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This is a follow-up question on an earlier question about the Stone–Čech compactification of limit ordinals (Compactifications of limit ordinals):

For a limit ordinal $\alpha$, what is the cardinality of its Stone–Čech compactification remainder $\alpha^*$?

Here are some of my thoughts:

  • We have $|\alpha^*| \leq 2^{2^{|\alpha|}}$ always.

  • If $\alpha$ has uncountable cofinality, then it's well known that $\beta \alpha = \alpha +1$, i.e. we have a 1-point remainder.

  • If $\alpha$ has countable cofinality, then considering the closure of a cofinal sequence in $\beta \alpha$, we find a copy of $\omega^*$ inside of $\alpha^*$, and hence $|\alpha^*| \geq |\omega^*|= 2^{2^{\aleph_0}}$. In particular, if $\alpha$ is a countable limit ordinal, the case is clear.

But I am stuck with uncountable limit ordinals of countable cofinality. For example, what is $|\aleph_{\omega}^*|$? By writing $$ \aleph_{\omega}=\aleph_0 + (\aleph_0 + \aleph_1) + (\aleph_0 + \aleph_1 + \aleph_2) + \cdots $$ one can find $\aleph_\omega$ many disjoint cofinal sequences, and hence $|\aleph_{\omega}^*| \geq \aleph_\omega \cdot 2^{2^{\aleph_0}}$. Is there a precise answer?

Styles
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Max
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  • By Iterated Forcing, ZFC is equiconsistent with ZFC+MA+$(2^{\aleph_0}>\aleph_{\omega}$,,,, MA is Martin's Axiom. ... MA implies that $2^k=2^{\aleph_0}$ for every infinite cardinal $k<2^{\aleph_0}. $ So it is consistent with ZFC that $2^{2^{\aleph_{\omega}}}=$ $\aleph_{\omega}\cdot 2^{2^{\aleph_0}}=$ $2^{2^{\aleph_0}}=$ $=|\aleph_0^*|.$ – DanielWainfleet Oct 28 '17 at 13:40
  • Of course it is also consistent that $2^{2^{\aleph_0}}=$ $2^{\aleph_1}=$ $\aleph_2<$ $\aleph_{\omega}$ ... (by GCH). ... & I dk whether $|\aleph_{\omega}^|$ is necessarily equal to $2^{2^{\aleph_{\omega}}}$.....In my previous comment the LAST expression should be $|\aleph_{\omega}^|,$ and NOT $|\aleph_0^*|$. – DanielWainfleet Oct 28 '17 at 13:48

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