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Justin Moore
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It is consistent that $2^{\aleph_0} > \aleph_\omega$. This is immediate from Cohen's work; force with all finite partial functions from $\aleph_{\omega}$ to $2$. More generally Easton has shown that the only constraints on exponentiation of regular cardinalcardinals is that it be monotonic and satisfy $\mathrm{cof}(2^\kappa) > \kappa$. For singular cardinals, the matter is much more subtle. For instance one of Shelah's celebrated results is that if $\aleph_\omega$ is a strong limit cardinal, then $2^{\aleph_\omega} < \aleph_{\omega_4}$ (and no, that $4$ is not a misprint).

It is consistent that $2^{\aleph_0} > \aleph_\omega$. This is immediate from Cohen's work; force with all finite partial functions from $\aleph_{\omega}$ to $2$. More generally Easton has shown that the only constraints on exponentiation of regular cardinal is that it be monotonic and satisfy $\mathrm{cof}(2^\kappa) > \kappa$. For singular cardinals, the matter is much more subtle. For instance one of Shelah's celebrated results is that if $\aleph_\omega$ is a strong limit cardinal, then $2^{\aleph_\omega} < \aleph_{\omega_4}$ (and no, that $4$ is not a misprint).

It is consistent that $2^{\aleph_0} > \aleph_\omega$. This is immediate from Cohen's work; force with all finite partial functions from $\aleph_{\omega}$ to $2$. More generally Easton has shown that the only constraints on exponentiation of regular cardinals is that it be monotonic and satisfy $\mathrm{cof}(2^\kappa) > \kappa$. For singular cardinals, the matter is much more subtle. For instance one of Shelah's celebrated results is that if $\aleph_\omega$ is a strong limit cardinal, then $2^{\aleph_\omega} < \aleph_{\omega_4}$ (and no, that $4$ is not a misprint).

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Justin Moore
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It is consistent that $2^{\aleph_0} > \aleph_\omega$. This is immediate from Cohen's work; force with all finite partial functions from $\aleph_{\omega}$ to $2$. More generally Easton has shown that the only constraints on exponentiation of regular cardinal is that it be monotonic and satisfy $\mathrm{cof}(2^\kappa) > \kappa$. For singular cardinals, the matter is much more subtle. For instance one of Shelah's celebrated results is that if $\aleph_\omega$ is a strong limit cardinal, then $2^{\aleph_\omega} < \aleph_{\omega_4}$ (and no, that $4$ is not a misprint).

It is consistent that $2^{\aleph_0} > \aleph_\omega$. This is immediate from Cohen's work; force with all finite partial functions from $\aleph_{\omega}$ to $2$.

It is consistent that $2^{\aleph_0} > \aleph_\omega$. This is immediate from Cohen's work; force with all finite partial functions from $\aleph_{\omega}$ to $2$. More generally Easton has shown that the only constraints on exponentiation of regular cardinal is that it be monotonic and satisfy $\mathrm{cof}(2^\kappa) > \kappa$. For singular cardinals, the matter is much more subtle. For instance one of Shelah's celebrated results is that if $\aleph_\omega$ is a strong limit cardinal, then $2^{\aleph_\omega} < \aleph_{\omega_4}$ (and no, that $4$ is not a misprint).

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Justin Moore
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YesIt is consistent that $2^{\aleph_0} > \aleph_\omega$. This is immediate from Cohen's workwork; force with all finite partial functions from $\aleph_{\omega}$ to $2$.

Yes. This is immediate from Cohen's work.

It is consistent that $2^{\aleph_0} > \aleph_\omega$. This is immediate from Cohen's work; force with all finite partial functions from $\aleph_{\omega}$ to $2$.

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Justin Moore
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