| Version 6 |
Version 5 |
| 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: |
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: |
| $$x\in A\Rightarrow x\in B.$$ |
$$x\in A\Rightarrow x\in B.$$ |
| The relation beetween $A$ and $B$ is then called {\em set inclusion}. |
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| Some examples: |
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, $A\subseteq B$. |
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, $A\subseteq B$. |
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| On the other hand, if $C=\{p,e,d,r,o\}$ neither $A$ is a subset of $C$ (because $t\in A$ but $t\not\in C$) nor $C$ is a subset of $A$ (because $p\in C$ but $p\not\in A$). |
On the other hand, if $C=\{p,e,d,r,o\}$ neither $A$ is a subset of $C$ (because $t\in A$ but $t\not\in C$) nor $C$ is a subset of $A$ (because $p\in C$ but $p\not\in A$). |
| The fact that $A$ is not a subset of $C$ is written as $A\not\subseteq C$. And then, in this example we also have $C\not\subseteq A$. |
The fact that $A$ is not a subset of $C$ is written as $A\not\subseteq C$. And then, in this example we also have $C\not\subseteq A$. |
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| If $X\subseteq Y$ and $Y\subseteq X$, it must be the case that $X=Y$. |
If $X\subseteq Y$ and $Y\subseteq X$, it must be the case that $X=Y$. |
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| Every set is a subset of itself, and the empty set is a subset of every other set. |
Every set is a subset of itself, and the empty set is a subset of every other set. |
| The set $A$ is called a \emph{proper subset} of $B$, if |
The set $A$ is called a \emph{proper subset} of $B$, if |
| $A\subset B$ and $A\neq B$ (in this case we do not use $A\subseteq B$). |
$A\subset B$ and $A\neq B$ (in this case we do not use $A\subseteq B$). |
