# $\kappa$-complete ultrafilter and bounded subsets of $\kappa$

If $U$ is a $\kappa$-complete nonprincipal ultrafilter on $\kappa$ then every bounded subset of $\kappa$ has measure $0$. This is because if we had $X \in U$ with $X$ having size less than $\kappa$, then since $X$ is a union of less than $\kappa$ many singletons and since $U$ is $\kappa$-complete then one of the singletons has to be in $U$ which contradicts $U$ being non principal.

There must be a better way to explain this simple fact? Thanks.

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Seems just fine! –  André Nicolas Sep 14 '12 at 21:56
Ok. For some reason I was thinking that there was a simpler way to explain it. I've used an alternative characterization of $\kappa$-complete non principal ultrafilters instead of the characterization in terms of closure under less than $\kappa$ many intersections. I guess that's Ok. –  user33714 Sep 14 '12 at 22:00
Seems OK to me. The "measure-theoretic" view is I think the natural one. –  André Nicolas Sep 14 '12 at 22:04

It’s already pretty straightforward, but if you want to work even more directly from the usual definition of $\kappa$-completeness, note that since $U$ is non-principal, it contains $\kappa\setminus\{\alpha\}$ for each $\alpha\in\kappa$. In particular, if $B$ is a bounded subset of $\kappa$, then $\kappa\setminus\{\beta\}\in U$ for each $\beta\in B$, and $|B|<\kappa$, so

$$\kappa\setminus B=\bigcap_{\beta\in B}\Big(\kappa\setminus\{\beta\}\Big)\in U\;,$$

and therefore $B\notin U$.

Added: It’s not difficult to show that the usual definition is equivalent to the one in terms of partitions of the underlying set (as you probably already know), and I don’t see much to choose between them in getting this result.

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Ah thank you. That's what I was looking for, just directly from the definition of $\kappa$-completeness. That way of seeing it seems very natural. –  user33714 Sep 14 '12 at 22:24

Ideals and filters are dual. This means that any explanation that works for one should work for the other.

To say that the union of less than $\kappa$ many sets of measure zero (read: in the dual ideal) is of measure zero, is the same as saying that the intersection of less than $\kappa$ many sets of measure one is of measure one.

This is true simply by taking complements.

Note that the fact that the filter is non-principal and $\kappa$-closed means that all the singletons are in the ideal, that is have measure zero.

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