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Boundary point

  1. May 18, 2010 #1
    Hello all,

    Suppose [tex]C\subseteq \mathbb{R}^{n}[/tex], if [tex]x \in \text{bd}\;C[/tex] where [tex]\text{bd}[/tex] denotes the boundary, a sequence [tex]\{x_{k}\}[/tex] can be found such that [tex]x_{k} \notin \text{cl}\;C[/tex] and [tex]\lim_{k\rightarrow \infty}x_{k} = x[/tex].

    The existence of such sequence is guaranteed by the definition of boundary point, of which a neighborhood contains at least one point of [tex]C[/tex] and at least one point of [tex]\mathbb{R}^{n}\C[/tex]. (Explicit structure of the sequence is no longer our task...if you want, find it yourself...?)

    Is my explanation right? Thanks.

    Last edited: May 18, 2010
  2. jcsd
  3. May 18, 2010 #2


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    If you mean by boundary that the point that can not be separated from R^n or the set C by open sets then this seems to work.

    What about saying that every open neighborhood of the point has non-empty intersection with the set C and with the complement of its closure? Does that say the same thing?
  4. May 24, 2010 #3
    As a general comment, that result is true for all 1st-countable spaces ( it is actually
    an iff) result, in that continuity is equivalent to sequential continuity.
  5. May 24, 2010 #4


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    Nope, it is not iff. The result that continuity is equivalent to seq. continuity even holds in a Sequential space. Every 1st countable space is a seq. space, but not conversely (wikipedia gives the example "take the real line R and identify the set Z of integers to a point.")

    See also this interesting article, which has the useful diagram

    http://img37.imageshack.us/img37/2064/sequentialspace.png [Broken]
    Last edited by a moderator: May 4, 2017
  6. May 24, 2010 #5
    Well, we may be talking about slightly different things (I may have misread --and misunderestimated : -- the statement of the problem) ). What I mentioned was the sequence lemma:

    For X a topological space and if A<X , then x is in Cl(A) iff there is a sequence
    of points in A that converge to x.

    In this case, this is an iff result with 1st countability.
  7. May 24, 2010 #6


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    Yes, I misunderstood you. I thought you claimed that "continuity is equivalent to sequential continuity" is equivalent to "the space being 1st countable".

    You are right that that result about closure and sequences is an iff result in a 1st countable space. We can do even better: this is an iff result in a sequential space (so my post was not completely off topic :) )
  8. May 25, 2010 #7


    Let C := R^n \ {0} and let x := 0.

    Then x is in the boundary of C, but no sequence satisfying your criteria exists.
  9. May 26, 2010 #8

    Thanks all comments so far.

    So shall I include a restriction that [tex]\text{cl} C \neq \mathbb{R}[/tex]?
  10. May 26, 2010 #9
    No, it still wouldn't work: e.g., let C = { x in R^n : |x| < 1 } \ {0}, the unit disk with the origin removed, and the same thing applies.

    Just replace the requirement "[tex]x_k \notin \overline{C}[/tex]" with "[tex]x_k \notin C[/tex]" (overline denoting closure).
  11. May 26, 2010 #10


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    I think you mean [tex]x_k\in C[/tex] instead of [tex]x_k\notin C[/tex]
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