A problem on neighbourhood bases of a metric space Let $(X,d)$ be a metric space and $ a \in X $. If $\{a\}$ is not open, then show that any neighbourhood base of $a$ is infinite.
 A: Suppose that $\{a\}$ is not open, and also that $a$ has a finite neighbourhood base, say $U_1,\ldots,U_k$ for some $k$. 
As every $U_i$ is open and contains $a$, there are $r_i > 0$ such that $B(a,r_i) \subset U_i$ for $i=1,\ldots,k$.
Then define $r_0 = \min(r_1,\ldots,r_k) > 0$. (Here we use the finiteness: the minimum exists and is still positive.) We know by assumption that $\{a\} \neq B(a,r_0)$ (or $\{a\}$ is open) so there is some $b \neq a$ with $d(a,b) < r_0$ (which is then a point in  all $B(a, r_i)$ and thus in all $U_i$). Let $s = d(a,b) > 0$. 
Then $B(a,s)$ is open, and does not contain $b$ by construction, while all $U_i$ do contain $b$, as we saw. So no $U_i$ exists with $U_i \subset B(a,s)$, so the $U_i$ do not form a neighbourhood base at $a$, and this is a contradiction. So there can be no finite neighbourhood base as required.
A: A metric space is $T_1$, meaning that for each two points $x,y\in X$ there is an open set $U$ containing $x$ but not $y$. Equivalently, every point is the intersection of all of its neighborhoods. But intersecting all neighborhoods of a point gives the same as intersecting all elements of a neighborhood base. So if $x$ had a finite neighborhood base, the singleton $\{x\}$ would be a neighborhood of $x$.
A: Hint: If $U$ is any open set containing the set $\{a\}$ then there is a $r > 0$ such that $B(a;r) \subset U$. Now, if you had a finite number of open sets forming a neighbourhood basis for $a$, then what about the open ball strictly inside the intersection of those finitely many open sets?
