$X$ have a countable basis at $x$, if $\exists \{U_n \in \mathcal{N}_x |n\in \Bbb{N}\}$ with the following property: $\forall U\in \mathcal{N}_x, \exists m\in \Bbb{N}$ such that $U_m\subseteq U$. If $X$ have countable basis at $\forall x\in X$, then $X$ is called first countable.
Munkres defined countable set as, finite or countably infinite. But the way we write set $\{U_n \in \mathcal{N}_x |n\in \Bbb{N}\}$, seems like we are only interested in countably infinite set. And what if $U=\{ U_n| n\in \Bbb{N}\}$ is finite. Then $U$ is multiset, somewhat different concept than set.
Munkres defined separable space, as $\exists D\subseteq X$ such that $D$ is countable and $\overline{D}=X$. Exact same definition of separable space is given $D$ as countably infinite in Baby Rudin book, chapter 2. Which definition should I use?
If space $X$ has a countable basis for its topology, then $X$ is said to satisfy the second countability axiom, or to be second countable.
$X$ is equipped with $\mathcal{T}_X$ topology. Second countability axiom says that, $\exists$ a countable basis $\mathcal{B}$ of $\mathcal{T}_X$, i.e. $\exists \mathcal{B} =\{B_n \subseteq X|n\in \Bbb{N}\}$ such that $\mathcal{B}$ is basis of given $\mathcal{T}_X$. Am I right? It is a general fact that, there need not be unique basis for a given $\mathcal{T}_X$ topology. So we can also have uncountably infinite basis $\mathcal{B}’$ of $\mathcal{T}_X$, don’t we?
Edit: Rephrasing definition of 2-countable to my taste
$(X,\mathcal{T}_X)$ is second countable if $\exists \mathcal{B}=\{B_n\subseteq X| n\in \Bbb{N}\}$ such that $\mathcal{B}$ is basis of $\mathcal{T}_X$.
Best Answer
The set $\{U_n:n\in\Bbb N\}$ can actually be finite as well, say, if $U_n=U_N$ for all $n>N$ for some fixed $N\in\Bbb N$.
I guess the same applies on Baby Rudin's definition of separatedness.
Yes, you're right in both of the questions of the last paragraph.
For example the real line $\Bbb R$ is a second countable space, a countable basis is the set of open intervals with rational endpoints.
It indeed admits, however, a bigger basis, containing all open intervals.
(And actually, $\mathcal T_X$ is also always a basis for itself.)