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X in dA => x isolated or strict limit (Topological space)

  1. Mar 22, 2008 #1
    I am trying to show the following proposition, but I can only do so in the special case where the basis for the topology is countable. I posted this in the calc & analysis forum and didn't get any responses - I think that this is a more appropriate forum judging by the other posts.

    Proposition. If x is in [tex]\partial A[/tex], then x is either an isolated point, or a strict limit point of both [tex]A[/tex] and [tex]A ^{c}[/tex]. ([tex]A[/tex] is a set in a topological space, [tex]\partial A[/tex] is the boundary)
    ---

    So far, my reasoning is this. Suppose x is not isolated - then lets show that it is a strict limit point of [tex]A[/tex] by constructing a sequence [tex]\left\{ x_{n}\right\}[/tex] in [tex]A[/tex] that converges to x.

    If the number of open sets containing x were countable, [tex]\left\{ E_{1}, E_{2}, E_{3}, ... \right\}[/tex], then I would put [tex]x_{1} \in \left(A \cap E_{1}\right) \backslash\left\{x\right\}[/tex], [tex]x_{2} \in \left(A \cap E_{1}\cap E_{2}\right) \backslash\left\{x\right\}[/tex], [tex]x_{3} \in \left(A \cap E_{1}\cap E_{2}\cap E_{3}\right)\backslash\left\{x\right\}[/tex], and so forth.

    We know these intersections can't be empty - if [tex]E_{i}[/tex] and [tex]E_{j}[/tex] are open and contain x, then [tex]E_{i}\cap E_{j}[/tex] is also an open set containing x. Since x is not isolated, [tex]E_{i}\cap E_{j}[/tex] must contain some [tex]y \in A[/tex] distinct from x. Thus for a countable number of open sets containing x, we can construct the proper convergent sequence to prove the result.

    However, this logic will not work if there are an uncountable number of open sets. I could extend my argument if only the basis is countable, but not to a topology without a countable basis.
     
    Last edited: Mar 22, 2008
  2. jcsd
  3. Mar 22, 2008 #2

    morphism

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    Why are you restricting yourself to sequences? There is nothing in the definition of a limit point requiring that it be a sequential limit. In fact, a point x is a limit point of A iff you can construct a net in A\{x} that converges to x.
     
  4. Mar 22, 2008 #3
    I'm not really sure what a net is, though from a quick googling it seems to be a generalization of a sequence, so I will look into it.

    However, the book I'm working from (which could be wrong...) defines a strict limit as follows (page 9),

    Definition. A point [tex]x \in \partial A[/tex] is a strict limit point of [tex]A \subset X[/tex] if there is a sequence [tex]\left\{x_{n}\right\}\subset A \backslash \left\{x\right\}[/tex] such that [tex]lim_{n\rightarrow \infty}x_{n}=x[/tex].

    ---

    If the book is wrong and you need a net to prove the proposition, I would be interested in finding a counterexample for it as currently written.
     
  5. Mar 22, 2008 #4
    Ok, I think I might have a counterexample to it as written, but I'm not totally convinced and need to mull it over a bit. Let me run it by yall.

    Let Y be a uncountable set and let [tex]\Re[/tex] be the the real line. Construct a basis for the topology of the combined sets as follows: 1) All singleton points in Y are in the basis. 2) The standard metric basis for [tex]\Re[/tex] is in the basis.

    Basically we are joining a set with the discrete topology together with the real line. I think this works. Edit: no it doesnt

    Now, if [tex]J[/tex] is the interval [tex]\left( -\infty,1\right)[/tex], then [tex]\overline{J}[/tex] is still [tex]\left( -\infty,1\right][/tex] (?), and [tex]\overline{J_{c}}[/tex] is [tex]Y \cup \left[1,\infty\right)[/tex]. Thus [tex]\partial J[/tex] is [tex]\left\{1\right\}[/tex], which is not an isolated point.

    Consider the uncountable number of open sets [tex]\left\{1, y\right\}[/tex] Edit: These actually aren't open )-: for all points [tex]y \in Y[/tex]. These all intersect each other only at the single point 1, so there is no way you could construct a sequence that would get them all.

    (Edit: changed from [0,1) to all reals.)
    Edit2: Wait actually there is a problem with this counterexample, it doesn't work.
     
    Last edited: Mar 22, 2008
  6. Mar 22, 2008 #5

    morphism

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    How about [itex][0, \omega_1][/itex], where [itex]\omega_1[/itex] is the first uncountable ordinal? [itex]\omega_1[/itex] is in the boundary of of [itex][0, \omega_1)[/itex]. However, it's not an isolated point (it's a limit ordinal), and it's not the limit of any sequence coming out of [itex][0, \omega_1)[/itex] (easy to prove).
     
  7. Mar 23, 2008 #6
    I'm not sure you mean by [tex][0, \omega_1)[/tex]. Could you elaborate?
     
    Last edited: Mar 23, 2008
  8. Mar 23, 2008 #7

    morphism

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    Are you familiar with order topologies? [itex][0, \omega_1)[/itex] denotes the set of all ordinals strictly less than [itex]\omega_1[/itex].
     
  9. Mar 23, 2008 #8
    I am new to topology and have not read about that yet. I'll look into it, thanks.

    So far my search for counterexamples has been mostly with variations on discrete topologies, I think I am close to coming up with an example of my own that works, so I'm going to keep thinking about my own ideas for a bit.
     
    Last edited: Mar 23, 2008
  10. Mar 23, 2008 #9

    morphism

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    I think I managed to come up with another example that doesn't involve ordinals or anything exotic. I can give you a hint if you want.
     
  11. Mar 23, 2008 #10
    Thanks, but not right now. Maybe in a day or so if my current strategy doesn't pan out.
     
  12. Mar 26, 2008 #11
    Heh, my counterexample has somehow managed to morph into something basically equivalent to [0,w1). I'd be interested in hearing a hint about the other example you mention.
     
  13. Mar 26, 2008 #12

    morphism

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    Try the cocountable topology on R, i.e. the topology whose proper closed sets are precisely the countable subsets of R.
     
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