Consider a topological space $(M,\mathscr{M})$ and $p,q\in M$. We suppose that for any two neighborhoods $N_p$ and $N_q$ we have that
$$\tag{1}
N_p\cap N_q\ne\emptyset
$$
But since this includes the smallest open sets that contain $q$ and $p$, does the condition in eq. $(1)$ not imply that any neighborhood of one of the two points is also a neighborhood of the other point? And this would imply that for $4$ distinct neighborhoods, we would also have
$(N_p^{(1)}\cap N_q^{(1)})\cap(N_p^{(2)}\cap N_q^{(2)})\ne\emptyset$?
Edit: the reason for the question is that I want to show that
$$
F=\{S\subseteq M|\exists N_p\exists N_q \text{ such that } N_p\cap N_q\subseteq S\}
$$
is a filter on $M$. Thus I need to show that I can find $4$ neighborhoods such that $(N_p^{(1)}\cap N_q^{(1)})\cap(N_p^{(2)}\cap N_q^{(2)})$ is in $F$, which means that $(N_p^{(1)}\cap N_q^{(1)})$ should be a neighborhood of $p$ and that $(N_p^{(2)}\cap N_q^{(2)})$ should be a neighborhood of $q$.
Best Answer
No. Let us consider $M=\{p,q\}$ with the topology $\{\emptyset, \{p\}, \{p,q\}\}$. Then $\{p\}\cap\{p,q\}=\{p\}$, but $\{p\}$ is not a neighbourhood of $q$.