I am given an exercise where I am supposed to show some properties of neighbourhoods.
I am given a metric space ($X$,$d$) and two points, $p$ and $q$, that both are members of $X$ ($p$,$q$ $\in$ $X$).
Firsly, I'm supposed to show that if both $N_1$ and $N_2$ is in the neighbourhood of $p$, then $N_1 \cap N_2$ is in the neighbourhood of $p$. I have tried to solve the problem, using this approach:
Given $N_1$, $N_2$ $\in$ $\mu_p$ (neighbourhood of $p$), then $p \ \in \ N_1^\circ$ and $p \ \in \ N_2^\circ$. Using that $N_1^\circ \cap N_2^\circ$ itself is open, in addition to the fact that $N_1^\circ \cap N_2^\circ \subseteq N_1\cap N_2$, such that $p \ \in \ (N_1\cap N_2)^\circ \ \Rightarrow \ N_1\cap N_2 \ \in \ \mu_p$.
Is this correct? And is there another way to solve the problem?
Further we are asked to show the following: Given two points, $p$ and $q$ ($p \neq q$), there exists a neighbourhood $\mu_p$ (of $p$) that doesn't include the point $q$.
I don't know exactly where to start here. I'm thinking about some way to show that there exists an $\epsilon \ > 0$, representing the set $N$'s interval, such that $d(p,q) > \epsilon$. The problem is that I don't know how to actually show this. Besides this, I guess there a plenty of ways of better ways to show this.
Can someone please help me?