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Reading this wikipedia article, I arrived at the fact that $\omega_1$ can exist without choice. Since the proof I know of the fact that $[0,\omega_1)$ is sequentially compact depends on the fact that the countable union of countable sets is countable, I arrived at the following question:

What topological properties of $[0,\omega_1)$ depend on the axiom of choice?

Aloizio Macedo
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  • Is $S_\omega=[0,\omega_1)$? I have never seen that notation (it may be common, idk). –  Aug 06 '15 at 19:04
  • Yes, it is. I will change the title. – Aloizio Macedo Aug 06 '15 at 19:05
  • A countable union of countable sets is countable whether you take the axiom of choice or not, no? – Alex G. Aug 06 '15 at 19:16
  • As far as I know, you need at least countable choice to show that a countable union of countable sets is countable. – Aloizio Macedo Aug 06 '15 at 19:18
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    @AlexG That is wrong. For expample you can look here. Note that the reals are always uncountable, due to Cantors theorem (no choice involved). Another place to look is here –  Aug 06 '15 at 19:20
  • Which properties did you have in mind? – Asaf Karagila Aug 06 '15 at 19:20
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    @Alex: Not even remotely. There might not even be an upper bound, in terms of cardinals, to what sort of size a countable union of countable set is (meaning that it is possible that for every set $A$, there is a set $B$ such that $B$ is the countable union of countable sets, and there is no injection from $A$ into $B$). – Asaf Karagila Aug 06 '15 at 19:22
  • @PaulPlummer and Asaf, thank you for the information and references. Math surprises me yet again! – Alex G. Aug 06 '15 at 19:26
  • @Asaf My motivation-property was sequential compactness, because I know that the argument I know won't work, but maybe there is another argument that does? But my question is intended to be as general as can be. However, it is to be noted that my knowledge of set theory is minimal, although my curiosity is big. – Aloizio Macedo Aug 06 '15 at 19:28
  • Maybe, to make the question more specific, you can ask as a reference request of results of these type, plus a results on sequential compactness without choice. Although maybe with Asaf's answer it might be to\ late –  Aug 06 '15 at 19:36

1 Answers1

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First of all, yes. $\omega_1$ exists without assuming the axiom of choice. However it can be the countable union of countable sets, which can cause some troubles. Let me just point out that $\omega_1$ is, by definition, the set $[0,\omega_1)$. So I'll just use the shorter notation to make it easier.

So if $\omega_1$ is the countable union of countable sets, then it is the limit of an $\omega$-sequence of countable ordinals. Therefore it will no longer be the case that every sequence of countable ordinals has a countable limit. Therefore it will not be sequentially compact.

To see why this is true, if $\omega_1=\bigcup A_n$, where each $A_n$ is countable, there is a unique ordinal isomorphic [uniquely] to $A_n$. Now let $\alpha_n$ be that ordinal, and let $\beta_n=\alpha_1+\ldots+\alpha_k$. Then $\beta_n$ are increasing, and $\sup\beta_n=\lim\beta_n=\omega_1$.

For metrizability, the answer is that the axiom of choice is not needed to prove that $\omega_1$ is not metrizable. This is a combination of Corollaries 4.2 and 4.4 in the following paper:

C. Good and I. J. Tree, Continuing horrors of topology without choice, Topology Appl. 63 (1995), no. 1, 79--90.

user642796
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Asaf Karagila
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    Great! Just a last inquiry... What about metrizability? Because with choice, an argument for proving $[0,\omega_1)$ is not metrizable is arriving at the fact that it is sequentially compact but not compact. – Aloizio Macedo Aug 06 '15 at 19:39
  • @Aloizio: Asaf will correct me if I’ve overlooked something, but it appears that if $\omega_1$ is the limit of an $\omega$-sequence of countably ordinals, then the space $\omega_1$ is the union of countably many clopen, countable subspaces. Each of these is metrizable, so $\omega_1$ ought to be metrizable as well. – Brian M. Scott Aug 07 '15 at 06:39
  • @Brian: Why can you uniformly choose a metric on each ordinal? – Asaf Karagila Aug 07 '15 at 07:02
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    @Aloizio: That's a good question, I'll think about it today and let you know. – Asaf Karagila Aug 07 '15 at 07:02
  • @Aloizio: Yes, that looks like a genuine problem. By the way, another argument (with choice) for showing that $\omega_1$ isn’t metrizable is to use the pressing-down lemma to show that $\omega_1\times\omega_1$ isn’t hereditarily normal. Unfortunately, the pressing-down lemma requires at least some choice; Asaf may know how much. – Brian M. Scott Aug 07 '15 at 14:02