Simple analysis question

  • Thread starter ayas
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Hello,
I know how to start but I don't know how to end that proof. It's supposed to be easy:
Let S a subset of Rn.
PROVE THAT the boundary of S is a closed set.
(I'll use d for delta, so dS is my convention for "the boundary of S").

So here I go:
dS is closed iff it contains all of its boundary points,
so dS is closed iff d(dS) is included in dS.
Let x be any point such that x belongs to d(dS).
So for any Ball B(r, x), r>0, (ie centered at x),
| B intersection dS is not empty
| and B interesection (dS)complement is not empty.
(the second line is equivalent to) B interesection (interiorOfS union exteriorOfS) is not empty

Now what's next??
Thanks for your suggestions. If you do have a suggestion, please don't skip a step or don't bypass a detail because it seems obvious (trust me, nothing is obvious to the one who doesn't know yet!)
 

Answers and Replies

  • #2
morphism
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Have you tried showing that the complement of dS is open?
 
  • #3
mathwonk
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the boundary is the complement of the interior which is easily shown to be open. qed. oops. my bad. that was just said.
 
  • #4
HallsofIvy
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There are several different ways of characterizing closed sets. One is "complement of an open set"- that's the one morphism and mathwonk are suggesting. Another is "a set that contains all its boundary points"- that's the one ayas is using.

ayas, have you considered an indirect proof? If bd(S) is not closed then there must be some member, x, of bd(bd(S)) that is NOT in bd(S). Since x is in bd(bd(S)) every neighborhood of x contains some member, y, of bd(S). Now there is a neighborhood of y that is a subset of the original neighborhood . . .
 

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