Ultrafilters come in the principal and the free variant. Elsewhere it is said that principal is equivalent to the ultrafilter having a least element, i.e. one that is contained in every other element of the ultrafilter. Obviously this then should not be true for free ultrafilters.

Here is my "proof" that all ultrafilters have a least element:

Let $\mathcal F$ be an ultrafilter and $A\in\mathcal F$.

  1. If $A\cap B = A$ for all $B\in \mathcal F$, then $A$ is minimal.
  2. Otherwise $\exists B\in\mathcal F$ such that $C=A\cap B \subsetneq A$. Since $\mathcal F$ is a filter, $C\in \mathcal F$. Now re-apply (inductively so to say) (1) to $C$.

Ultimately the above process will single out the least element.

Since free ultrafilters do not have a least element, something is wrong with this proof. What is it? Does it have to do with the kind of induction which may involve an uncountable number of steps?


All you proved is that there is a sequence $A_n\in\cal F$ such that $A_{n+1}\subseteq A_n$. You did not prove that this sequence has an intersection inside $\cal F$, which may or may not be the case. And even then, you did not prove that the intersection is minimal, etc. So this is certainly not a transfinite recursion, since those require you to deal with the limit steps as well.

But even if you did deal with the limit step somehow, all you can conclude is that there is a limit ordinal $\delta$, and a decreasing sequence of sets $A_i\in\cal F$ for $i<\delta$ such that $\bigcap A_i\notin\cal F$.


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