Image of Ultrafilter is Ultrafilter

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Theorem

Let $X, Y$ be two sets, $f: X \to Y$ a mapping and $\mathcal F$ an ultrafilter on $X$.


Then the image filter $f \left({\mathcal F}\right)$ is an ultrafilter on $Y$.


Proof

From Image Filter is Filter, we have that $\mathcal F$ is a filter on $Y$.

Let $\mathcal G$ be a filter on $Y$ such that $f \left({\mathcal F}\right) \subseteq \mathcal G$.

We have to show that $f \left({\mathcal F}\right) = \mathcal G$.

Let $U \in \mathcal G$.

Assume that $U \notin f \left({\mathcal F}\right)$.

By the definition of $f \left({\mathcal F}\right)$ this implies that $f^{-1}(U) \not \in \mathcal F$.

By definition of ultrafilter:

$V := X \setminus f^{-1} \left({U}\right) \in \mathcal F$

Because $V \subseteq f^{-1} \left({f \left({V}\right)}\right)$ it follows that $f^{-1}\left({f \left({V}\right)}\right) \in \mathcal F$ and thus also $f \left({V}\right) \in f \left({\mathcal F}\right)$.

By assumption we have $f \left({\mathcal F}\right) \subseteq \mathcal G$, thus $f \left({V}\right) \in \mathcal G$.

But:

$f \left({V}\right) \cap U = f \left({X \setminus f^{-1} \left({U}\right)}\right) \cap U = \varnothing \notin \mathcal G$

Thus $\mathcal G$ is not a filter, a contradiction to our assumptions.

Thus:

$U \in f \left({\mathcal F}\right)$ and therefore $f \left({\mathcal F}\right) = \mathcal G$

Hence $f \left({\mathcal F}\right)$ is an ultrafilter.

$\blacksquare$