Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
5  
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
1  
By the way, an ultrafilter with the property you describe is called a uniform ultrafilter. –  Trevor Wilson Oct 13 '13 at 20:51
1  
@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

1 Answer 1

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.

share|improve this answer
1  
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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