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Let $U$ be a free ultrafilter on a set $X$.

I want to prove that the cardinality of every element $u\in U$ is equipotent to $X$. Is that true? Or does it lack some hypothesis?

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It lacks some hypothesis. It's true if $X$ is countable. Otherwise, take a countable subset $C \subset X$, and a free ultrafilter $\mathscr{U}$ on $C$. Then $\mathscr{U}$ generates a free ultrafilter on $X$ that contains countable elements. – Daniel Fischer Oct 13 '13 at 20:46
By the way, an ultrafilter with the property you describe is called a uniform ultrafilter. – Trevor Wilson Oct 13 '13 at 20:51
@DanielFischer: That sounds like it settles the question. Why not make it an answer? – Henning Makholm Oct 13 '13 at 20:53
@HenningMakholm Now: because Asaf already answered it. Before that: because maybe the OP would say "Oh, right, there's that additional hypothesis I forgot to mention". – Daniel Fischer Oct 13 '13 at 20:55
up vote 3 down vote accepted

No, this is not provable because it can be false.

Consider an ultrafilter on $\Bbb N$, say $\cal F$. You can show that $\{A\subseteq\Bbb R\mid A\cap\Bbb N\in\cal F\}$ is a free ultrafilter on the real numbers.

The property that you are looking for is called uniform ultrafilter.

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It should perhaps be noted explicitly that uniform ultrafilters always exist if $X$ is infinite. Namely, $\{A\subseteq X \mid~ |X\setminus A|<|X|\}$ is a filter and must extend to an ultrafilter by the ultrafilter lemma. – Henning Makholm Oct 13 '13 at 21:29

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