\(\DeclarePairedDelimiterX{\Set}[2]{\{}{\}}{#1 \nonscript\;\delimsize\vert\nonscript\; #2}\) \( \DeclarePairedDelimiter{\set}{\{}{\}}\) \( \DeclarePairedDelimiter{\parens}{\left(}{\right)}\) \(\DeclarePairedDelimiterX{\innerproduct}[1]{\langle}{\rangle}{#1}\) \(\newcommand{\ip}[1]{\innerproduct{#1}}\) \(\newcommand{\bmat}[1]{\left[\hspace{2.0pt}\begin{matrix}#1\end{matrix}\hspace{2.0pt}\right]}\) \(\newcommand{\barray}[1]{\left[\hspace{2.0pt}\begin{matrix}#1\end{matrix}\hspace{2.0pt}\right]}\) \(\newcommand{\mat}[1]{\begin{matrix}#1\end{matrix}}\) \(\newcommand{\pmat}[1]{\begin{pmatrix}#1\end{pmatrix}}\) \(\newcommand{\mathword}[1]{\mathop{\textup{#1}}}\)
Natural Arithmetic
Number of Disjoint Unions
Number of Set Products
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Set Numbers and Arithmetic


How does the number of elements change with unions, and products.


There are a few nice relations.1 Recall from Finite Sets that the union and product of finite sets is finite. Also, the power of a finite set is finite.

Let $A$ and $B$ be finite sets with $A \cap B = \varnothing$. Then $\num{A \cup B} = \num{A} + \num{B}$.
Let $A$ and $B$ be a finite sets Then $\num{A \times B} = \num{A} \cdot \num{B}$.
Let $A$ and $B$ be a finite sets Then $\num{A^B} = \num{A}^{\num{B}}$.
Let $A$ be a finite set. Then $\num{\powerset{A}} = 2^{\nu m{A}}$.

  1. Proofs will appear in future editions. ↩︎
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