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Proving a Certain Set is Closed in the Uniform Topology

  1. Jan 12, 2017 #1
    1. The problem statement, all variables and given/known data
    Let ##Q = \{(x_1,x_2,...,) \in \mathbb{R}^\omega ~|~ \lim x_n = 0 \}##. I would like to show this set is closed in the uniform topology, which is generated by the metric ##\rho(x,y) = \sup d(x_i,y_i)##, where ##d## is the standard bounded metric on ##\mathbb{R}##.


    2. Relevant equations


    3. The attempt at a solution

    Let ##x = (x_n) \in \overline{Q} - Q##. Then ##\lim x_n = \ell \neq 0##, and WLOG take ##\ell > 0##. In my attempt, I mistakenly thought that ##\prod (x_n - \ell, x_n + \ell)## was a basis element for the product topology containing ##x##, concluding from this that it must be open in the uniform topology. From there I proved that if ##z = (z_n) \in \prod (x_n -\ell, x_n + \ell)##, then it cannot converge to ##0##, which gives the desired contradiction.

    My question is, is ##\prod (x_n -\ell, x_n + \ell)## in fact open in the uniform topology?
     
  2. jcsd
  3. Jan 13, 2017 #2

    haruspex

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    Is ##\overline{Q} ## only convergent sequences?
     
  4. Jan 14, 2017 #3
    Yes. ##\overline{Q}## a subset of ##\mathbb{R}^\omega## which consists of all convergent real sequences.
     
  5. Jan 14, 2017 #4

    haruspex

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    But there is nothing in the problem statement that allows you to restrict attention that subset.
     
  6. Jan 14, 2017 #5
    I don't quite follow...

    I have a topological space ##\mathbb{R}^\omega##, which I defined in a previous post; the space is endowed with the uniform topology, defined in the original post. Now, I have a subset ##Q## which I am trying to show is closed in this topology. Yes, you are right that I am restricting my attention to this particular subset, but I am doing so purposefully.
     
  7. Jan 14, 2017 #6

    haruspex

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    My problem is that I so far see no connection between the thing to be proved and neither the ##\overline Q## subset nor the product topology you mention.
    At the same time, I see no reference made to the metric nor to any usual definition of closure. If you are appealing to some standard results, please quote them.

    I understand that (xn-l) is a sequence, but I have no idea what you mean by ##\prod (x_n - \ell, x_n + \ell)##. ## (x_n - \ell, x_n + \ell) ## is a pair of numbers.## ((x_n - \ell), (x_n + \ell)) ## would be a pair of sequences. For a product topology we need pairs of sets.

    Edit: I believe I have a fairly short proof from first principles.
     
    Last edited: Jan 14, 2017
  8. Jan 16, 2017 #7
    Points in ##\mathbb{R}^\omega## can be regarded as points in a space, or they can be thought of as real convergent sequences in ##\mathbb{R}##. When I write ##(x_n)## or ##(x_n-\ell)##, I am referring to either one of these interpretations. When I write ##\prod (x_n - \ell, x_n + \ell)##, on the other hand, I am writing a subset of the product space ##\mathbb{R}^\omega##. This is in fact a basis element in the box topology. What I am trying to show is that it is open in the uniform topology, but I have been unsuccessful.

    Every approach I have tried requires that I construct a basis element that containing no sequences that converge to zero. You say that you have a solution. Would you mind offering a hint?
     
  9. Jan 16, 2017 #8

    haruspex

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    The definition of closure in a metric space with which I am familiar is that the set contains all its limit points. So I took a sequence (of sequences) which, according to the given metric, converges, and showed ithat the sequence it converges to converges to zero.
     
  10. Jan 17, 2017 #9
    Unfortunately, I do not have that characterization of closure in a metric space. I only have the definition, which states that the closure of ##A## is the intersection of all closed sets containing ##A##, and the general characterization that ##x \in \overline{A}## if and only if every open neighborhood of ##x## intersects ##A##.
     
  11. Jan 17, 2017 #10

    haruspex

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    Ok, but that only defines closure in terms of other closed sets, so cannot be fundamental.
    See bullet 2 at
    http://mathworld.wolfram.com/ClosedSet.html

    Ok, assuming the overline here means closure of, that matches the definition at bullet 4 of
    http://mathworld.wolfram.com/SetClosure.html
    In a metric space, if every open neighbourhood of x intersects set S then it follows that there exists an infinite sequence in S that converges to x. Thus x is a limit point of S. Since a closed set is one that is its own closure, the definition then says that a closed set contains all its limit points.
     
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