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Logical Point in Topological Problem

  1. May 4, 2017 #1
    1. The problem statement, all variables and given/known data
    Consider ##\mathbb{R}^\omega## in the uniform topology. Show that ##x## and ##y## lie in the same component if and only if ##x-y = (x_1-y_1,x_2-y_2,...)## is a bounded sequence.

    2. Relevant equations

    The uniform topology is induced by the metric ##p(x,y) := \sup_{i \in \mathbb{N}} d(x_i,y_i)##, where ##d(x_i,y_i) = \min \{|x_i-x_j|,1\}##.

    3. The attempt at a solution

    My first question is, by bounded sequence does the author mean a bounded sequence in ##\mathbb{R}## and with respect to the metric on ##\mathbb{R}##, which would just be the absolute value function?

    My next question pertains to the hint. When the author say it suffices to show such-and-such, do the mean that the special case

    ##x \sim 0## if and only if ##x-0## is a bounded sequence

    implies the more general case

    ##x \sim y## if and only if ##x-y## is bounded;

    and therefore I need to prove the implication to show that it is an actual reduction, right?
     
  2. jcsd
  3. May 4, 2017 #2

    Dick

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    Yes, the usual metric in ##\mathbb{R}##.

    You didn't quote the hint part, but it's pretty easy to show those two are equivalent, isn't it? The metric is translation invariant.
     
    Last edited: May 4, 2017
  4. May 5, 2017 #3
    Okay. Perhaps you could critique this attempt at proving they are equivalent.

    Let (1) refer to the special case, and (2) the more general case. First, note that ##f_y : \mathbb{R}^\omega \to \mathbb{R}^\omega## defined by ##f_y(x) = x+y## is a homeomorphism, whose inverse is ##f^{-1}_y = f_{-y}##. First we will prove that (1) implies (2). Suppose that the (1) is true. Then ##x-y = (x-y) - 0## is, according to (1), bounded if and only if ##(x-y) \sim 0##. This happens if and only if ##x-y## and ##0## are in the same connected subspace ##C##. Since ##f_y## is continuous, the image ##f_y(C)## will also be connected in ##\mathbb{R}^\omega##, and by definition ##f_y(x-y) = x## and ##f_y(0) = y## are in this connected subspace, i.e., ##x \sim y##. And clearly if ##x## and ##y## are in the same connected subspace, we we can use the inverse function ##f_{-y}## to show ##x-y## and ##0## are in the same connected subspace Then $x \sim y$ if and only if x-y = (x-y) - 0 is bounded. But according to (1), this happens if and only if (x-y) \sim 0

    Obviously proving (2) implies (1) starts almost the same way that the above proof ends, so we are finished.
     
  5. May 5, 2017 #4
    Also, I am having trouble proving the claim "##x## and ##0## lie in the same of ##\mathbb{R}^\omega## if and only if ##x-0=x## is bounded. I could use a hint on how to prove ##\implies## (forward) direction.
     
  6. May 5, 2017 #5

    Dick

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    Here's a notion to start from, let ##B## be the set of all bounded sequences. Can you show ##B## and ##B^c## are both open? Hint: consider open balls of radius 1 around points in each.
     
  7. May 10, 2017 #6
    I am not sure how this would help. The only conclusion I could draw from ##B## being clopen is that ##\mathbb{R}^\omega## is not a connected space.
     
  8. May 10, 2017 #7

    Dick

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    No, not the only conclusion. It also tells you ##B## is a union of connected components. Your next job would be to show it consists of a single component (i.e. ##B## is connected).
     
  9. May 15, 2017 #8
    Oh. Does your solution rely on clopen sets being the union of connected components? If so, I don't yet have this fact.
     
  10. May 15, 2017 #9

    Dick

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    It's sort of obvious, isn't it? The whole set is a union of its connected components - connected components are clopen. If you have doubts about this you might want to prove any missing parts as a exercise. Can it contain only part of a component?
     
    Last edited: May 16, 2017
  11. May 16, 2017 #10
    I am pretty certain this is false. Consider the rationals endowed with the standard (order) topology. Its components are singletons, which are closed but not open.
     
  12. May 16, 2017 #11

    Dick

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    Good catch. Sure. But in this case the components are open as well. Can you show that?
     
  13. May 17, 2017 #12
    Everything I have read suggests that this is wrong, that connected components are not necessarily clopen. In the case of ##\Bbb{Q}##, singletons are not open. Also, see JHance's comment here https://math.stackexchange.com/ques...e-components-of-the-set-of-irrational-numbers which has gone uncorrected for two years, suggesting, again, that connected components are not generally clopen.
     
  14. May 17, 2017 #13

    Dick

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    I acknowledged that that is correct. My point is that IN THIS CASE the components ARE OPEN. Can you prove it?
     
  15. May 17, 2017 #14
    Suppose that ##\{x\} \subset \Bbb{Q}## is open. Since ##x \in \{x\}##, there exists a ##\epsilon > 0## such that ##(x-\epsilon,x + \epsilon) \subseteq \{x\}##, which cannot happen since the ##\epsilon##-nbhd captures rationals outside of ##\{x\}##.

    Moreover, if the components, which we know are singletons, were indeed open, then the order topology on ##\Bbb{Q}## would be identical to the discrete topology, and therefore every set is clopen. This would mean, if I am not mistaken, that ##S = \{x \in \Bbb{Q} ~|~ x^2 < 2\}## is closed, which is clearly absurd since it has a limit point of ##\sqrt{2}##.
     
  16. May 17, 2017 #15

    Dick

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    No, no, no. I mean can you prove that components are open in ##\mathbb{R}^\omega##?
     
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