assignments/main_text.tex

@@ -1,6 +1,7 @@
 \documentclass{template/homework}
 
 \usepackage{enumitem}
+\usepackage{tcolorbox}
 \usepackage{subfiles}
 
 \title{MIT OCW Real Analysis}

assignments/week1/1.tex

@@ -4,11 +4,16 @@
 \section*{Week 1}
 
 \exercise*[0.3.6]
-\begin{enumerate}[label=\emph{\alph*)}]
-    \item Wanting to show:
-        $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$
+\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}
 
-        In order to prove this equivalence, we have to prove the implication both ways. We use two lemmas for this.
+\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}
@@ -28,6 +33,7 @@
             \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 $x \in B)$.
@@ -38,12 +44,52 @@
 
         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.
+    \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}
 
-\hrule
+
+\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
 
 \end{document}