Frechet Urysohn but not first countable space Consider the following equivalence relation on $\mathbb{R}$ defined by:
$x\sim y \Longleftrightarrow{x=y \ \ or  \ \ x, y \in{} \mathbb{Z}} $
Prove that  $\mathbb{R}/\sim$ is a Frechet-Urysohn space, but it's not first countable.
I have proved that for $x \in{} \mathbb{Z}$, $[x]=\mathbb{Z}$, and for $x \in{} \mathbb{R}/ \mathbb{Z}$, we have $[x]=\left\{{x}\right\}$.
I start my proof of this form: Let be $A\subseteq \mathbb{R}/\sim$ and $p \in{} \bar{A}$.


*

*If $p \in{} A$, it's trivial to show that there exist a sequence of elements of A that converges to p.


But I have not been able of proof this when $p \notin A$
 A: Let $X=\Bbb R/\!\!\sim$, and let $z$ be the point of $X$ corresponding to $\Bbb Z$. It’s helpful to realize that $X\setminus\{z\}$ is homeomorphic to $(0,1)\times\Bbb Z$: it’s just the union of a countably infinite family of pairwise disjoint copies of $(0,1)$, each of which is clopen in $X\setminus\{z\}$.
For $p\in(\operatorname{cl}_XA)\setminus A$, you have two cases: $p=z$, and $p\ne z$. In the second case the point of $\Bbb R$ corresponding to your $p$ must lie strictly between $n$ and $n+1$ for some $n\in\Bbb Z$; in terms of my first paragraph, it’s in $\{n\}\times(0,1)$. Here the topology of $X$ is just like that of $\Bbb R$: as long as you stay between $n$ and $n+1$, you’re essentially just working in a metric space homeomorphic to $(0,1)$.
The first case takes a little more work, because the set of points of $\Bbb R$ corresponding to $A$ might be something like
$$\left\{n+\frac1n:n\in\Bbb Z^+\right\}\;,$$
a set that doesn’t have any single integer as a limit point. However, it does have points arbitrarily close (in $\Bbb R$) to the set $\Bbb Z$, and that’s the observation that you need. Show that if $z\in(\operatorname{cl}_XA)\setminus A$, then there is a sequence $\langle a_n:n\in\Bbb Z^+\rangle$ in $A$ such that if $r_n$ is the point of $\Bbb R$ corresponding to $a_n$, there is a $k\in\Bbb Z$ such that $|r_n-k|<\frac1n$.
Once you’ve done all this, it should be clear that $X$ is first countable at every point of $X\setminus\{z\}$, since at points of $X\setminus\{z\}$ it looks locally just like $\Bbb R$. Thus, you should try to prove that it’s not first countable at $z$. Here’s a HINT:

For each doubly-infinite sequence $r=\langle r_n:n\in\Bbb Z\rangle$ of positive real numbers let
$$B(r)=\bigcup_{n\in\Bbb Z}(n-r_n,n+r_n)\;,$$
and let $\tilde{B}(r)$ be the corresponding subset of $X$. Let $\mathscr{B}$ be the family of all such sets $\tilde{B}(r)$. Show that $\mathscr{B}$ is a local base at $z$. Then show that if $\mathscr{U}=\{U_n:n\in\Bbb N\rangle$ is any family of open nbhds of $z$ in $X$, there is a $\tilde{B}(r)\in\mathscr{B}$ that does not contain any of the sets $U_n$, so that $\mathscr{U}$ cannot be a local base at $z$. This will show that $X$ is not first countable at $z$.

Added: If $\mathscr{U}=\{U_n:n\in\Bbb N\}$ is a family of open nbhds of $z$, for each $n\in\Bbb N$ there is an $r^{(n)}=\left\langle r_k^{(n)}:k\in\Bbb N\right\rangle$ such that $\tilde{B}\left(r^{(n)}\right)\subseteq U_n$. For each $n\in\Bbb N$ let $r_n=\frac12r_n^{(n)}$, and let $r=\left\langle r_n^{(n)}:n\in\Bbb N\right\rangle$. Now show that $\tilde{B}\left(r^{(n)}\right)\nsubseteq\tilde{B}(r)$ for each $n\in\Bbb N$, and conclude that $U_n\nsubseteq\tilde{B}(r)$ for each $n\in\Bbb N$. This shows that $\mathscr{U}$ is not a local base at $z$, and since $\mathscr{U}$ was an arbitrary countable family of open nbhds of $z$, we can conclude that $X$ is not first countable at $z$.
