I am self-learning Real Analysis from the text Understanding Analysis by Stephen Abbott. I would like to ask, if this constitutes a valid proof of the fact the closure of the union of two sets equals the union of the respective closures?
$$\overline{A \cup B} = \overline{A} \cup \overline{B}$$
Proof.
Suppose $\displaystyle x\in \overline{A\cup B} .$ Let $\displaystyle L$ be the set of limit points $\displaystyle A\cup B$. Then, either $\displaystyle x\in A\cup B$ or $\displaystyle x\in L$. If $\displaystyle x\in A\cup B$, then $\displaystyle x\in A$ or $\displaystyle x\in B$ or $\displaystyle x$ belongs to both. So, $\displaystyle x\in \overline{A} \ \cup \overline{B}$. If $\displaystyle x\in L$, then $\displaystyle x$ is a limit point of atleast one of the sets, $\displaystyle A$ or $\displaystyle B$. ($\star$)
Thus, $\displaystyle x\in \overline{A} \cup \overline{B}$.
Altogether, $\displaystyle \overline{A\cup B} \subseteq \overline{A} \cup \overline{B}$.
In the opposite direction, suppose $\displaystyle x\in \overline{A} \cup \overline{B}$. Then, $\displaystyle x\in \overline{A}$ or $\displaystyle x\in \overline{B}$ or $\displaystyle x$ belongs to both. Since $\displaystyle A\subseteq A\cup B$, $\displaystyle \overline{A} \subseteq \overline{A\cup B}$. Similarly, $\displaystyle \overline{B} \ \subseteq \overline{A\cup B}$. Consequently, $\displaystyle x\in \overline{A\cup B}$.
Edit:
$\star$ If $x$ is a limit point of $A \cup B$, then for all $\epsilon>0$, by definition, the open interval $(x-\epsilon,x+\epsilon)$ intersects $A \cup B$ in some point other than $x$. Consequently, $V_\epsilon(x)$ intersects atleast one of $A$, $B$ in some point other than $x$. Thus, $x$ is limit point of atleast one of $A$, $B$.
Best Answer
You don't have to use limit points at all: $\overline{A}$ and $\overline{B}$ are both closed and their union is closed and contains $A \cup B$. So $\overline{A \cup B} \subseteq \overline{A} \cup \overline{B}$ is immediate (the closure is the minimal closed superset).
The reverse follows from $A \subseteq A \cup B$ so $\overline{A} \subseteq \overline{A \cup B}$ and ditto for $B$ and taking these two inclusions together we get $\overline{A} \cup \overline{B} \subseteq \overline{A \cup B}$, and we have equality.
Your proof under $\star$ is wrong. To have $p \in L=(A \cup B)'$ means $$\forall \varepsilon>0: \exists q: (q \neq p) \land (q \in (A \cup B) \cap (p-\varepsilon, p+\varepsilon))$$ while $p \in A' \cup B'$ means
$$(\forall \varepsilon>0: \exists q: (q \neq p) \land (q \in (A \cap (p-\varepsilon, p+\varepsilon)))\\ \lor (\forall \varepsilon>0: \exists q: (q \neq p) \land (q \in (B \cap (p-\varepsilon, p+\varepsilon)))$$ so you first have to decide whether every pointed neighbourhood of $p$ is going to intersect $A$ all the time or $B$ all the time. You make it seem like you can just go from the former to the latter, while this is a logical fallacy.
To illustrate how a valid proof using limit points would go:
let $x \in \overline{A \cup B}$. Then either $x \in A \cup B$ so $x \in A$ or $x \in B$ and in either case it's clear that $x \in \overline{A} \cup \overline{B}$, so assume $x \notin A \cup B$ but $x \in (A \cup B)'$. Suppose that $x \notin \overline{A}$, so in particular $x \notin A$ and $x \notin A'$ so then we can find a $\varepsilon>0$ so that $(x-\varepsilon, x+\varepsilon) \cap A = \emptyset$. But if $\delta>0$ is arbitrary, let $\delta' =\min(\delta, \varepsilon)>0$ and note (as $x \in A \cup B)'$, the open set $(x-\delta', x+\delta')$ must intersect $A \cup B$ in some $q$. Then as $\delta' \le \varepsilon$, $q \notin A$ by the choice of $\varepsilon$ and so $(x-\delta, x+\delta)\supseteq (X-\delta', x+\delta')$ actually intersects $B$ and as $\delta$ was arbitrary, $x \in B'$ and so $x \in \overline{B}$ as required. So $x \in \overline{A \cup B}$ implies $x \in \overline{A} \cup \overline{B}$. The reverse inclusion is already done OK in the OP's post.