\(\DeclarePairedDelimiterX{\Set}[2]{\{}{\}}{#1 \nonscript\;\delimsize\vert\nonscript\; #2}\) \( \DeclarePairedDelimiter{\set}{\{}{\}}\) \( \DeclarePairedDelimiter{\parens}{\left(}{\right)}\) \(\DeclarePairedDelimiterX{\innerproduct}[1]{\langle}{\rangle}{#1}\) \(\newcommand{\ip}[1]{\innerproduct{#1}}\) \(\newcommand{\bmat}[1]{\left[\hspace{2.0pt}\begin{matrix}#1\end{matrix}\hspace{2.0pt}\right]}\) \(\newcommand{\barray}[1]{\left[\hspace{2.0pt}\begin{matrix}#1\end{matrix}\hspace{2.0pt}\right]}\) \(\newcommand{\mat}[1]{\begin{matrix}#1\end{matrix}}\) \(\newcommand{\pmat}[1]{\begin{pmatrix}#1\end{pmatrix}}\) \(\newcommand{\mathword}[1]{\mathop{\textup{#1}}}\)
Needs:
Set Dualities
Set Partitions
Needed by:
None.
Links:
Sheet PDF
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Set Exercises

Why

Here are some exercises on sets.1

\begin{exercise} Let $A, B, C$ denote sets. Show $((A \cap B) \cup C = A \cap (B \cup C)) \iff (C \subset A)$ Observe that the condition does not involve $B$. \end{exercise}

\begin{exercise}

\[ A - B = A \cap B'. \]

\end{exercise}

\begin{exercise}

\[ A \subset B \text{ if and only if} A - B = \varnothing. \]

\end{exercise}

\begin{exercise}

\[ A - (A - B) = A \cap B. \]

\end{exercise}

\begin{exercise}

\[ A \cap (B - C) = (A \cap B) - (A \cap C). \]

\end{exercise}

\begin{exercise}

\[ (A \cap B) \subset ((A \cap C) \cup (A \cap C')). \]

\end{exercise}

\begin{exercise}

\[ ((A \cup C) \cap (B \cup C')) \subset (A \cup B). \]

\end{exercise}

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  1. Future editions will give the hypotheses more clearly. ↩︎
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