Can closed balls also be a neighborhood basis? I am reading about bases in topology and have encountered the concept of a (local) neighborhood basis. The example they give is for the topology on metric spaces: the set of all open balls around $x$ (or the set of all small open balls around $x$, or all open balls with radius $\frac{1}{n}$ for all $n \in \mathbb{N}$ around $x$). What confuses me is why they only give open balls as examples. Would the set of all closed balls around $x$ also be an example of a neighborhood basis? I think that it satisfies the definition, but I can't find any source mentioning it as an example.

Edit: (adding the definition)
A neighbourhood system $\mathcal{N}(x)$ for a point $x$ is the collection of all neighbourhoods of the point $x$.
A neighbourhood basis or local basis for a point $x$ is a subset $\mathcal{B} \subseteq \mathcal{N}(x)$, such that for all $V \in \mathcal{N}(x)$, there exists some $B \in \mathcal{B}$ such that $B \subseteq V$. That is, for every neighbourhood $V$, we can find a neighbourhood $B$ in the neihbourhood basis that is contained in $V$.

Edit 2: (definition of neighbourhood)
If $(X, \mathcal{T})$ is a topological space and $x$ is a point in $X$, a neighbourhood of $x$ is any set $U \in \mathcal{T}$ containing $x$.

You are correct, in a metric space you could use the closed balls (with positive radius) centered at a point $x$ as a neighborhood basis for $x$.  Then define: a set $E$ is open iff, for every $x \in E$ there exists an element $U$ of the neighborhood basis at $x$ so that $x \in U \subseteq E$.  Even if your neighborhood bases do not consist of open sets, you still get a topology by doing this.
To prove this works (using open balls or closed balls for the neighborhood basis give you the same topology):
The main thing you need is: every open ball centered at $x$ contains a closed ball centered at $x$, and every closed ball centered at $x$ contains an open ball centered at $x$.
