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I know this is a duplicate but the other two haven't helped me much.

Fist attempt: Tried proving through double inclusion, but wasn't sure of how to convey being an element of one implied being an element of the other in either direction, although I suspect from left to right would be the easier of the two.

Second attempt: Tried proving equality directly, using the fact that the boundary of a set is equal to the set difference of its closure and interior, but struggled proving closure of union is a union of closures, or the interior of a union is a union of interiors.

I'm getting pretty frustrated and any help would be greatly appreciated!

Asaf Karagila
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Julian
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2 Answers2

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It is false.

Consier $\Bbb R$ with usual topology, $A=[0,2]$, $B=[1,3]$.

$$\partial(A\cup B)=\{0,3\}$$ $$\partial A\cup\partial B=\{0,1,2,3\}$$

Just think that some of the border of $A$ can be in the interior of $B$.

ajotatxe
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    There can also be points in the boundaries of both sets which are not in the boundary of the union. Take $A=[0,1]$ and $B=[1,2]$. Then $1\in\partial A$ and $1\in\partial B$, but $1\notin\partial(A\cup B)$. This typically happens when you "glue" regions together along their boundaries. – MPW Nov 05 '14 at 21:20
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Your intuition is fine with respect to the left to right inclusion: $$ \partial(A\cup B)\subseteq \partial(A)\cup \partial(B) .$$ See a complete formal proof in this MSE link.

If $\overline{A} \cap B = \emptyset = A \cap \overline{B}$, then the other inclusion holds too. See a proof in this other MSE link.

Community
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ir7
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