I am currently working on Rudin's real analysis problems. With the following definition from Rudin, I would like to prove a lemma.
Definition: A point $\textit{p}$ is a $\textit{limit point}$ of the set $\textit{E}$ if every neighborhood of $\textit{p}$ contains a point $\textit{q}\neq p$ such that $\textit{q}\in\textit{E}$.
Lemma: Let $A_1, A_2, A_3,...$ be subsets of a metric space $X$. Let $(\cup_{i=1}^{\infty}A_i)'$ denote the set of limit points of $\cup_{i=1}^{\infty}A_i$. Then $(\cup_{i=1}^{\infty}A_i)'$=$\cup_{i=1}^{\infty}A_i^{'}$.
Here's my proof:
$(\cup_{i=1}^{\infty}A_i)'$ = { $p\in X$ | $\forall r \in \mathbb{R}_{>0}$, $\exists q\in N_r(p), q \neq p, q\in \cup_{i=1}^{\infty}A_i$ }
= $(\cup_{i=1}^{\infty}A_i)'$ = { $p\in X$ | $\forall r \in \mathbb{R}_{>0}$, $\exists q\in N_r(p), q \neq p, q \in A_1$ or $q \in A_2$ or ...}
= $A_1^{'} \cup A_2^{'} \cup...$
= $\cup_{i=1}^{\infty}A_i^{'}$
I find my own proof somewhat suspicious. Please let me know what you all think about it. Thanks!
If $q \in \mathbb{Q}$, then $\{q\}' = \emptyset$.
But $\mathbb{R} = \mathbb{Q}' = (\bigcup_{q \in \mathbb{Q}} \{q\})' \neq \bigcup_{q \in \mathbb{Q}} \{q\}' = \emptyset$.
Use some fixed enumeration of $\mathbb{Q}$, if you want a proper sequence. Same difference.