To prove $\omega_1 \cup \{\omega_1\} $ is not first countable but $\omega_1$ is The topological space $X=\omega_1\cup \{\omega_1\}$ is not first countable but $\omega_1$ is. Here $\omega_1$ denotes the set of all countable ordinals and $\{\omega_1\}$ is the first uncountable ordinal. Cardinality of $\omega_1$ is $\aleph_1.$And also $\omega_1$ is called the limit ordinal meaning it is the limit point of the set $\omega_1.$
Here I want to prove two things. $1)$ $\omega_1$ is first countable. and $2)$ $\{\omega_1\}$ is the limit point of $\omega_1$. But since no sequence converges to $\{\omega_1\}$, using this I could say that our space $X$ is not first countable.
$1)$ The proof of this I got from here with just one question: it says we need to prove that such an $S(y)$ exists,and I'm thinking $S(y)=\cap\{u\in \omega_1:u\gt y\}.$ This is correct$?$ But here is a chance that the countable base around $y$ may be finite and not countably infinite.
$2)$ If possible let us assume that $\omega_1$ is not a limit point of the set $\omega_1.$ $U$ be a nbd of $\omega_1$ that does not intersect $\omega_1.$ Then what are the elements of $U$ like? The ones coming before $\omega_1$  cannot be uncountable because $\omega_1$  is the first. Then they must be countable. This is absurd so no such $U$ exists. Thus $\{\omega_1\}$ is a limit point.
Are my proofs correct? Thank you.
 A: A neighbourhood base for an ordinal $\alpha \in \omega_1$, either $\{\alpha\}$, when $\alpha$ is an isolated point, which happens when it is a successor ordinal or $0$ : if $\alpha=\beta+1$, then $\{\alpha\}=(\beta, \alpha+1)$,so open as an open interval.$0$ being the minimal element has $[0,1)=\{0\}$ as an open neighbourhood.Isolated points have a finite local base, which is no problem as first countable means every point has a local base that is at most countable, it need not be infinite per se.
If $\alpha \in \omega$ is a limit ordinal, it also has a local base that is countable, namely all sets of the form $(\beta, \alpha+1)$, where $\beta < \alpha$, of which there are but countably many as $\alpha$ is a countable ordinal.
If $U_n$ were a countable base at $\omega_1 \in \omega_1 +1$, then as it is the max of the set, every $U_n$ must contain (by the definition of order topology) a set of the form $(\alpha_n, \omega_1]$, where $\alpha_n$ is some countable ordinal. But then $\alpha =\cup_n \alpha_n$ is also a countable ordinal (a countable union of countable ordinals), which is the sup of the $\alpha_n$. But then $(\alpha+1, \omega_1]$ is an open neighbourhood of $\omega_1$ that contains no $U_n$.
So no countable family of open neighbourhoods of $\omega_1$ can form a local base.
A: (1) Your construction of $S(y)$ is correct (though of course you still need to actually prove that it works).  A finite set is countable, so there is no issue with having a finite local base instead of a countably infinite local base.  That is, a space is first-countable if every point has a countable local base, which includes the possibility that the local base is finite.
(2) Your argument shows that if $U$ is a neighborhood of $\omega_1$ that does not intersect $\omega_1$, then $U$ contains no ordinals less than $\omega_1$.  But you have not explained why this gives a contradiction.  Maybe the singleton set $\{\omega_1\}$ is open in $X$.
