $\mathbb{R}$ with the usual topology and glue the integers together, one obtains a sequential space $\mathbb{R}/ \sim$ that is not first countable. This MSE link provides the necessary proof that the space is not first countable. It remains to show that $\mathbb{R} / \sim$.

Let $\mathbb{R}$ have the usual topology and let $\sim$ be an equivalence relation on $\mathbb{R}$ so that

$$x \sim y \iff x =y \,\, \mathrm{or} \, \, x,y \in \mathbb{Z}$$

Let $A \subset \mathbb{R} / \sim$ be a sequentially closed subset. I want to show that $A$ is closed. I can do this by showing that $A^c$ is open. As $A$ is sequentially closed, then if $\{[x_n]\}_{n=1}^\infty$ is a sequence in $A$, and $[x_n] \to [x]$, then $[x] \in A$ as well. As $[x_n] \to [x]$, we mean that

$$\forall \, \mathrm{open} \, \, U, [x] \in U, \exists \, N \in \mathbb{N}, \forall n \geq N, [x_n] \in U$$

Now, we know that since $U$ is $\mathbb{R} / \sim$ - open, then $q^{-1}(U)$ is $\mathbb{R}$-open. And when $n \geq N$, $x_n \in q^{-1}(U)$. I'm tempted to suspect that I can somehow show that $x_n \to x$ and then use $q$ to send this over to $\mathbb{R} / \sim$ and draw a conclusion. Any thoughts?

It would equally suffice if I could prove every quotient of a sequential space is sequential.

Answer 1

To do this in a hands-on way (but there is actually a straightforward proof quotients of sequential spaces are sequential):

Let $z$ be the collapsed point $q(\Bbb Z)$, $X=q(\Bbb R)$ your quotient space; infinite bouquet of circles. The open subspace $X\setminus\{z\}$ is homeomorphic via $q$ with $\Bbb R\setminus\Bbb Z$.

Say $A$ does not contain $z$. I claim $q^{-1}(A)$ is locally bounded away from the integers: for any $n\in\Bbb Z$ there is a $0<\delta_n<1$ so that $(n-\delta,n+\delta)$ never touches $q^{-1}(A)$, for any $n\in\Bbb Z$. Were this not so, allowing $\delta_n=1/m,\,m\in\Bbb N$ and choosing counterexample points $x_m\in A\cap(n-1/m,n+1/m)$, for some fixed $n$, finds a sequence $(q(x_m))_{m\in\Bbb N}$ lying in $A$ convergent (careful! Why do I need $n$ to be fixed?) to $z\notin A$, a contradiction.

So: $q^{-1}(A)\subseteq\bigsqcup_{n\in\Bbb Z}[n+\delta_n,n+1-\delta_{n+1}]$ (viewing those intervals as empty if $\delta_n>1-\delta_{n+1}$ for our somewhat arbitrary choices). $q^{-1}(A)$ is closed in $\Bbb R$ if and only if it is closed in each of these disjoint, isolated closed intervals (why?). But, this is true by the fact each such interval is a sequential space, are $q$-saturated and that $q$ is a topological embedding of these intervals into $X$; $q^{-1}(A)$ intersected with one of these intervals would be sequentially closed in them (the proof is easy to write down without details, but you do need all the adjectives I used), thus closed. Then $\Bbb R\setminus q^{-1}(A)$ is open and contains $\Bbb Z$, so it is an open saturated set; therefore its $q$-image is open, $X\setminus A$ is open, $A$ is closed in $X$.

If $A$ does contain $z$, we must show its complement $U$ is open and as $U\subseteq X\setminus\{z\}$ - which is an open subspace of $X$ - it suffices to show its homeomorph $q^{-1}(U)\subseteq\Bbb R\setminus\Bbb Z$ is open in $\Bbb R$, or to show that each $q^{-1}(U)\cap(n,n+1)$ is open. By the same point that $(n,n+1)$ embeds into $X$ and is $q$-saturated, $q^{-1}(A)\cap(n,n+1)$ is sequentially closed in $(n,n+1)$, thus closed (relatively!) so $q^{-1}(U)\cap(n,n+1)$ is open (relatively and globally) as desired.