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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?

Martin Sleziak
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Olórin
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  • In general, $\operatorname{pred}(\alpha)$ doesn't exist - take for example $\alpha = \omega$. A simple definition of $\omega_{1}$ is $\omega_{1} = {\alpha \in \operatorname{Ord} \mid \alpha \text{ is countable} }$. Also $b$ is not countable by definition, but by the regularity of $\omega_{1}$. – Stefan Mesken Aug 01 '16 at 22:40
  • @Stefan I'm guessing the $\text{Ord}$ part is where it gets intense – Olórin Aug 01 '16 at 22:43
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    @JohnSwoon No, that definition is perfectly tame (although it takes a bit of work to show that such a set exists, and in particular is not a proper class). Specifically, using Powerset and Separation you can get the set of all well-orderings of $\mathbb{N}$ - and then, via Replacement, construct a minimal ordinal longer than all of these. That ordinal is $\omega_1$, and you can easily show that it is exactly the set of all countable ordinals. – Noah Schweber Aug 01 '16 at 22:51
  • @NoahSchweber Hmm the problem I have is that $\alpha$ in my definition is an element of an uncountable well order (don't actually have an example), whereas in this new definition $\alpha$ is like a set because countability is a property of a set. Anyways it is not important at the moment I will come back to these definitions – Olórin Aug 01 '16 at 23:16
  • @JohnSwoon Ordinals are sets - specifically, the class $ON$ of ordinals is a class with the following properties: (a) Each $\alpha\in ON$ is a set of ordinals, and is well-ordered by "$\in$". (b) Every well-ordering is in order-preserving bijection with exactly one $\alpha\in ON$ (with the order on $\alpha$ taken to be "$\in$"). See https://en.wikipedia.org/wiki/Ordinal_number. – Noah Schweber Aug 01 '16 at 23:57

3 Answers3

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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.

Community
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Taumatawhakatangihangakoauauot
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  • +1, I like this proof very much! The fact it relies on - that in a second-countable space, any base can be "thinned" to a countable base - is much harder to prove than the OP's actual question, but it's a much deeper fact, and worth the OP's time. – Noah Schweber Aug 01 '16 at 23:01
  • Hi I have checked your proof and everything works out. So $\mathcal{B}$ does not exist, so the original space is not second countable. The only problem I have is that the set is ${(\alpha, \beta): \alpha< \beta < \omega_1}$. Does it make sense to use $\omega_1$ as an element, when we are using $\omega_1$ as a topological space equipped with the order topology? Would it be also correct to write the set as: ${(\alpha, \beta): \alpha< \beta < \infty}$ – Olórin Aug 01 '16 at 23:46
  • @JohnSwoon I'm not sure I understand your last point. $\omega_1$ is not an element of any set, but I'm using $\omega_1$ to describe the sorts of "open intervals" I take in $\mathcal D$. Writing $"\alpha < \beta < \omega_1$ is a little more succinct than saying "$\alpha < \beta$ are countable ordinals", and saying "$\alpha < \beta < \infty$" can be misleading since there are ordinals much greater than $\omega_1$. – Taumatawhakatangihangakoauauot Aug 02 '16 at 04:24
  • @NoahSchweber Thanks for your comment. I like this fact since it allows one to "directly" prove that certain spaces are not second-countable, instead of indirectly showing it by demonstrating some property that second-countable spaces cannot have. (Of course, it's important to be able to do the "indirect" kind of proofs, too.) – Taumatawhakatangihangakoauauot Aug 02 '16 at 04:26
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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 $\{\{\alpha+1\}:\alpha<\omega_1\}$?

Noah Schweber
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  • Do you mean to use [second countable space $\implies$ ccc]? So suppose ${{\alpha+1}: \alpha < \omega_1}$ is an uncountable pairwise-disjoint open set, how does it imply the negation of the reverse? Since the negation $\neg$ [second countable space $\implies$ ccc] = [Not second countable space $\land$ Not ccc]. All in all, does not ccc imply not second countable? – Olórin Aug 01 '16 at 22:52
  • @JohnSwoon I've edited for clarity. The answer to the last sentence in your comment is yes: if a space is not c.c.c., then it is not second countable (although of course this requires a proof). – Noah Schweber Aug 01 '16 at 22:53
  • ${a+1}$ is open for every $a\in \omega_1.$ So any base $B $ must satisfy $ B\supset {{a+1}: a\in \omega_1} $ so $ B $ is uncountable. – DanielWainfleet Aug 03 '16 at 10:03
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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 $\alpha\notin M.$

Daniel
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bof
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