-
Notifications
You must be signed in to change notification settings - Fork 4
/
03-00-Subset.tex
50 lines (43 loc) · 1.78 KB
/
03-00-Subset.tex
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
\documentclass[12pt]{article}
\usepackage{pmmeta}
\pmcanonicalname{Subset}
\pmcreated{2013-03-22 11:52:38}
\pmmodified{2013-03-22 11:52:38}
\pmowner{Wkbj79}{1863}
\pmmodifier{Wkbj79}{1863}
\pmtitle{subset}
\pmrecord{13}{30472}
\pmprivacy{1}
\pmauthor{Wkbj79}{1863}
\pmtype{Definition}
\pmcomment{trigger rebuild}
\pmclassification{msc}{03-00}
\pmclassification{msc}{00-02}
\pmrelated{EmptySet}
\pmrelated{Superset}
\pmrelated{TotallyBounded}
\pmrelated{ProofThatAllSubgroupsOfACyclicGroupAreCyclic}
\pmrelated{Property2}
\pmrelated{CardinalityOfAFiniteSetIsUnique}
\pmrelated{CriterionOfSurjectivity}
\pmdefines{set inclusion}
\endmetadata
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{graphicx}
%%%%\usepackage{xypic}
\begin{document}
Given two sets $A$ and $B$, we say that $A$ is a subset of $B$ (which we denote as $A\subseteq B$ or simply $A\subset B$) if every element of $A$ is also in $B$. That is, the following implication holds:
$$x\in A\Rightarrow x\in B.$$
The relation between $A$ and $B$ is then called {\em set inclusion}.
Some examples:
The set $A=\{d,r,i,t,o\}$ is a subset of the set $B=\{p,e,d,r,i,t,o\}$ because every element of $A$ is also in $B$. That is, $A\subseteq B$.
On the other hand, if $C=\{p,e,d,r,o\}$, then neither $A \subseteq C$ (because $t\in A$ but $t\not\in C$) nor $C \subseteq A$ (because $p\in C$ but $p\not\in A$). The fact that $A$ is not a subset of $C$ is written as $A\not\subseteq C$. Similarly, we have $C\not\subseteq A$.
If $X\subseteq Y$ and $Y\subseteq X$, it must be the case that $X=Y$.
Every set is a subset of itself, and the empty set is a subset of every other set. The set $A$ is called a proper subset of $B$, if $A\subset B$ and $A\neq B$. In this case, we do not use $A\subseteq B$.
%%%%%
%%%%%
%%%%%
%%%%%
\end{document}