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Clarification: proof that perfect subsets of R^k uncountable

  1. Feb 12, 2016 #1
    From Baby Rudin
    "Thm: Let P be a non-empty, perfect subset of R^k. Then P is uncountable.

    Pf: Since P has limit points, P must be infinite. Suppose P is countable, list the point of P {x1 ...xn }. Construct a sequence of nbhds. as follows. Let V1 be any nbhd of x1 . Suppose Vn has been constructed so that Vn ∩ P is non-empty. Since every point of P is a limit point of P, there is a nbhd Vn+1 such that it's closure is in Vn , xn is NOT in the closure of Vn+1, and Vn+1 ∩ P is non-empty. Let Kn=closure(Vn)∩P, then Kn is closed, bndd., therefore compact. The Kn are clearly nested compact sets, thus their arbitrary intersection is non-empty. But xn∉Kn+1 implies that ∩nKn is empty, thus a contradiction. "

    My confusion is in the claim of the contradiction. How does xn∉Kn+1 imply the intersection is empty? My thoughts are that, since we assumed P to be countable, xn must be an element of any inductive subset of P indexed by n+1. Is my logic behind this statement true? It just seemed a little artificial to me as I have not seen a proof which uses induction to arrive at a contradiction in this manner before.
     
  2. jcsd
  3. Feb 12, 2016 #2

    fresh_42

    Staff: Mentor

    What do you mean by this? ##X = \{x\}## has limit points, ##\{x\}##.
    What does it mean for P to be perfect? Beside there is a point in P and P has nice topological separation properties that is all you have.
     
  4. Feb 12, 2016 #3

    Krylov

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    This is not true.
    A set is perfect when it is closed and each of its point is a limit point.
     
  5. Feb 12, 2016 #4

    fresh_42

    Staff: Mentor

    I've read closed and no isolated points.
    Why is a constant sequence not allowed? Likely I've mistranslated limit point by limit. It's kind of a standard pit.
     
  6. Feb 12, 2016 #5

    Krylov

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    Because ##x## is a limit point of a set ##A## in a metric space if, by definition, every neighborhood of ##x## meets ##A## in a point other than ##x##.

    All limit points of ##A## are in ##\overline{A}##, but not all points in ##\overline{A}## are necessarily limit points of ##A##.
     
    Last edited: Feb 12, 2016
  7. Feb 12, 2016 #6
    If you denote ##K=\cap_{n\ge 1} K_n##, then the condition ##x_1\notin K_2## implies that ##x_1\notin K##. Similarly, condition ##x_2\notin K_3## implies that ##x_2\notin K##, and, more generally, for every ##n\ge 1## the condition ##x_n\notin K_{n+1}## implies that ##x_n\notin K##. But this means that ##P\cap K=\varnothing## (because ##P=\cup_{n\ge 1} \{x_n\}##). On the other hand, by construction ##K\subset P## (because ##K_n\subset P## for all ##n\ge 1##).

    So ##K=\varnothing##.
     
  8. Feb 12, 2016 #7
    Thanks, that's much clearer now! :)
     
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