Proving $A’ \cup B’=(A \cup B)’$

Question

Duplicate to: This

My approach: Let $x \in A' \cup B' \implies x \in A'$ or $x \in B'$. Hence, in either case, $(N'_\epsilon (x) \cap A) \cup (N'_\epsilon (x) \cap B)= N'_\epsilon (x) \cap (A \cup B) \neq \emptyset\ , \forall \epsilon>0.$

Therefore, $A' \cup B' \subset (A \cup B)' $

Again, $x \in (A \cup B)' \implies N'_\epsilon(x) \cap (A\cup B) \neq \emptyset \implies$ either $N'_\epsilon(x) \cap A \neq \emptyset $ or $N'_\epsilon(x) \cap B \neq \emptyset $. Therefore, $x \in A' \cup B' $

Hence, we are done.

Kindly verify this.

Answer 1

The proof is wrong: you just conclude that either $N'_\epsilon(x)\cap A\ne\emptyset$ or $N'_\epsilon(x)\cap B\ne\emptyset$, but this doesn't show that either $x\in A'$ or $x\in B'$.

Indeed, showing that $x\in A'\cup B'$ requires proving that,

(for every $\epsilon>0$, $N'_\epsilon(x)\cap A\ne\emptyset$) or (for every $\epsilon>0$, $N'_\epsilon(x)\cap B\ne\emptyset$)

which is different from what you proved.

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Suppose $x\in (A\cup B)'$, but $x\notin A'$. Let's show that necessarily $x\in B'$. Fix $\epsilon_0>0$ such that $N'_{\epsilon_0}(x)\cap A=\emptyset$.

Hence, for every $0<\delta<\epsilon_0$, $N'_\delta(x)\cap A=\emptyset$. From the assumption $x\in(A\cup B)'$, we deduce that $N'_\delta(x)\cap(A\cup B)\ne\emptyset$, so $N'_\delta(x)\cap B\ne\emptyset$.

Let $\epsilon>0$; take $\delta<\min\{\epsilon,\epsilon_0\}$. Then $$ \emptyset\ne N'_\delta(x)\cap B\subseteq N'_\epsilon(x)\cap B $$ This proves $x\in B'$.