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I have posted a counterexample given in a solution below, but when I attempted the problem I did something different.

My Attempt

Take complements of both sides

$$ \int(A\cup B)\stackrel{?}{=}\int(A)\cup \int(B) \\ \big(\int(A\cup B)\big)^c\stackrel{?}{=}\big(\int(A)\cup \int(B)\big)^c \\ \overline{ (A\cup B)^c}\stackrel{?}{=}\overline{(A)^c}\cap \overline{(B)^c} $$ I got this step from this answer Prove that the closure of complement, is the complement of the interior $$\overline{ (A)^c\cap (B)^c}\stackrel{?}{=}\overline{(A)^c}\cap \overline{(B)^c}$$

My question

Could I prove from where I stopped that the two sides are not equal? I tried to sketch a ven diagram but both sides seem to give the same intersected set.


Counterexample

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  • $\begingroup$ From your example, you can get a counterexample to $\overline{A\cap B}=\overline A\cap\overline B$ too. $\endgroup$ – Lord Shark the Unknown Jul 28 '18 at 11:48
  • $\begingroup$ Let $A=(0,1)$ and $B=(1,2)$ thus $\overline{A\cap B}=\emptyset$ and $\overline A\cap\overline B=1$ $\endgroup$ – john fowles Jul 28 '18 at 12:01
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Proposition. If the boundaries of A and B are disjoint, then
int A$\cup$B = int A $\cup$ int B.

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The correct answer to, "does $\operatorname{int}(A\cup B)$ equal $\operatorname{int}(A)\cup\operatorname{int}(B)$?" is:

Perhaps yes, perhaps no. That depends on which sets $A$ and $B$ are.

So you have no hope of starting out from nothing and then prove that $\operatorname{int}(A\cup B) \ne \operatorname{int}(A)\cup\operatorname{int}(B)$. That conclusion would be as wrong as it is to claim $\operatorname{int}(A\cup B) = \operatorname{int}(A)\cup\operatorname{int}(B)$.

For a claim (with parameters) that is sometimes true and sometimes false, the best you can do is to prove that it is not always true (and subsequently that it is not always false). You can't hope to do that by doing general manipulations starting from nothing, because that way -- if you do them correctly -- the only thing you will ever reach are things that always have such-and-such truth value.

Proving something not always true generally takes a concrete counterexample.

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