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 $V\subseteq U$
Additionaly, I have the following proposition in my lectures notes
Let $(X,\tau)$ be a topological space
(1) If $\mathfrak{B}$ is a basis of $\tau$ and if $p\in X$, the set $\mathfrak{B}_p=\{B \in \mathfrak{B}| p \in B\}$ is a neighborhood basis at $p$.
(2) If for each $p\in X$ a neighborhood basis at $p $ : $\mathfrak{B}_p$ is defined, the set $\mathfrak{B}=\{B \in \mathfrak{B}_p | p \in X\}$ is a basis of $\tau$
and the definition of countable basis and countable neighborhood basis
Let $(X,\tau)$ be a topological space. If $\tau$ admits a basis whose cardinality is countable, $\tau$ is said to have a countable basis
If each $p \in X$ admits a countably-infinite neighborhood basis, $\tau$ is said to have a countably-infinite neighborhood basis
In those cases $(X, \tau)$ is called a countable-basis topological space or countable-neighborhood-basis topological space respectively.
*The term countable includes finite sets: a set is countable if it is finite or countably-infinite.
Then they say that by the second statement (2) of the initial proposition, countable basis implies countable neighborhood basis(that is if $\mathfrak{B}$is countable, then $\mathfrak{B}_p$ is countable), but the converse is not true. ….(3)
A couple of related questions
- Are neighborhood bases always a subset of a basis as stated in (1) : $\mathfrak{B}_p=\{B \in \mathfrak{B}| p \in B\}$ ?, or this is just one possibility?
- I attempt to give an explanation of (3):
Since by (1)(they used (2), but I do not agree, I Think I need (1)), $\mathfrak{B}_p \subseteq \mathfrak{B}$, $\mathfrak{B}$ is countable by hypothesis and a subset of a countable set is countable, it follows that $\mathfrak{B}_p $ is countable.
The converse is not true, because we could have an uncountable number of points in X, for which the union of countable neighbordhood bases might not be countable
Is this explanation correct?, as you see it depends heavily on the definition of $\mathfrak{B}_p$ as a subset of $\mathfrak{B}$, so it wouldn't be general if a neighborhood basis could be defined without taking the elements from a basis.
Best Answer
If $\mathscr{B}$ is a base for a topology $\tau$ on a set $X$, and $p\in X$, the set $\mathscr{B}_p=\{B\in\mathscr{B}:p\in B\}$ is always a nbhd base at $p$. However, it is entirely possible to have a base $\mathscr{B}$ for $\tau$ and a nbhd base $\mathscr{B}_p$ at $p$ such that $\mathscr{B}_p\cap\mathscr{B}=\varnothing$. For instance, let $X=\Bbb R$, and let $\tau$ be the usual topology on $\Bbb R$. Let
$$\mathscr{B}=\{(p,q):p,q\in\Bbb Q\text{ and }p<q\}\;,$$
the set of open intervals with rational endpoints; $\mathscr{B}$ is a base for $\tau$. Let
$$\mathscr{B}_0=\{(-x,x):0<x\in\Bbb R\setminus\Bbb Q\}\;,$$
the set of symmetric open intervals around $0$ with irrational endpoints. Clearly $\mathscr{B}_0\cap\mathscr{B}=\varnothing$, but $\mathscr{B}_0$ is nevertheless a nbhd base at $0$.
The best way to explain (3) is to give an actual example showing that it is possible to have a countable nbhd base at each point but no countable base for the whole topology. Probably the simplest example is to let $X$ be an uncountable set and $\tau=\wp(X)$ the discrete topology on $X$. For each $x\in X$ the collection $\big\{\{x\}\big\}$ is a nbhd base at $x$, and it’s certainly countable! However, any base for $\tau$ must include every one of the singleton sets $\{x\}$ for $x\in X$ and must therefore be uncountable.