Proving that functions send ultrafilter basis to ultrafilter basis

Question

I'm currently revisiting a proof of Tychonoff's theorem via ultrafilters. The definitions we were working with are as follows,

• $\mathcal{B}$ is a basis for a filter $\mathcal{F}$ on a set $X$ if $\mathcal{F} = \{A : B \subset A, \text{for some $B \in \mathcal{B}$}\}$.
• an ultrafilter on a set $X$ is a filter $\mathcal{F}$ that is a maximal element of the set of filters on $X$, ordered by inclusion.

As an intermediate step, a lemma was left for the reader to prove:

Lemma. Let $f : X \to Y$ be a function. Then,

• if $\mathcal{B}$ is a basis for a filter on $X$, then $f(\mathcal{B})$ is a basis for a filter on $Y$

• if $\mathcal{B}$ is a basis for a filter on $Y$, then $f^{-1}(\mathcal{B})$ is a basis for a filter on $X$, provided that $f^{-1}(B) \neq \emptyset$ for all $B \in \mathcal{B}$.

• if $\mathcal{B}$ is a basis for an ultrafilter on $X$, then $f(\mathcal{B})$ is a basis for an ultrafilter on $Y$

I have managed to prove the first two stamentents, but I am struggling with the last one. Certainly $f(\mathcal{B})$ is a basis for a filter on $Y$, but why is this an ultrafilter?

I've attempted to assume the contrary and take a finer filter than the one generated by $f(\mathcal{B})$, in order to take preimages and contradict that $\mathcal{B}$ generates an ultrafilter, but I haven't been able to complete the proof.

Answer 1

For most purposes, the most useful characterization of ultrafilters is not as maximal filters but as filters with the property that for any $A\subseteq X$, either $A$ or $X\setminus A$ is in the filter. (Prove this is equivalent to maximality, if you haven't seen this! The key point is that if neither $A$ nor $X\setminus A$ is in the filter, you can add $A$ to get a larger filter.)

So, now suppose $\mathcal{B}$ is a basis for an ultrafilter on $X$. This means that for any $A\subseteq X$, there exists $B\in \mathcal{B}$ such that either $B\subseteq A$ or $B\subseteq X\setminus A$. So now, to prove that $f(\mathcal{B})$ is a basis for an ultrafilter, let $A\subseteq Y$ be arbitrary. We want to show there is $B\in\mathcal{B}$ such that either $f(B)\subseteq A$ or $f(B)\subseteq Y\setminus A$. Note that these inclusions are equivalent to $B\subseteq f^{-1}(A)$ or $B\subseteq X\setminus f^{-1}(A)$. So, we know there exists such a $B$ because $\mathcal{B}$ is a basis for an ultrafilter on $X$.

Answer 2

Assume $\mathcal{B}$ is a base for an ultrafilter $\mathcal{U}$.

By point 1 we know that $f[\mathcal{B}]$ generates a filter too, call it $\mathcal{F}$. Each $F \in \mathcal{F}$ contains some $f[B]$ for $B \in \mathcal{B}$ so $f^{-1}[F]$ is non-empty (it contains $B$) and so $f^{-1}[\mathcal{F}]$ generates a filter on $X$ by point 2.

If $B \in \mathcal{B}$ then $f[B] \in f[\mathcal{B}]$, so $f[B] \in \mathcal{F}$ and as $B \subseteq f^{-1}[f[B]]$, we have that $f^{-1}[f[B]] \in \mathcal{U}$, and as this also holds for all supersets of $f[B]$ we see that $f^{-1}[\mathcal{F}] = \mathcal{U}$. But this immediately implies that $\mathcal{F}$ is maximal and so $f[\mathcal{B}]$ is an ultrafilter base.