WIP on exercise 4, very close

assignments/week3/main.tex

@@ -25,10 +25,104 @@ since $x < k < m < h < y$. \qed
 \exercise*
 \begin{tcolorbox}
     Let $E \subset (0,1)$ be the set of all real numbers with decimal representation using only the digits 1 and 2:
-    \begin{equation}
+    \begin{equation*}
         E := \{x \in (0,1) : \forall j \in \N, \exists d_j \in \{1,2\} \text{ such that } x = 0.d_1d_2...\}
-    \end{equation}
+    \end{equation*}
     Prove that $|E| = |\mathcal{P}(\N)|$.
 \end{tcolorbox}
 
+As a hint to this exercise: \textit{Consider the function $f : E \rightarrow \mathcal{P}(\N)$ such that if $x \in E,
+x = 0.d_1d_2...$,
+\begin{equation*}
+    f(x) = \{j \in \N : d_j = 2\}.
+\end{equation*}}
+
+In order to prove that 2 sets are of equal cardinality, we need to prove that there is a bijective function between
+the 2 sets. In this case, the aforementioned hint function does the trick. Non-formally speaking, it is exactly what
+we are looking for: it is a (weird) representation of the power set of natural numbers, in that for every decimal,
+represented by a natural number, it is decided if that decimal is a 2 or a 1. This is similar to the actual power set
+of the natural numbers, in which for every natural number it is decided whether the number is in a subset or not.
+
+Now for a formal proof. To show that $f$ is bijective, we need to show that it is surjective and injective.
+\begin{description}
+    \item[Injectivity of $f$] In order to show that $f$ is injective, we have to show that for every $x \in E$, there
+        is a unique $y \in \mathcal{P}(\N)$ for the function, by showing that $f(a) = f(b) \implies a = b$.
+
+        So, let's assume that for some $a,b \in E$, $f(a) = f(b)$. So, there two sets of natural numbers
+        $\{a_1,a_2,...,a_n\} = \{b_1,b_2,...,b_m\}$. Equality in sets means that every element that is present in the
+        one set, is present in the other, and vice versa. No element that is present in either set, is missing in the
+        other. So, in this case, both sets will represent the same sequence of digits that are 2. Because the only
+        other option for digits is 1, that means the complete digital representation of $a$ and $b$ are known,
+        unique and the same. This concludes the proof for injectivity.
+    \item[Surjectivity of $f$] To prove surjectivity, we need to prove that for any arbitrary $y \in \mathcal{P}(\N)$,
+        there exists a corresponding $x \in E$ such that $f(x) = y$.
+
+        So, take an arbitrary $y = \{y_1, y_2, ... , y_n\}$, where each $y_i \in \N$ and thus $y \in \mathcal{P}(\N)$.
+        Then, corresponding $x \in E$ can be constructed easily as follows. Take a decimal number $0.d_1d_2...$ and 
+        turn every decimal $d_i$ for which $i \in y$ into a 2, and every other decimal into a 1. Since every decimal
+        can only be a 1 or 2, this handles every decimal correctly. Also, $f(x)$ will be in $\mathcal{P}(\N)$.
+\end{description}
+
+Since, $f$ is 1-to-1 and onto, $f$ is bijective. Then, because there exists a bijective function from
+$E$ to $\mathcal{P}(\N)$, $|E| = |\mathcal{P}(\N)|$. \qed
+
+\exercise*
+\begin{tcolorbox}
+    \begin{enumerate}[label=\emph{(\alph*)}]
+        \item Let $A$ and $B$ be two disjoint, countably infinite sets. Prove that $A \cup B$ is countably infinite.
+        \item Prove that the set of irrational numbers, $\R\backslash\Q$, is uncountable. You may use the facts
+            discussed in the lectures that $\R\backslash\Q$ is infinite and $\R$ is uncountable without proof.
+    \end{enumerate}
+\end{tcolorbox}
+\begin{enumerate}[label=\emph{(\alph*)},wide]
+    \item So let $A$ and $B$ be two disjoint, countably infinite sets. Since these sets are countably infinite,
+        a bijective function to $\N$ exists for both functions separately. It is then straightforward to map both these
+        function together to $\Z$ instead, in the following way. Let $f$ be the bijective function such that
+        $f : A \rightarrow \N$ and let $g$ be the bijective function such that $g : B \rightarrow \N$. Then, we can
+        define a new function $h : A \cup B \rightarrow \Z$ as
+        \begin{align*}
+            h(x) &= f(x) \text{ if } x \in A \\
+                 &= -g(x) \text{ if } x\in B.
+        \end{align*}
+        Since $A \cap B = \emptyset$, this function is unambiguously defined. Since $\Z$ is countably infinite,
+        $A \cup B$ is countably infinite as well. \qed
+
+    \item Because of part (a), we know that if we have two disjoint, countably infinite sets and join them, the result
+        is still countably infinite. The opposite must then also be true: if we have a countably infinite set and
+        we divide it into two disjoint subsets, both of which are infinite, then they still must be countable.
+
+        So then, for $\R\backslash\Q$, we know that $\R$ is uncountably infinite. So when we split it into rational
+        and irrational subsets, from which we know that $\Q$ is countably infinite, $\R\backslash\Q$ must be at least
+        and at most uncountably infinite. \qed
+\end{enumerate}
+
+\exercise*
+\begin{tcolorbox}
+    Let $A$ be a subset of $\R$ which is bounded above, and let $a_0$ be an upper bound for $A$. Prove that $a_0 =
+    \sup A$ if and only if for every $\epsilon > 0$, there exists $a \in A$ such that $a_0 - \epsilon < a$.
+\end{tcolorbox}
+
+Let $A \subset \R$, with $A$ bounded above by $a_0$.
+So, we have to prove the implication both ways. First, let's prove that the implication to the right $(\rightarrow)$.
+
+Assume that $a_0 = \sup A$, so for all $a \in A$, $a \leq a_0$. Also, let $\epsilon > 0$. If $a_0 \in A$,
+then we pick $a_0$ as $a$ and get $a_0 - \epsilon < a_0$, which holds $\forall \epsilon > 0$. If $a_0 \notin A$,
+then we choose $a$ as the average of $a_0$ and $a_0 - \epsilon$, which is definitely smaller than $a_0$. We are
+allowed to pick this as $a$, because we assume without loss of generality that $a \geq \inf A$. Then we get
+\begin{align*}
+    a_0 - \epsilon &< \frac{a_0 + a_0 - \epsilon}{2} \\
+                   &< a_0 - \frac{\epsilon}{2} \implies \\
+         -\epsilon &< -\frac{\epsilon}{2}.
+\end{align*}
+Since $\epsilon > 0$, this always holds.
+
+Now for the implication to the left $(\leftarrow)$.
+
+Assume now that the right side is true, i.e. let's assume that $\forall \epsilon > 0$, there exists $a \in A$ such that
+$a_0 - \epsilon < a$. Again, let us first investigate the case where $a_0 \in A$. Well certainly still, if $a_0$ is 
+an upper bound for $A$ and it is also part of the set itself, it must be the supremum. Then, let's assume that
+$a_0 \notin A$. Now, for all positive $\epsilon$, we know there exists an $a \in A$ such that $a \neq a_0$ and
+$a_0 - \epsilon < a$. So, $a_0 < a + \epsilon$. $a_0$ is an upper bound and we can find $a$'s such that $a+\epsilon$
+is always bigger than $a_0$, so $a_0$ must be the supremum. If $a_0$ wasn't the supremum, then there must be some $b$
+such that $a < b < a_0 < a + \epsilon$. WIP
 \end{document}