Is this quotient space first countable? Let $\tau$ be the usual metric topology on $\mathbb{R}$. Consider the partition $\{\mathbb{Z}\}\cup\{\{x\}: x\in \mathbb{R}\setminus\mathbb{Z}\}$.
Equip this space with the quotient topology. 
The question is asking to show if the quotient map is open (resp. closed) and also if the quotient space is first countable.
I have managed to show that the quotient map, $f$, is not open since $f^{-1}(f(0,2)) = (0,2)\cup\mathbb{Z}$. But $(0,2)\cup\mathbb{Z}$ is clearly not open in $\mathbb{R}$ so $f(0,2)$ is not open in the quotient space.
However the quotient map is closed since $f^{-1}(f(K))$ is either $K$ itself (and so closed in $\mathbb{R}$) or $K\cup \mathbb{Z}$ (also closed in $\mathbb{R}$). So every closed image is closed in the quotient topology.
However, I am not sure how to show if the quotient is first countable (that is, has a countable nhood-base). How can I do this? If the quotient map were open, then is would be easier.
 A: This space is not first countable: let's prove it by contradiction.
Suppose by contradiction there exists a countable local base of neighbourhoods of $\{ \mathbb{Z} \}$, and let $\{ U_k \}_{k \in \Bbb{Z}}$ be an enumeration of it.
For all $n,k \in \Bbb{Z}$ denote by
$$(a_n^k, b_n^k) = f^{-1} (U_k) \cap \left( n-\frac{1}{3} , n + \frac{1}{3} \right)$$
We can suppose without loss of generality that $$f^{-1}(U_k) = \bigcup_{n \in \Bbb{Z}} (a_n^k, b_n^k)$$
where $n \in (a_n^k, b_n^k)$ (I say that this is "$U_k$ locally at $n$").
Now we build a special neighbourhood of $\{ \mathbb{Z} \}$ which does not contain any of the $U_k$ by a diagonalization argument. Cosider the following:
$$U= f \left( \bigcup_{k \in \Bbb{Z}} (a_k^k + \varepsilon_k ,b_k^k - \varepsilon_k) \right)$$
where $(a_k^k - \varepsilon_k ,b_k^k + \varepsilon_k)$ is an open interval containing $k$ which is slightly tighter than $(a_k^k, b_k^k)$ (and $\varepsilon_k>0$ is very small).
By construction, this is an open neighbourhood of $\{ \mathbb{Z} \}$ , and it does not contain any of the $U_k$s because for all $k \in \Bbb{Z}$ "$U$ is smaller than $U_k$ locally at $k$" (I hope you understand this: please, if you don't, ask details).
This is a contradiction: hence our space is not first countable.
