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 $A\in\mathcal F$.
- If $A\cap B = A$ for all $B\in \mathcal F$, then $A$ is minimal.
- 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?