Prove that $\overline{A\cup B} = \overline{A}\cup\overline{B}$ and $\overline{A\cap B}\subseteq\overline{A}\cap\overline{B}$

My attempt: $x\in\overline{A\cup B}$ iff for every open set $U$ containing $x$, $U\cap\ (A\cup B)\neq\varnothing$. This happens iff $(U\cap A)\cup (U\cap B)\neq\varnothing$, and this happens iff $U\cap A\neq\varnothing$ or $U\cap B\neq\varnothing$, and this happens iff $x\in\overline{A}$ or $x\in\overline{B}$, and this is true iff $x\in\overline{A}\cup\overline{B}$. So $\overline{A\cup B} = \overline{A}\cup\overline{B}$

If $x\in A\cap B$, then $x\in A$ and $x\in B$, which implies $x\in\overline{A}$ and $x\in\overline{B}$ (since closure of $E$ is the smallest closed set containing $E$), so $x\in\overline{A}\cap\overline{B}$. This shows that $A\cap B\subseteq\overline{A}\cap\overline{B}$, and since $\overline{A}$ and $\overline{B}$ are both closed, $\overline{A}\cap\overline{B}$ is closed. But by definition, $\overline{A\cap B}$ is a subset of every closed set containing $A\cap B$, so $\overline{A\cap B}\subseteq\overline{A}\cap\overline{B}$.

Do my proofs look correct?

  • $\begingroup$ Actually it is $\overline{A\cup B} = \overline{A} \cap \overline{B}$ and $\overline{A}\cap\overline{B} \subseteq\overline{A\cap B}$ $\endgroup$ – Swapniel Feb 27 '14 at 6:32
  • $\begingroup$ what do you mean @swapniel99 $\endgroup$ – Ittay Weiss Feb 27 '14 at 6:35
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    $\begingroup$ the proofs are excellent. $\endgroup$ – Ittay Weiss Feb 27 '14 at 6:36
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    $\begingroup$ @swapniel99 you may be misinterpreting the overline to mean set complementation. It is meant to indicate the closure. $\endgroup$ – Ittay Weiss Feb 27 '14 at 6:50
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    $\begingroup$ The result about intersection has already been discussed here: math.stackexchange.com/questions/599460/… The approach given in an answer there uses monotonicity of closure (i.e, $X\subseteq Y$ $\implies$ $\overline X\subseteq \overline Y$). I'd say this approach is a little bit simpler. $\endgroup$ – Martin Sleziak Feb 27 '14 at 8:41

The reasoning is not quite correct. The statement

$x\in\overline{A\cup B}$

can be expressed as

for all open $U$ we have $(U\cap A)\ne\emptyset$ or $(U\cap B)\ne\emptyset$


$x\in\overline{A}$ or $x\in\overline{B}$


for all open $U$ is $U\cap A\ne\emptyset$, or for all open $U$ is $U\cap B\ne\emptyset$

The second statement clearly implies the first one, so $$x\in\overline A\vee x\in\overline B\implies x\in\overline{A\cup B}$$ In order to show the other direction, I recommend that you try and prove the contrapositive. So you start by assuming that

there is an open $U$ disjoint to $A$, and there is an open $V$ disjoint to $B$

and then seek to obtain from this the negation of the first statement. Here you will actually need some topology.

Ironically you used topology in the second proof where it can most efficiently be done the way you proved the $\overline{A}\cup\overline B\subseteq\overline{A\cup B}$ direction.


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