(Baby Rudin, Chapter 2, Exercise 7a)
I am trying to prove:
Let $A_{1}, A_{2}, A_{3}, …$ be subsets of a metric space. If $B_{n} = \bigcup^n_{i=1} A_{i}$, prove that $\bar{B}_{n} = \bigcup^n_{i=1} \bar{A_{i}}$ for $n \in \mathbb{N}$.
My attempt:
Proof. First, we show that $\overline{\cup_{i =1}^{n} A_i} \subseteq \cup_{i = 1}^{n} \bar A_i$. Let $x \in \overline{\cup_{i =1}^{n} A_i}$. Then, either $x \in \cup_{i =1}^{n} A_i$ or $x$ is a limit point of $\cup_{i =1}^{n} A_i$. If $x \in \cup_{i =1}^{n} A_i$, we have that $x \in A_i$ for some $i$ $\implies x \in \bar A_i$ for some $i$ $\implies x \in \cup_{i = 1}^{n} \bar A_i$. If $x$ is a limit point of $\cup_{i =1}^{n} A_i$, and if $N(x, r)$ is an arbitrary neighborhood of $x$, then $\exists q \neq x$ such that $\color{blue}{q \in N(x, r)}$ and $q \in \cup_{i =1}^{n} A_i$. Then, $q \in A_i$ for some $i \implies q \in \bar A_i$ for some $i \implies q \in \cup_{i = 1}^{n} \bar A_i$.
(The proof for the other direction is pretty similar.)
My question: Can I now conclude that $\overline{\cup_{i =1}^{n} A_i} \subseteq \cup_{i = 1}^{n} \bar A_i$? I am guessing that I cannot since I still have to show that if $x$ is a limit point of $\cup_{i =1}^{n} A_i$ then $x \in \cup_{i = 1}^{n} \bar A_i$ (I've shown that $q \in \overline{\cup_{i =1}^{n} A_i} \implies q \in \cup_{i = 1}^{n} \bar A_i$ instead.) Is my guess correct? What arguments are missing from this proof still? I think will need to use the statement in blue, but I don't know how. Thanks for taking the time to read this.
- Proof Attempt 2 based on @copper.head's suggestions:
Suppose $B_{n} = \bigcup^n_{i=1} A_{i}$. First, we claim that
\begin{equation*}
\overline{A_1 \cup A_2} \subset \overline{A_1} \cup \overline{A_2}
\end{equation*}
Let $x \in A_1 \cup A_2$. Then, $x \in A_1 \textrm{ or } x \in A_2 \implies x \in \overline{A_1} \textrm{ or } x \in \overline{A_2} \implies x \in \overline{A_1} \cup \overline{A_2}$. Thus, $A_1 \cup A_2 \subset \bar{A_1} \cup \bar{A_2}$. (Just to be clear, it was mentioned in Chapter 1 that Rudin uses $\subset$ and $\subseteq$ interchangeably.) Note that since $\bar{A_1}$ and $\bar{A_2}$ are closed, $\bar{A_1} \cup \bar{A_2}$ is closed as well. Then, $\overline{A_1 \cup A_2} \subset \bar{A_1} \cup \bar{A_2}$.
Continuing inductively, suppose $\overline{\cup_{i = 1}^{n}A_i} \subset \cup_{i = 1}^{n} \bar{A_i}$. Let $y \in \bigcup_{i = 1}^{n+1} A_i$. Then, $y \in A_i$ for some $i \in [1, n+1]$.
My question: How can I prove that $\cup_{i = 1}^{n+1}A_i \subset \bigcup_{i = 1}^{n+1} \overline{A_i}$ in a way that utilizes the induction hypothesis (I know how to prove the same without using the induction hypothesis)? Then, $\overline{\cup_{i = 1}^{n}A_i} \subset \bigcup_{i = 1}^{n} \overline{A_i}$ will follow immediately.
Please know that I am aware of this and this but I am asking for specific ways of completing my proof.
Best Answer
Here is another way to look at this.
Notice $\bigcup^n_{k=1}A_k \subset \bigcup^n_{k=1}\overline{A_k}$ and so
$$\overline{\bigcup^n_{k=1}A_k } \subset \bigcup^n_{k=1}\overline{A_k}$$ Here we have used the fact that $A\subset B$ implies that $\overline{A}\subset\overline{B}$ and also that $\overline{B}=B$ when $B$ is closed.
On the other hand, $$\overline{A_k}\subset\overline{\bigcup^n_{k=1}A_k}$$ Thus $$\bigcup^n_{k=1}\overline{A_k}\subset\overline{\bigcup^n_{k=1}A_k}$$