How to show that $\omega_1$ is not secound countable?

Question

I wish to produce a direct proof that $\omega_1$ is not second countable

Recall the definition of $\omega_1$

$\omega_1 = \{\alpha \in W| \{x \in W| x < \alpha\} \text{ is countable }, W \text{ is an uncountable well order }\}$

More commonly:

$\omega_1 = \{\alpha| \text{pred}(\alpha) \text{is countable}\}$, where $\text{pred}(\alpha)$ is the predecessor set

(Note: I only know these two definitions, which are quite clumsy, if anyone knows simpler definitions of $\omega_1$ I would also appreciate it)

Then,

proposition: $\omega_1$ with its order topology induced by $\leq$ is not second countable

How do I prove this?

Here's my attempt:

Suppose that $\omega_1$ is second countable, then it has a countable basis $\mathcal{B}$.

Take the supremum of all the elements in the basis $\mathcal{B}$, call it $b = \sup\mathcal{B}$.

Then $\text{pred}(b)$ is a countable set (by def) and $\text{pred}(b) \subset \omega_1$, so it is bounded above by some $\alpha \in \omega_1$ (fact).

Then clearly the immediate successor to $\alpha$, $\alpha+1$, is not covered by a basis element, contradiction.

Can someone check whether this is valid?

Answer 1

The main problem with this proof is that you are assuming that all of the sets in $\mathcal B$ are countable. Of course, there are uncountable open subsets of $\omega_1$ ($\omega_1$ itself, for example). So you have to somehow disallow this possibility.

Luckily, it can be done, thanks to the following fact:

If a topological space $X$ is second-countable, then every base for $X$ includes a countable subset which is also a base for $X$.

(An outline of a proof of the more general statement can be found in a previous answer on mine.)

So all you have to do to fix up your proof is to start with a base for $\omega_1$ consisting entirely of countable sets. For example, $$\mathcal D = \{ ( \alpha , \beta ) : \alpha < \beta < \omega_1 \}$$ where by $(\alpha , \beta )$ it is meant the "usual" open interval $\{ \xi : \alpha < \xi < \beta \}$. Now by the quoted fact if $\omega_1$ were second-countable, there is a countable $\mathcal B \subseteq \mathcal D$ which is also a base for $\omega_1$. Then rest of your proof then follows.

Answer 2

This is not valid, unfortunately, since maybe some element of $\mathcal{B}$ is unbounded (e.g. maybe $\omega_1$ itself is in $\mathcal{B}$).

Instead:

• Show that if $X$ is a space which has an uncountable family of pairwise-disjoint open sets, then $X$ is not second countable. (That is: if $X$ is not c.c.c., then $X$ is not second countable, or equivalently if $X$ is second countable then $X$ is c.c.c.)

• What can you say about $\{\{\alpha+1\}:\alpha<\omega_1\}$?

Answer 3

Assume for a contradiction that $\mathcal B$ is a countable base for the order topology of $\omega_1.$ That the set $M=\{\min B:B\in\mathcal B\}$ is countable. Consider an ordinal $\alpha\in\omega_1\setminus M.$ Since $[\alpha,\omega_1)$ is a neighborhood of the point $\alpha,$ there is a set $B\in\mathcal B$ such that $\alpha\in B\subseteq[\alpha,\omega_1).$ But then $\min B=\alpha,$ contradicting the fact that $\alpha\notin M.$