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real_analysis/assignments/week1/1.tex

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\documentclass[../main_text.tex]{subfiles}
\begin{document}
\section*{Week 1}
\exercise*[0.3.6]
\begin{enumerate}[label=\emph{\alph*)}]
\item Wanting to show:
$A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$
In order to prove this equivalence, we have to prove the implication both ways. We use two lemmas for this.
\begin{lemma}
\label{lem:set1}
$A \cap (B \cup C) \implies (A \cap B) \cup (A \cap C)$
\end{lemma}
Let $x \in A \cap (B \cup C)$.
By the definition of set intersection, $x \in A$ and $x \in B \cup C$.
By the definition of set union, $x \in A$ and $(x \in B$ or $x \in C)$.
From propositional logic we know that for propositions $P$, $Q$ and $R$ the following holds:
$P \land (Q \lor R) \iff (P \land Q) \lor (P \land R)$.
So, substituting for this particular case yields $(x \in A$ and $x \in B)$ or $(x \in A$ and $x \in C)$.
Using the definition of set intersection again gets $x \in A \cap B$ or $x \in A \cap C$.
Using the definition of set union again gives $x \in (A \cap B) \cup (A \cap C)$. \qed
\begin{lemma}
\label{lem:set2}
$(A \cap B) \cup (A \cap C) \implies A \cap (B \cup C)$
\end{lemma}
Let $x \in (A \cap B) \cup (A \cap C)$.
By the definition of set union, $x \in (A \cap B)$ or $x \in (A \cap C)$.
By the definition of set intersection, $(x \in A$ or $x \in B)$ and $(x \in A$ or $x \in B)$.
Using the same propositional logical equivalence as in Lemma \ref{lem:set1}, this gives
$x \in A$ and $(x \in B$ or $x \in C)$.
Wrapping up, we use the definition of set union to get $x \in A$ and $x \in B \cup C$ and
the definition of intersection to get $x \in A \cap (B \cup C)$. \qed
Using Lemma \ref{lem:set1} and \ref{lem:set2}, we get the desired equivalence of
$A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$. \qed
\item Wanting to show:
$A \cup (B \cap C) = (A \cup B) \cap (a \cup C)$
This proof is so similar to a) that it feels like a waste of time and will therefore be left to the reader.
\end{enumerate}
\hrule
\end{document}