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This is a proof-verification request.

Let $\Omega$ be the set of countable ordinals, $\omega_1$ the first uncountable ordinal, and $\Omega^*=\Omega\cup\{\omega_1\}$. Remarkable properties of these sets are enumerated as follows:

  • $\Omega$ and $\Omega^*$ are uncountable;
  • $\Omega$ and $\Omega^*$ are well-ordered;
  • for any $x\in\Omega$, the initial segment $I_x\equiv\{\omega\in\Omega\,|\,\omega<x\}$ is countable and hence the final segment $J_x\equiv\{\omega\in\Omega^*\,|\,\omega>x\}$ is uncountable.

Endow $\Omega^*$ with the order topology.

Claim: For every open set $U$, either $U\cup\{\omega_1\}$ or $U^c\cup\{\omega_1\}$ contains an uncountable closed set.

I will use the following, easy-to-prove

Lemma: $E\subseteq\Omega$ is uncountable if and only if $\forall x\in\Omega$, $\exists y\in E$ such that $x<y$.

Proof of the claim: Let $U$ be open. Then $U^c$ is closed. If $U^c$ is uncountable, then desired result immediately follows. Now, suppose that $U^c$ is countable and let $V\equiv U^c\setminus\{\omega_1\}$. Then, $V\subseteq\Omega$ is countable. By the lemma, there exists some $x_0\in\Omega$ such that $$V\subseteq I_{x_0}\cup\{x_0\},$$ which implies that $J_{x_0}\subseteq V^c=U\cup\{\omega_1\}$. Pick any $y_0\in J_{x_0}\cap\Omega$ (note that $J_{x_0}$ cannot contain only $\omega_1$, otherwise its complement, which is $I_{x_0}\cup\{x_0\}$ would be uncountable, which is impossible). Then, $C\equiv J_{y_0}\cup\{y_0\}$ is uncountable, closed (because its complement is the open and countable initial segment $I_{y_0}$), and $C\subseteq U\cup\{\omega_1\}$. $\blacksquare$

Could you please let me know whether you think this proof is correct? Thank you for your time.

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This is a terrible proof. First of all, you're not doing topology here. This is order theory. Whether your set is open or closed is irrelevant. What you're saying is that a subset of $\Omega^\ast$ is cofinal if and only if it is uncountable, which is true.

However, in your proof, you do not show both directions. You have stated a biconditional (if and only if). Yet you seem to be proving only one direction. Also, it is not proper to cite a lemma in the proof of the lemma (when you state "by the Lemma..."). The proper verbology is "by hypothesis". Of course, you need to be clear which hypothesis you're proving when you prove a biconditional.

Also, nowhere in your proof do you even mention the set E. You cannot prove something about a set without ever talking about it. You've never defined U, and U seems to depend on E in some way that isn't apparent in the proof.

First, assume that E is uncountable and then show how that implies that it is cofinal in $\Omega^\ast$. Then show the converse. Assume that $E$ is cofinal and then show how that implies that it is uncountable.

And, as far as notation is concerned, in order theory, typically a limit ordinal is the union of all smaller ordinals. In other words, the way you've defined the sets, $\Omega$ and $\omega_1$ are actually equal. And $\Omega^\ast$ is really just $\omega_1+1$, the successor ordinal of $\omega_1$.

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  • $\begingroup$ You seem to think that I'm trying the prove the lemma. Actually, the lemma is just something I state without proof and the “proof of the claim” pertains to the claim stated two lines above the lemma. Topology creeps in insomuch as the claim pertains to sets that are open in the order topology, which is generated by sets of the form $\{I_x|x\in\Omega^*\}$ and $\{J_x|x\in\Omega^*\}$. I'm sorry if you find the structure of my question confusing, I wrote it down in a rush. $\endgroup$ – triple_sec Jul 5 '14 at 20:37
  • $\begingroup$ I see. Yes, I did misunderstand. It is a correct proof. The notation is awkward, but it is logically sound. $\endgroup$ – Keith Penrod Jul 10 '14 at 13:45
  • $\begingroup$ That's all right. Thank you for your time and double-checking. $\endgroup$ – triple_sec Jul 10 '14 at 17:11

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