assignments/week3/main.tex

@@ -121,7 +121,7 @@ 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\footnote{Proven in
-earlier exercise}. 
+earlier exercise.}. 
 
 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 $a_0 - \epsilon < a$. Let us assume then that this implies that $a_0 \neq \sup A$ and
@@ -136,11 +136,11 @@ Since the implication holds both ways, the equivalence is proven. \qed
 \begin{tcolorbox}
     \begin{enumerate}[label=\emph{(\alph*)}]
         \item Let $a,b \in \R$ with $a < b$. Prove that the sets $(-\infty, a), (a,b)$ and $(b, \infty)$ are open.
-        \item Let $A$ be a set (not necessarily a subset of $\R$), and for each $\lambda \in A$, let $U_\lambda \subset
-            \R$. Prove that if $U_\lambda$ is open for all $\lambda \in A$ then the set
+        \item Let $\Lambda$ be a set (not necessarily a subset of $\R$), and for each $\lambda \in \Lambda$,
+            let $U_\lambda \subset \R$. Prove that if $U_\lambda$ is open for all $\lambda \in \Lambda$ then the set
             \begin{equation*}
-                \bigcup_{\lambda \in A} U_\lambda = 
-                \{x \in \R : \exists \lambda \in A \text{ such that } x \in U_\lambda\}
+                \bigcup_{\lambda \in \Lambda} U_\lambda = 
+                \{x \in \R : \exists \lambda \in \Lambda \text{ such that } x \in U_\lambda\}
             \end{equation*}
             is open.
         \item Let $n \in \N$, and let $U_1,...,U_n \subset \R$. Prove that if $U_1,...,U_n$ are open then the set
@@ -157,9 +157,28 @@ Since the implication holds both ways, the equivalence is proven. \qed
         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.
 
+        Let $a,b \in \R$ such that $a < b$. We want to show that for all $x \in (a,b)$ there exists $\epsilon > 0$
+        such that $(x - \epsilon, x + \epsilon) \subset (a,b)$. Since for all $y \in (a,b)$ such that 
+        $x - \epsilon < y < x + \epsilon$ this statement will recursively hold, we only need to prove that there 
+        exists $\epsilon > 0$ such that $a < x - \epsilon$ and $x + \epsilon < b$. Then we can pick a fitting
+        $\epsilon$ in the following way, depending if $x$ is closer to $a$ or to $b$, formalized as follows.
 
-    \item
-    \item
+        If $x - a = b - x \implies 2x = b + a \implies x = \frac{b+a}{2}$, then $x$ is precisely between $a$ and $b$
+        and we can pick $\epsilon$ to be $\frac{b - a}{4}$. Then $x + \epsilon < b$ and $x - \epsilon > a$.
+
+        When $x - a < b - x$, then $x$ will be closer to $a$ then to be and $\epsilon$ is bounded more by $x$'s
+        proximity to $a$ than to $b$, i.e. $\epsilon < x - a$. So we can pick $\epsilon = \frac{x-a}{2} < x - a$.
+        Then $x - \epsilon = x - \frac{x - a}{2} = \frac{x + a}{2}$. Since $x > a$, $\frac{x + a}{2} > a$. For the
+        other side, $x + \epsilon = x + \frac{x - a}{2} < x + \frac{b - x}{2} = \frac{x + b}{2} < b$ since $x < b$.
+        So for both sides, we have shown that there exists an $\epsilon$ such that both
+        $x - \epsilon, x + \epsilon \in (a,b)$. All elements inbetween $x - \epsilon$ and $x + \epsilon$ will also
+        definitely be in $(a,b)$.
+        
+        The argument when $b - x < x - a$ is very similar and will be left to the reader. Then, for all $x \in (a,b)$,
+        the statement is proven. \qed
+    \item Non-formally speaking, in this exercise we want to prove that any union of open sets in $\R$ is open itself.
+    \item Similarly to the previous exercise, non-formally speaking we want to prove that any intersection of open sets
+        in $\R$ is open itself.
     \item
 \end{enumerate}