saturated subset



topology (point-set topology, point-free topology)

see also differential topology, algebraic topology, functional analysis and topological homotopy theory


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The following terminology is used in topology:


(saturated subset)

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

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

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



Let f:XYf \colon X \longrightarrow Y be a function. Then a subset SXS \subset X is ff-saturated (def. 1) precisely if its complement X\SX \backslash S is so.


A continuous function

f:(X,τ X)(Y,τ Y) f \;\colon\; (X, \tau_X) \longrightarrow (Y,\tau_Y)

whose underlying function f:XYf \colon X \longrightarrow Y is surjective exhibits τ Y\tau_Y as the corresponding quotient topology precisely if ff sends open and ff-saturated subsets in XX (def. 1) to open subsets of YY. By lemma 1 this is the case precisely if it sends closed and ff-saturated subsets to closed subsets.



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

  2. CXC \subset X be a closed subset which is ff-saturated;

  3. UCU \supset C an open subset containing CC

then there exists a smaller open subset VV still containing CC

UVC U \supset V \supset C

and such that VV is ff-saturated.


We claim that

VX\(f 1(f(X\U))) 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\UX \backslash U is closed, since UU is assumed to be open;

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

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

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

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

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

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


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