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Denote by $\mathfrak{c}^+$ the cardinal successor of continuum. Can we prove in $\mathsf{ZFC}$ that

$(\mathfrak{c}^+)^{\aleph_0} = \mathfrak{c}^+$?

I guess not. Of course this question is trivial for $2^{\mathfrak{c}}$ yet I don't want to assume any sort of $\mathsf{GCH}$.

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Yes we can and it is an easy application of the Hausdorff formula.

Specifically, since $\mathfrak{c}^+$ is regular, any function from $\omega$ to $\mathfrak{c}^+$ is bounded. Any such function is then determined by an upper bound below $\mathfrak{c}^+$ and a function from $\omega$ into $\mathfrak{c}$, and this shows that $(\mathfrak{c}^+)^{\aleph_0}\leq \mathfrak{c}^+\cdot \mathfrak{c}^{\aleph_0}=\mathfrak{c}^+$. Equality then follows by Cantor-Schröder-Bernstein.

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  • $\begingroup$ Aww. I was gonna post that, and a student came in to ask something. :( $\endgroup$ – Asaf Karagila Feb 16 '15 at 13:26
  • $\begingroup$ @AsafKaragila Priorities, man. $\endgroup$ – Miha Habič Feb 16 '15 at 16:13

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