Q:

Show that given quotient space does not satisfy $1A\mathbb{N}$.

Let $\sim$ be equivalence relation on $\mathbb{R}$ such that:
$x \sim y \iff (x = y$ or $x, y \in \mathbb{Z})$
Show that $X = \mathbb{R}/_{\sim}$ does not satisfy axiom $1A\mathbb{N}$.
Hint: consider $[0]$.
So neighborhoods of $[0]$ are neighborhoods of $p^{-1}([0]) = \mathbb{Z}$, so if I take 
$B_1 = \bigcup_{k \in \mathbb{Z}}(k-1/1, k+1/1)$,
$B_2 = \bigcup_{k \in \mathbb{Z}}(k-1/2, k+1/2)$,
$B_3 = \bigcup_{k \in \mathbb{Z}}(k-1/3, k+1/3)$,
$B_n = \bigcup_{k \in \mathbb{Z}}(k-1/n, k+1/n)$,
I can't see why does this not work as a countable basis for $[0]$, because I think that if I take any neighborhood of $[0]$, I can find some $B_i$ contained in it.
So my question is: what am I missing and how to prove that the given space doesn't satisfy that axiom.

A:

In my answer here ,I show that $\mathbb{N} \subset \mathbb{R}$ does not a countable base of neighbourhoods, so we cannot have a countable family of open sets $U_n, n \in \mathbb{N}$, such that 
$\mathbb{N} \subset U_n$ for all $n$ and for any open $O$ with $\mathbb{N} \subset O$, we have $U_n \subset O$.
But a supposed local base $\{B_n: n \in \mathbb{N}\}$ at $[0]$ when pulled back under $q$, so $\{q^{-1}[B_n]: n \in \mathbb{N}\}$ would exactly be such a non-existent family of neighbourhoods: 
This holds, because if $\mathbb{N} \subset O$ is open, $q^{-1}[q[O]] = O$ is open and so $q[O]$ is open and contains $[0]$.
So then $[0] \in B_n \subset q[O]$ for some $n$, would exist and then $q^{-1}[[0]] = \mathbb{N} \subset q^{-1}[B_n] \subset q^{-1}[q[O]] = O$. So the $q^{-1}[B_n]$ would form a countable base of neighbourhods of $\mathbb{N}$. And we cannot have a countable one as proved. 
Your supposed solution fails (you mean to take $q[B_n]$ of course, otherwise they're not in the quotient space), if you look at my proof:
Define $$O = q[\cup_{k \in \mathbb{Z}} (k-\frac{1}{2(|k|+1)} ,k+\frac{1}{2(|k|+1)})]$$ 
Then $B_n \subset O$ fails at the interval around $n$, at least.
It doesn't really matter whether we identify $\mathbb{Z}$ or $\mathbb{N}$ to one class.

