Are the members of neighbourhood system base at a point open sets?

Question

I know that in general Topology a neighbourhood system base at a point contains neighbourhoods of that point. My question is - are those neighbourhoods necessarily open sets?

Answer 1

No. For example let $X= \mathbb R^n$ with the usual topology induced by the $||*||_2$ -norm and $x_0 \in X$. For $r>0$ let $B_r(x_0):=\{x \in X: ||x-x_0|| \le r\}$.

Then $\{B_r(x_0): r>0\}$ is a basis of neighborhoods of $x_0$, and all sets in this basis are not open.

Answer 2

Usually, a neighbourhood of a point $p$ is defined to be a set containing an open set that contains $p$. So (in the usual topology of the real line) $J=[-1,1]$ is a neighbourhood of $0$, as it contains the open set $(-1,1)$ that contains $0$. However, $J$ is not open in the real line. So no, the neighbourhoods in a neighbourhood system need not be open sets.

Of course, you can talk about "open neighbourhood systems" instead, and only consider the open sets that contain a given point.