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.

Isn't this wrong? I think you are referring to De Morgan's law. — prog_SAHIL, Aug 07 '18 at 15:30
@prog_SAHIL I thought the same, but the proclaimed duplicate gives more context. I still think that the context should be included here. — Arnaud Mortier, Aug 07 '18 at 15:36
@ArnaudMortier, Yes. It just starts like 'prove this' without any proper context as to what we are dealing with. — prog_SAHIL, Aug 07 '18 at 18:22

score 1 · Accepted Answer · answered Aug 07 '18 at 15:47

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.

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 $x\in B'$.

answered Aug 07 '18 at 15:47

egreg

238,574

A beautiful proof indeed. The answer is similar to that of the included duplicate. – Subhasis Biswas Aug 07 '18 at 16:16
@SubhasisBiswas I didn't look at that; the important part of the answer was the analysis of your attempt. – egreg Aug 07 '18 at 16:20
My methodology was the following : let us consider the following infinite intersections of neighbourhoods and either of the set, $[N'{\epsilon_1}(x) \cap N'{\epsilon_2}(x) \cap ...N'_{\epsilon_n}(x) \cap A]=A_k$ where $ n \rightarrow \infty$. Now, everything reduces to $A_k \cup B_k \neq \emptyset$. Due to the consideration of infinite intersection, either one of the intersection, say $A_k$, has to be zero (in some cases, both are non-zero) , hence the point $x$ must be in $B_k$, i.e, $x$ is in $B'$. Where did I go wrong? – Subhasis Biswas Aug 07 '18 at 16:27
@SubhasisBiswas It's the same error: you need to show that either every $N'\epsilon(x)$ intersects $A$ or every $N'\epsilon(x)$ intersects $B$. It's not sufficient to show that each $N'_\epsilon(x)$ intersects either $A$ or $B$. – egreg Aug 07 '18 at 16:31
are you implying that there is a possibility of "periodic exclusion and inclusion" of $N'_\epsilon(x)$ with $A$ and $B$ individually? – Subhasis Biswas Aug 07 '18 at 16:34
@SubhasisBiswas Precisely. It doesn't happen, but your argument doesn't show it. – egreg Aug 07 '18 at 16:36
Let us continue this discussion in chat. – Subhasis Biswas Aug 07 '18 at 16:37

Proving $A' \cup B'=(A \cup B)'$

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