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Boundary and closure

  1. Dec 1, 2013 #1
    My first analysis/topology text defined the boundary of a set S as the set of all points whose neighborhoods had some point in the set S and some point outside the set S. It also defined the closure of a set S the union of S and its boundary.

    Using this, we can prove that the closure of S is the smallest closed set containing S. We can also prove that the boundary of S is the intersection of the closure of S and the closure of the complement of S.

    I was wondering, if we define the closure of S to be the smallest closed set containing S, and the boundary of S to be the intersection of the closure of S and the closure of the complement of S, will we have the machinery necessary to work backwards and prove the first two definitions, i.e. are the two definitions of boundary/closure equivalent?

    BiP
     
  2. jcsd
  3. Dec 1, 2013 #2
    Yes, they are! And your proof will show it!

    Given a subset [itex]S[/itex] of your space [itex]X[/itex]...
    - Let [itex]\partial S=\{x\in X: \enspace \text{ for any neighborhood } N \text{ of } x, \enspace N\cap S\neq\emptyset \text{ and } N\cap (X\setminus S)\neq\emptyset \}[/itex].
    - Let [itex]\bar S = S \cup \partial S[/itex].
    - Let [itex]\tilde S[/itex] be the smallest closed set containing [itex]S[/itex]. (You have to prove that there is such a set. But this is easy; it's just the intersection of every closed set containing [itex]S[/itex].)

    Notice that I haven't used the words "closure" or "boundary" anywhere above.

    It sounds like you know how to prove that [itex]\bar S = \tilde S[/itex] and [itex]\partial S = \bar S \cap \overline{X\setminus S}[/itex].

    Having shown that, you can define:
    - The closure of [itex]S[/itex] (denoted [itex]cl(S)[/itex]) is either [itex]\bar S[/itex] or [itex]\tilde S[/itex], whichever definition you like. [We now know they're equivalent.]
    - The boundary of [itex]S[/itex] is either [itex]\partial S[/itex] or [itex]cl(S)\cap cl(X\setminus S)[/itex], whichever definition you like. [We now know they're equivalent.]
     
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