Contents

Definition

The following terminology is used in topology:

Definition

(saturated subset)

Let $f \;\colon\; X \longrightarrow Y$ be a function of sets. Then a subset $S \subset X$ is called an $f$-saturated subset (or just saturated subset, if $f$ is understood) if $S$ is the pre-image of its image:

$\left(S \subset X \,\, f\text{-saturated} \right) \,\Leftrightarrow\, \left( S = f^{-1}(f(S)) \right) \,.$

If $S$ is not necessarily $f$-saturated, then $f^{-f}(f(S))$ is called its saturation. Notice that $S \subset f^{-1}(f(S))$.

Properties

Lemma

Let $f \colon X \longrightarrow Y$ be a function. Then a subset $S \subset X$ is $f$-saturated (def. ) precisely if its complement $X \backslash S$ is so.

Proposition
$f \;\colon\; (X, \tau_X) \longrightarrow (Y,\tau_Y)$

whose underlying function $f \colon X \longrightarrow Y$ is surjective exhibits $\tau_Y$ as the corresponding quotient topology precisely if $f$ sends open and $f$-saturated subsets in $X$ (def. ) to open subsets of $Y$. By lemma this is the case precisely if it sends closed and $f$-saturated subsets to closed subsets.

Lemma

Let

1. $f \;\colon\; (X, \tau_X) \longrightarrow (Y, \tau_Y)$ be a closed map.

2. $C \subset X$ be a closed subset which is $f$-saturated;

3. $U \supset C$ an open subset containing $C$

then there exists a smaller open subset $V$ still containing $C$

$U \supset V \supset C$

and such that $V$ is $f$-saturated.

Proof

We claim that

$V \coloneqq X \backslash \left( f^{-1}\left( f\left( X \backslash U \right) \right) \right)$

has the desired properties. To see this, observe first that

1. the complement $X \backslash U$ is closed, since $U$ is assumed to be open;

2. hence the image $f(X\backslash U)$ is closed, since $f$ is assumed to be a closed map;

3. hence the pre-image $f^{-1}\left( f\left( X \backslash U \right)\right)$ is closed, since $f$ is continuous, therefore its complement $V$ is indeed open;

4. this pre-image $f^{-1}\left( f\left( X \backslash U \right) \right)$ is saturated, as all pre-images clearly are, and hence also its complement $V$ is saturated, by lemma .

Therefore it now only remains to see that $U \supset V \supset C$.

The inclusion $U \supset V$ means equivalently that $f^{-1}\left( f\left( X \backslash U \right)\right) \supset X \backslash U$, which is clearly the case.

The inclusion $V \supset C$ meas that $f^{-1}\left( f\left( X \backslash U \right) \right) \,\cap \, C = \emptyset$. Since $C$ is saturated by assumption, this means that $f^{-1}\left( f\left( X \backslash U \right)\right) \,\cap \, f^{-1}(f(C)) = \emptyset$. This in turn holds precisely if $f\left( X \backslash U \right) \,\cap \, f(C) = \emptyset$. Since $C$ is saturated, this holds precisely if $X \backslash U \cap C = \emptyset$, and this is true by the assumption that $U \supset C$.

Applications

Last revised on May 21, 2017 at 14:14:24. See the history of this page for a list of all contributions to it.