Finished exercise 6 from week 3, just a few to go

assignments/week3/main.tex

@@ -177,9 +177,40 @@ Since the implication holds both ways, the equivalence is proven. \qed
         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.
+        In order to make this formal, we will assume that $U_\lambda$ is open for all $\lambda \in \Lambda$ and
+        follow the definition as presented.
+
+        So, let us assume that $U_\lambda$ is open for all $\lambda \in \Lambda$. This means that for every $x_\lambda
+        \in U_\lambda$, there exists an $\epsilon > 0$ such that $(x_\lambda - \epsilon, x_\lambda + \epsilon) \subset
+        U_\lambda$. To prove that $\bigcup_{\lambda \in \Lambda} U_\lambda$ is open, we need to show that the same
+        property holds for all $y$ in this set. But since the union between some sets is defined as the set that holds
+        all the elements that any of these sets hold, this is trivial: for any $y \in \bigcup_{\lambda \in \Lambda}$
+        for which we want to know what $\epsilon$ we need to show that the union is open around that $y$, we just pick
+        the corresponding $\epsilon$ for the subset $U_\lambda$ which was open. Since all elements in that $U_\lambda$
+        are also in the union, this must certainly be the case. \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.
+        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.
+
+        Formally, let $n \in \N$ and let $U_1,...,U_n \subset \R$. We assume that all $U_1,...,U_n$ are open.
+
     \item
 \end{enumerate}
 
+\exercise*
+\begin{tcolorbox}
+    Prove that
+    \begin{equation*}
+        \lim_{n \rightarrow \infty} \frac{1}{20n^2 + 20n + 2020} = 0.
+    \end{equation*}
+\end{tcolorbox}
+In order to prove that this limit holds, we need to show that a function $\{x_n\}$ converges to $x$, 
+i.e. if for all $\epsilon > 0$, $\exists M \in \N$ such that $\forall n \geq M$ the following inequality holds:
+$|x_n = x| < \epsilon$.
+
+Let $\epsilon > 0$. We choose $M \in \N$ such that $\frac{1}{M} < \epsilon$ (Archimedean Property). Then for all
+$n \geq M$, $|\frac{1}{20n^2+20n+2020} - 0| = \frac{1}{20n^2 + 20n + 2020} \leq \frac{1}{n^2 + n} \leq
+\frac{1}{n} \leq \frac{1}{M} < \epsilon$. \qed
+
 \end{document}