1
$\begingroup$

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.

$\endgroup$
  • 5
    $\begingroup$ 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
3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.