Prove: $2^{\aleph_0}\not=\aleph_{\epsilon_0}$ I would like a hint for this problem
2026-04-03 11:51:46.1775217106
Prove: $2^{\aleph_0}\not=\aleph_{\epsilon_0}$
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HINT: If $\delta$ is a limit ordinal, then the cofinality of $\aleph_\delta$ is the same cofinality of $\delta$. Note that $\varepsilon_0$ is countable.