The definition of basis and neighborhood basis are:
Let $(X,\tau)$ be a topological space, a base of $\tau$ is a subset $\mathfrak{B}$ of $\tau$ such that each open set $A \in \tau$ is union of elements of $\mathfrak{B}$
If $p \in X$, a subset $\mathfrak{B}_p\subseteq U_p=\{U \in \tau | p \in U\}$ of neighborboods of $ p$ is called a neighborhood basis of $ p$ if for each $U \in U_p$ there exist an $V\in \mathfrak{B}_p$ such that $V\subseteq U$
I have the following example in my lecture notes:
If X is a set such that $|X|>\aleph_0$ and $\tau$ is the discrete topology on $X$, now $(X,\tau)$ does not have a countable basis. In fact, let $\mathfrak{B}$ be a basis of $\tau$: Because $\{p\} \in \tau$ for each $p \in X$, the set $\{p\}$ must be union of elements of $\mathfrak{B}$. Therefore $\{p\}\in \mathfrak{B}$ and $|\mathfrak{B}|\geq|X|>\aleph_0$.
Furthermore, each neighborhood of $p \in X$ contains the open set $\{p\}$, so it is a finite neighborhood basis of $p$
I don't understand quite well the concept of neighborhood basis so to understand the last paragraph, I came up with a concrete example:
Let $X=\{1,2,3\}$ and $(X,\tau)$ a topological space with discrete topology.
Then $\tau$ equals the power set $\tau=\{ \{1\},\{2\},\{3\},\{1,2\},\{1,3\},\{2,3\},X,\emptyset \}$
If I choose $p=1$ the set of neighborhoods of $p$ are: $U_p= \{ \{1\},\{1,2\},\{1,3\},X \}$
so that $\mathfrak{B}_p \subseteq U_p$ , $\mathfrak{B}_p=\{\{1\}\}$ is a neighborhood basis because it is contained in each element of $U_p$ in agreement with the definition of neighborhood basis
and
$\mathfrak{B} \subseteq \tau$ ,$\mathfrak{B}=\{ \{1\},\{2\},\{3\} \}$ is a basis
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Is this example correct? and does it correctly explain why $\{p\}$ is a finite neighborhood basis of $p$ in the initial example?
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Assuming this example is correct,
just like when having a basis, each element of the topology can be expressed as union of elements of the basis, I was expecting that when having a neighborhood basis, each element of the set of neighborhoods of a point $p$ could be expressed as union of elements of the neighborhood basis, but it looks like it is not the case, since with the basis $\mathfrak{B}_p=\{\{1\}\}$ , I can't express the elements$ \{1,2\},\{1,3\}$and $X $ of $U_p$ as union of elements of $\mathfrak{B}_p$.
How do I make sense of this?
Best Answer
Yes, your example is correct.
The basis and the neighbourhood basis, despite the similar name have two different scopes.
The idea behind the basis of the topology is that you want something simpler from which it is possible to reconstruct the full topology.
The neighbourhood basis does not help you in finding all the possible neighbourhoods of a point. It's quite the opposite goal. It wants to select a subset of neighbourhoods which allows to study the property of what's happening around this point only. So it's a set of neighbourhoods which can be finer and finer, and the finer they become the more they forget about the rest of the topology, which is not of interest when studying some property of a single point.
In your discrete example, when you look at the property of the element $1$ you may want to forget about all the rest. That's when the neighbourhood basis is just $\{1\}$. In the discrete topology, all the other points are as if disconnected from $1$, so you don't want them around when studying properties of the point $1$ only.
If the topology is not discrete then things can be more interesting. If you want to study the topological properties of some object around a point $x_0\in\mathbb{R}$ you don't need to look at all the open set which contain $x_0$. Often it is enough to look at small intervals around it. Indeed, a neighbourhood basis around $x_0$ (in the Borel topology) can be a sequence of open shrinking interval which is collapsing on $x_0$.