Image Filter is Filter

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Theorem

Let $X, Y$ be sets.

Let $\mathcal P \left({X}\right)$ and $\mathcal P \left({Y}\right)$ be the power sets of $X$ and $Y$ respectively.

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

Let $\mathcal F \subset \mathcal P \left({X}\right)$ be a filter on $X$.


Then the image filter of $\mathcal F$ with respect to $f$:

$f \left[{\mathcal F}\right] := \left\{ {U \subseteq Y: f^{-1} \left[{U}\right] \in \mathcal F}\right\}$

is a filter on $Y$.


Proof

From the definition of a filter we have to prove four things:

$(1): \quad f \left[{\mathcal F}\right] \subset \mathcal P \left({Y}\right)$
$(2): \quad Y \in f \left[{\mathcal F}\right], \varnothing \notin f \left[{\mathcal F}\right]$
$(3): \quad U, V \in f \left[{\mathcal F}\right] \implies U \cap V \in f \left[{\mathcal F}\right]$
$(4): \quad U \in f \left[{\mathcal F}\right], U \subseteq V \subseteq Y \implies V \in f \left[{\mathcal F}\right]$


By construction we have:

$f \left[{\mathcal F}\right] \subseteq \mathcal P \left({Y}\right)$

Since $f^{-1} \left[{\varnothing}\right] = \varnothing \notin \mathcal F$ we know that:

$\varnothing \notin f \left[{\mathcal F}\right]$

Therefore:

$f \left[{\mathcal F}\right] \ne \mathcal P \left({Y}\right)$

which implies $(1)$.


Because $f^{-1} \left[{Y}\right] = X \in \mathcal F$, we have:

$Y \in f \left[{\mathcal F}\right]$

Since we've already shown $\varnothing \notin f \left[{\mathcal F}\right]$, this implies $(2)$.


Let $U, V \in f \left[{\mathcal F}\right]$.

From Preimage of Intersection under Mapping:

$f^{-1} \left[{U \cap V}\right] = f^{-1} \left[{U}\right] \cap f^{-1} \left[{V}\right] \in \mathcal F$

(since $\mathcal F$ is a filter).

Thus $U \cap V \in f \left[{\mathcal F}\right]$, and so $(3)$ holds.


Finally, let $U \in f \left[{\mathcal F}\right]$ and $V \subseteq Y$ such that $U \subseteq V$.

Then from Image of Subset under Relation is Subset of Image/Corollary 3:

$f^{-1} \left[{U}\right] \subseteq f^{-1} \left[{V}\right]$

Since $f^{-1} \left[{U}\right] \in \mathcal F$ and $\mathcal F$ is a filter it follows that:

$f^{-1} \left[{V}\right] \in \mathcal F$

which implies:

$V \in f \left[{\mathcal F}\right]$

and thus $(4)$.

$\blacksquare$