# Image of Ultrafilter is Ultrafilter

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## Theorem

Let $X, Y$ be sets.

Let $f: X \to Y$ be a mapping.

Let $\FF$ be an ultrafilter on $X$.

Then the image filter $f \sqbrk \FF$ is an ultrafilter on $Y$.

## Proof

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

Let $\GG$ be a filter on $Y$ such that $f \sqbrk \FF \subseteq \GG$.

We have to show that $f \sqbrk \FF = \GG$.

Aiming for a contradiction, suppose there exists $U \in \GG$ such that $U \notin f \sqbrk \FF$.

By the definition of $f \sqbrk \FF$ this implies that $f^{-1} \sqbrk U \not \in \FF$.

By definition of ultrafilter:

$V := X \setminus f^{-1} \sqbrk U \in \FF$

Because $V \subseteq f^{-1} \sqbrk {f \sqbrk V}$ it follows that:

$f^{-1} \sqbrk {f \sqbrk V} \in \FF$

Thus also:

$f \sqbrk V \in f \sqbrk \FF$

By assumption we have:

$f \sqbrk \FF \subseteq \GG$

thus:

$f \sqbrk V \in \GG$

But:

$f \sqbrk V \cap U = f \sqbrk {X \setminus f^{-1} \sqbrk U} \cap U = \O \notin \GG$

Thus $\GG$ is not a filter, which is a contradiction to our assumptions.

Thus there can be no $U \in \GG$ such that $U \notin \FF$.

Therefore:

$f \sqbrk \FF = \GG$

Hence $f \sqbrk \FF$ is an ultrafilter.

$\blacksquare$