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saturated subset

Context

Topology

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

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

Introduction

Basic concepts

Universal constructions

Extra stuff, structure, properties

Examples

Basic statements

Theorems

Analysis Theorems

topological homotopy theory

Contents

Definition

The following terminology is used in topology:

Definition

(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)).

Properties

Lemma

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.

Proposition

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.

Lemma

Let

  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.

Proof

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.

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.