I imagine this is true as it is easy to prove that any principal ultrafilter on $\mathbb{N}$ is uncountable, and nonprincipal ultrafilters seem in a way bigger. (My proof is $\mathcal{P}(\mathbb{N})= \bigcup_{i\in\omega}\mathcal{U}_{i}$ where each $\mathcal{U}_{i}$ is the principal ultrafilter generated by $i$. It is a countable union, so each $\mathcal{U}_{i}$ can't be countable as $\mathcal{P}(\mathbb{N})$ is not countable. Here im also assuming all principal ultrafilters over $\omega$ have the same size, but this seems reasonable.)

Anyway, to clarify and summarize: i'm looking for a proof (or a proof of the negation) that any nonprincipal ultrafilter $\mathcal{U}$ on $\omega$ contains uncountably many infinite elements.

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    $\begingroup$ If $\mathscr{U}$ is an ultrafilter on $\omega$, then for any $A\subseteq\omega$ exactly one of $A$ and $\omega\setminus A$ belongs to $\mathscr{U}$. $\endgroup$ – Brian M. Scott Apr 11 '15 at 21:33
  • $\begingroup$ You are asking (I think) if there are uncountably many infinite elements in $\mathscr{U}$. $\endgroup$ – hardmath Apr 11 '15 at 21:42

Let $\mathcal F$ be an ultrafilter on an infinite set $S$. Then for any subset $X\subseteq S$ precisely one of $X\in \mathcal F$ or $X^c \in \mathcal F$ holds. So, half of the sets in $\mathcal P (S)$ are in $\mathcal F$. Clearly the cardinality of $\mathcal F$ is thus the same as the cardinality of $\mathcal P (S)$, which is uncountable since $S$ is infinite.


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