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Updated sets.md

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Luc Bijl 2023-12-02 13:43:29 +01:00
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> *Definition*: suppose $A$ and $B$ are sets. Then $A$ is called a subset of $B$, if for every element $a \in A$ there also is $a \in B$. Then $B$ contains $A$ and can be denoted by $A \subseteq B$.
The extra line under the symbol implies properness. A subset $A$ of a set $B$ which is not the empty set $\empty$ nor the full set $B$ is called a proper subset of $B$, denoted by $A \subsetneq B$. For example $\mathbb{N} \subsetneq \mathbb{Z}$.
The extra line under the symbol implies properness. A subset $A$ of a set $B$ which is not the empty set $\varnothing$ nor the full set $B$ is called a proper subset of $B$, denoted by $A \subsetneq B$. For example $\mathbb{N} \subsetneq \mathbb{Z}$.
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> *Definition*: if $B$ is a set, then $\wp(B)$ denotes the set of all subsets $A$ of $B$. The set $\wp(B)$ is called the power set of $B$.
Suppose for example that $B = {x,y,z}$, then $\wp(B) = \{\empty,\{x\},\{y\},\{z\},\{x,y\},\{x,z\},\{y,z\},\{x,y,z\}\}$.
Suppose for example that $B = {x,y,z}$, then $\wp(B) = \{\varnothing,\{x\},\{y\},\{z\},\{x,y\},\{x,z\},\{y,z\},\{x,y,z\}\}$.
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> *Proposition*: let $B$ be a set with $n$ elements. Then its power set $\wp(B)$ contains $w^n$ elements.
??? note "*Proof*:"
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<summary><em>Proof</em>:</summary>
Let $B$ be set with $n$ elements. A subset $A$ of $B$ is completely determined by its elements. For each element $b \in B$ there are two options, it is in $A$ or it is not. So, there are $2^n$ options and thus $2^n$ different subsets $A$ of $B$.
Let $B$ be set with $n$ elements. A subset $A$ of $B$ is completely determined by its elements. For each element $b \in B$ there are two options, it is in $A$ or it is not. So, there are $2^n$ options and thus $2^n$ different subsets $A$ of $B$.
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