Are first and second countability preserved under intersection of topologies? For a given set $X$ endowed with two topologies $\mathcal{T}$ and $\mathcal{T}'$, i.e. such that $(X,\mathcal{T})$ and $(X,\mathcal{T}')$ are two topological spaces defined on
the same $X$, it is easy to see that $\mathcal{T}\cap\mathcal{T}':=\{A:A\in\mathcal{T}\land A\in\mathcal{T}'\}$ is also  a topology of $X$. Then, my two questions are:


*

*If $(X,\mathcal{T})$ and $(X,\mathcal{T}')$ are first countable, is it necessarily the case that $(X,\mathcal{T}\cap\mathcal{T}')$ is also first countable?

*If $(X,\mathcal{T})$ and $(X,\mathcal{T}')$ are second countable, is it necessarily the case that $(X,\mathcal{T}\cap\mathcal{T}')$ is also second countable?


This is what I know. I do know that first and second countability are preserved under the finite product of topologies. In fact, I also know that, if we define the sum of the topologies $\mathcal{T}$ and $\mathcal{T}'$ of $X$ as $\mathcal{T}+\mathcal{T}':=\{A\cap A':A\in\mathcal{T}\land A'\in\mathcal{T}'\}$, which can be proven to be the minimal topology that contains $\mathcal{T}\cup\mathcal{T}'$, then the sum of first/second countable topologies is first/second countable. This last fact is due to: (1) if $\mathcal{B}(x)$ and $\mathcal{B}'(x)$ are local bases for a given $x\in X$ in their respective topologies, then $\mathcal{B}''(x):=\{B\cap B':B\in\mathcal{B}(x)\land B'\in\mathcal{B}'(x)\}$ is a local basis of $x$ in $\mathcal{T}+\mathcal{T}'$ (thus, first countability is preserved under addition); and (2) if $\mathcal{B}$ and $\mathcal{B}'$ are global bases of their respective topologies, then
$\mathcal{B}'':=\{B\cap B':B\in\mathcal{B}\land B'\in\mathcal{B}'\}$
is a global basis of $\mathcal{T}+\mathcal{T}'$ (which implies that second countability is also preserved under addition).
This is what I'm struggling with. Let's say that $(X,\mathcal{T})$ and $(X,\mathcal{T}')$ are second countable, so there are countable global bases $\mathcal{B}=\{B_n:n\in\mathbb{N}\}$ and $\mathcal{B}'=\{B'_n:n\in\mathbb{N}\}$ for each topology. Now, to prove that $\mathcal{T}\cap\mathcal{T}'$ is second countable (if it is provable), we would like to construct a countable basis $\mathcal{B}''$ of $\mathcal{T}\cap\mathcal{T}'$ from $\mathcal{B}$ and $\mathcal{B}'$. We know that $\forall \emptyset\neq A\in\mathcal{T}\cap\mathcal{T}'$ and every $x\in A$ we have that $\exists B_x\in\mathcal{B}, B'_x\in\mathcal{B}'$ such that $x\in B_x\subseteq A$ and $x\in B'_x\subseteq A$. The thing is that, from what I know, it could so happen that $\nexists B_x''\in\mathcal{T}\cap\mathcal{T}':x\in B''_x\subseteq B_x\cup B'_x$. And that's why I'm struggling with proving this (and the same spiel with first countability). Because of this, I'm also considering that maybe there are some counterexamples of these two statements out there. Any help would be very much appreciated.
 A: Consider the following two topologies on the set $\mathbb{N}^2$.  The first topology is the finest topology that makes $(m,n)\to(m,0)$ as $n\to\infty$ for each $m$ (so neighborhoods of $(m,0)$ must contain $(m,n)$ for all but finitely many $n$ and all other points are isolated).  The second topology is the finest topology that makes $(m,0)\to(1,1)$ as $m\to\infty$.  It is easy to see both of these topologies are second-countable.  However, I claim that their intersection is not first-countable.
To prove this, note that any neighborhood of $(1,1)$ in the intersection must contain $(m,0)$ for all but finitely many $m$ and then contain $(m,n)$ for all but finitely many $n$ for each such $n$.  Suppose $(U_n)_{n\in\mathbb{N}}$ is any countably family of neighborhoods of $(1,1)$; we will find a neighborhood $U$ that does not contain in any of them.  First pick a strictly increasing sequence $(k_n)$ such that $(k_n,0)\in U_n$ for each $n$ (this is possible since each $U_n$ contains $(m,0)$ for infinitely many different $m$).  Then pick $l_n>1$ such that $(k_n,l_n)\in U_n$ for each $n$.  Now let $U=\mathbb{N}^2\setminus\{(k_n,l_n):n\in\mathbb{N}\}$.  It is easy to see that $U$ is open, and it does not contain any $U_n$ since it omits each $(k_n,l_n)$.
(Here is how I came up with this example.  The intersection of two topologies on $X$ can be described as the quotient topology for the fold map $X\coprod X\to X$, where the first copy of $X$ has one topology and the second copy of $X$ has the other topology.  So, what you're really looking for here is an example of a first/second-countable space with a quotient that is not first/second-countable.  A space is a quotient of a first-countable space iff it is sequential, so you want to consider examples of sequential spaces that are not first-countable.  The intersection topology considered above is one of the most basic such examples.  It's then just a matter of seeing whether you can write it as a quotient of a first-countable space where the quotient map can be viewed as a fold map for some set.)
