Say, we have $n$ non-principal ultrafilters $\mathcal{U}_1,...,\mathcal{U}_n$ on an infinite set $X$. Obviously they all contain all cofinite subsets of $X$. But can they all contain some common element which is not cofinite, i.e. is there anything non-cofinite in their intersection?

This seems obvious there should be such an element (I think), but the proof... I don't know :(


Write $X$ as the disjoint union of $n+1$ infinite subsets $X=X_1\sqcup\ldots\sqcup X_{n+1}$. For each $i=1,\ldots,n$, we can choose some $j_i\in\{1,\ldots,n+1\}$ such that $X_{j_i}\in\mathcal{U}_i$. Then $X_{j_1}\cup\ldots\cup X_{j_n}$ is a non-cofinite set that is in each ultrafilter.

  • $\begingroup$ But what if only $X_i\in\mathcal{U}_i$ for $i<n$ and $X_1\cup X_n\in\mathcal{U}_n$? Then neither $X_1$ nor $X_n$ are in, say, $\mathcal{U}_2$. $\endgroup$ – Jules May 22 '16 at 13:43
  • $\begingroup$ I'm not sure I understand. In the situation you describe, $X_1\cup\ldots\cup X_1$ is in every $\mathcal{U}_i$. $\endgroup$ – Julian Rosen May 22 '16 at 14:56
  • $\begingroup$ Oops, I meant $X_1\cup\ldots\cup X_n$ is in every $\mathcal{U}_i$. $\endgroup$ – Julian Rosen May 22 '16 at 15:10
  • $\begingroup$ Not sure if I understand then. I mean $X_1\in\mathcal{U}_1$, but $X_i\notin\mathcal{U}_1 \forall i>1$. And so on $\endgroup$ – Jules May 22 '16 at 15:18
  • 1
    $\begingroup$ In the scenario that $X_i\in\mathcal{U}_i$ for $i=1,2,\ldots,n$, define $Y=X_1\cup\ldots\cup X_n$. Then for every $i$, $Y$ contains $X_i$, so necessarily $Y\in\mathcal{U}_i$. This means $Y$ is in the intersection $\mathcal{U}_1\cap\ldots\cap\mathcal{U}_n$. Also, $Y$ is not cofinite because it is disjoint from the infinite set $X_{n+1}$. $\endgroup$ – Julian Rosen May 22 '16 at 20:53

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