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 $x$.
 A: 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 $x$.
A: The problem is that there is no commonly accepted definition of the conecpt of a neighborhood.
Some authors define it as in your Edit 2. That is, a neigborhood of a point $x \in X$ is an open subset $U$ of $X$ such that $x \in U$.
Other authors define a neighborhood of $x \in X$ to be any subset $N \subset X$ such that there exists an open $ U \subset X$ with the property $x \in U \subset N$.
Clearly, the second variant is more general than the first. Using this variant, one must explicitly speak about open neighorhoods if one wants to refer to the neigborhoods in the sense of variant 1. But there also exist closed neigborhoods, compact neighborhoods etc.
Personally I prefer variant 2 because it is more flexible. And you need this interpretation to state that that in a metric space $X$ the set of all closed balls around $x$ form a neighborhood basis.
Let us write  $\mathcal N_1(x)$ for the neighbourhood system of $x$ in the sense of variant 1 and $\mathcal N_2(x)$ for that in the sense of variant 2. Clearly $\mathcal N_1(x) \subset \mathcal N_2(x)$, but in general $\mathcal N_1(x) \ne \mathcal N_2(x)$, i.e. $\mathcal N_1(x) $ is a proper subset of $\mathcal N_2(x)$.
Let us come to your question. Your interpretation of a neigborhood is that of variant 1. Since closed balls are in general no open sets, they do in general not belong to $\mathcal N_1(x)$, but only to $\mathcal N_2(x)$. Therefore in general they do not form a neighbourhood basis for the  point $x$ in the sense of Edits 1 and 2.
However, if you work with $\mathcal N_2(x)$, then the closed balls do form neighbourhood basis for $x$. See GEdgars answer.
