assignments/week3/main.tex

@@ -74,7 +74,7 @@ $E$ to $\mathcal{P}(\N)$, $|E| = |\mathcal{P}(\N)|$. \qed
             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]
+\begin{enumerate}[label=\emph{(\alph*)}]
     \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
@@ -152,7 +152,7 @@ Since the implication holds both ways, the equivalence is proven. \qed
     \end{enumerate}
 \end{tcolorbox}
 
-\begin{enumerate}[label=\emph{(\alph*)},wide]
+\begin{enumerate}[label=\emph{(\alph*)}]
     \item Since $\R$ is open, it is clear that $(-\infty, a)$ and $(b, \infty)$ are open to the left and right
         respectively as well. Also, their respective right and left side are present in $(a,b)$ as well, so we will
         only prove it for this case. The other cases follow logically.
@@ -191,11 +191,29 @@ Since the implication holds both ways, the equivalence is proven. \qed
     \item Similarly to the previous exercise, non-formally speaking we want to prove that any intersection of open sets
         in $\R$ is open itself. This is not as trivial as in the previeous exercise however: since every set that is
         added as an intersection poses another restriction, we don't have the immediate guarantee that every $\epsilon$
-        from the subsets will also hold for the intersection set.
+        from the subsets will also be a well-defined element for the intersection set.
 
-        Formally, let $n \in \N$ and let $U_1,...,U_n \subset \R$. We assume that all $U_1,...,U_n$ are open.
+        Now formally. Let $n \in \N$ and let $U_1,...,U_n \subset \R$. We assume that all $U_1,...,U_n$ are open.
+        Then we will prove that $\bigcap_{m = 1}^n U_m$ is open by induction over $n$.
 
-    \item
+        For the base case, let $n = 1$. Then the intersection set is equivalent to $U_1$. Since $U_1$ is open, then
+        so is the intersection set.
+
+        For the inductive step, we assume that the intersection set is open for a certain $n = h$, i.e. there exists
+        an $\epsilon > 0$ such that $(x - \epsilon, x + \epsilon)$ is open for every $x \in \bigcap_{m = 1}^h U_m$.
+        Now, we will add one additional open set to this intersection, $U_{h+1}$, such that $n$ become $h+1$. Note
+        that $h + 1 \in \N$. Let the new intersection set be denoted as $\bigcap'$, and the old one as $\bigcap$.
+        Then in order to find an $\epsilon > 0$ for every $x \in \bigcap'$ such that $(x - \epsilon, x + \epsilon)$,
+        we take the smallest of $\epsilon$'s for that $x$ compared between $\bigcap$ and $U_{h+1}$. Since
+        $x \in \bigcap'$, we know that $x \in \bigcap$ and $x \in U_{h+1}$. Then the smallest accomponying $\epsilon$
+        always gives a well-defined open set inside of $\bigcap'$ because $|(x + \epsilon_1) - (x - \epsilon_1)| < 
+        |(x + \epsilon_2) - (x - \epsilon_2)|$ if $\epsilon_1 < \epsilon_2$, and thus $\bigcap'$ is open itself. \qed
+    \item No, $\Q$ is not open in $\R$. This is because we can't find an $\epsilon > 0$ such that for every $q \in \Q$,
+        $(q - \epsilon, q + \epsilon) \subset \Q$. We know that $\Q$ is dense in $\R$, but as we have proven in
+        exercise 1, the converse is also true. For every real numbers, we can find a real number inbetween that is not
+        a rational number. So, we cannot pick an $\epsilon > 0$ such that there is an interval around $x$ that itself
+        is completely contained in $\Q$. For every $\epsilon$ we pick, we can always find a real number $r$ such that
+        $x < r < x + \epsilon$ and $x - \epsilon < r < x$. So, $\Q$ is not open. \qed
 \end{enumerate}
 
 \exercise*