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\section*{Week 1}

\exercise*[0.3.6]
\begin{tcolorbox}
    Prove:
    \begin{enumerate}[label=\emph{\alph*)}]
        \item $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$
        \item $A \cup (B \cap C) = (A \cup B) \cap (a \cup C)$
    \end{enumerate}
\end{tcolorbox}

\begin{enumerate}[label=\emph{\alph*)}, wide]
    \item 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 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}


\exercise[0.3.11]
\begin{tcolorbox}
    Prove by induction that $n < 2^n$ for all $n \in \N$.
\end{tcolorbox}

For this proof we will use induction. For this, we have to prove the base case, i.e. $n = 1$,
and the inductive step, $n < 2^n \implies n + 1 < 2^{n+1}$.

First, let's prove the base case. When $n = 1$, we get $1 < 2^1$, which is certainly true.

Then, for the inductive step. We assume that the proposition holds for any $m \in \N$.
So, $m < 2^m$. Multiplying both sides with 2 gives $2m < 2^{m+1}$.
Since $m \geq 1$, $m + 1 < 2m$, and thus $m + 1 < 2m < 2^{m+1}$.

Since both the base case and inductive step hold, we can close the induction, proving the proposition. \qed


\exercise[0.3.12]
\begin{tcolorbox}
    Show that for a finite set $A$ of cardinality $n$, the cardinality of $\mathcal{P}(A)$ is $2^n$.
\end{tcolorbox}

The power set of a set $A$, $\mathcal{P}(A)$, is defined as the set of all possible subsets of $A$.
This is very similar to an inclusion/exclusion problem.
It is built up by all the possible combinations of the different elements being either inside a certain subset or not.
For all possible subsets of $A$, we have that for every element $x \in A$ there are 2 possibilities,
either $x$ is in the subset or it isn't.
This means that for every additional element, the number of subsets increases by a factor of 2, with a minimum of 1,
in case of $A = \emptyset$. We will prove this formally now, using induction. 

For this, the base case is a set of 1 element (but the theorem also holds for the empty set, where $n=0$).
Let us assume that $A := \{ \pi \}$.
Then the cardinality of $\mathcal{P}(A)$ is $2^1$, with $\mathcal{P}(A)= \{ \emptyset, \{ \pi \}\}$.

For the inductive step, we assume that for any set $B$ of cardinality $m$, the cardinality of the power set of $B$
is $2^m$. Then, we will add an element $x \notin B$ to $B$ to increase its cardinality by 1, to $m + 1$,
creating a new set $C$.
Note that all the possible subsets of $B$ are still viable subsets of $C$, since $B \subset C$.
In order to create the new subsets, we can simply keep all the subsets of $B$, duplicate them and take the union with
the new element $x$, so now we also have all combinations of the old sets with possibly $x$ being in them.
Since this doubles the number of subsets, the cardinality of $\mathcal{P}(C)$ is $2^{m+1}$.

Both the base case and the inductive step hold, which closes the induction and proves the proposition. \qed

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