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I'm stuck on trying to get this proof started.

I want to prove that $\delta(S_1 \cup S_2)\subset \delta S_1\cup\delta S_2$, where $S$ is some set.

I don't need a full proof, just a hint to get started.

Here $\delta$ denotes the boundary of a set.

After considering some responses, this is what I have.

I will prove that $\delta(S_1 \cup S_2)\subset \delta S_1\cup\delta S_2$. Let $x_0$ be an interior point of $\delta(S_1 \cup S_2)$. By definition, we have that there exists some $\epsilon$-neighborhood of $x_0$ such that $ V = (x_0-\epsilon,x_0+\epsilon)\subset \delta(S_1 \cup S_2)$. Since $x_0$ is an interior point of $\delta(S_1 \cup S_2)$, that implies that the $\epsilon$-neighborhood $V$ contains some point $y\in S_1$ or $y\in S_2$ and $V$ contains some point $z\in(S_1\cup S_2)^c$, by definition of a boundary point. If $y\in S_1$, then that implies that $x_0\in \delta S_1$. Similarly, if $y\in S_2$, then $x_0\in \delta S_2$. Therefore, we have that $V$ is contained in $\delta S_1\cup \delta S_2$, which gives that $\delta(S_1\cup S_2)\subset \delta S_1\cup \delta S_2$, as desired.

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  • $\begingroup$ What is $\delta$? $\endgroup$
    – Clement C.
    Commented Sep 24, 2014 at 0:34
  • $\begingroup$ Boundary of $S$. $\endgroup$
    – Pubbie
    Commented Sep 24, 2014 at 0:37
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    $\begingroup$ Regarding your edit, you need to show that an arbitrary point in $\delta(S_1 \cup S_2)$ is contained in $\delta(S_1)\cup\delta(S_2)$. An arbitrary point of $\delta(S_1 \cup S_2)$ need not be an interior point. Indeed, a boundary need not contain any interior points: consider, for example, the boundary of the interval $(0,1) \subseteq \mathbb R$, which consists of just the two points $0$ and $1$. $\endgroup$
    – user169852
    Commented Sep 24, 2014 at 1:04
  • $\begingroup$ Ah. Then I'm not exactly sure how to go about it without assuming it is an interior point of $\delta (S_1\cup S_2)$. $\endgroup$
    – Pubbie
    Commented Sep 24, 2014 at 1:09

1 Answer 1

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It suffices to show that if a point is on the boundary of $S_1\cup S_2$, then it is on the boundary of $S_1$ or $S_2$.

It might be simpler to tackle to contrapositive, namely: if a point is not on the boundary of $S_1$ or $S_2$, then it is not on the boundary of $S_1\cup S_2$. If a point is neither on the boundary of $S_1$ nor $S_2$, then either it must be in the interior of one of the sets, or in the exterior of both. Now you just need to show that in each of these cases, some neighbourhood of said point is either completely contained in $S_1\cup S_2$, or completely disjoint from it.

Hint: if a point is in the interior of $S_1$, its also in the interior of $S_1\cup S_2$. If you alternatively have two neighbourhoods of a point completely disjoint from each $S_1,S_2$ respectively, then their intersection will be a neighbourhood also, and disjoint from the union $S_1\cup S_2$.

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