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Uniform Metric vs Box Topology

  1. May 11, 2005 #1
    Let [itex]\rho[/itex] be the uniform metric on [itex]\mathbb{R}^\omega[/itex]
    For reference, for two points:
    [tex]x = (x_i)[/tex] and [tex]y = (y_i)[/tex] in [tex]\mathbb{R}^\omega[/tex]

    [tex]\rho(x,y) = \sup_i\{ \min\{|x_i - y_i|, 1\}\}[/tex]

    Now, define:

    [tex]U(x,\epsilon) = \prod_i{(x_i - \epsilon, x_i + \epsilon)} \subset \mathbb{R}^\omega[/tex]

    I need to show 2 things:
    1) For [itex]0 < \epsilon < 1[/itex], the ball of radius [itex]\epsilon[/itex] about [itex]x[/itex] in the uniform metric, [itex]B_\rho(x,\epsilon) \neq U(x,\epsilon)[/itex]

    2)[tex]B_\rho(x,\epsilon) = \bigcup_{\delta < \epsilon}{U(x,\delta)}[/tex]

    Now, first off, I can't understand how 1 can be true. It makes no sense to me. Am I misunderstanding the metric? To further obfuscate the issue, 2 seems to contradict 1. Afterall, doesn't:

    [tex]\bigcup_{\delta < \epsilon}{\prod_i{(x_i-\delta, x_i+\delta)}}} = \prod_i{\bigcup_{\delta<\epsilon}{(x_i-\delta,x_i+\delta)}}}[/tex]

    The union inside the product of the latter being [itex](x_i-\epsilon, x_i+\epsilon)[/itex]? Thus, isn't part 2 suggesting that, in fact, [itex]B_\rho(x,\epsilon) = U(x,\epsilon)[/itex]

    Am I missing something big here? I feel like it's obvious the problem is incorrect, so I can't understand how the author ( Munkres' Topology First Edition, section 2-9 #6 for reference ) could make the mistake. Please, someone throw me a bone here, I'm just stumped
  2. jcsd
  3. May 11, 2005 #2
    This is actually a good problem because it exposes some of the subtlety that can be in the supremum. I claim that there is a [tex]y\in U(x,\epsilon)[/tex] such that [tex]\rho(x,y) =\epsilon[/tex] so [tex]y\not\in B_\rho (x,\epsilon)[/tex]. Namely [tex]y_i =x_i +\frac{i}{i+1}\epsilon [/tex]. You should be able to show that such a [tex]y[/tex] satisfies what I said above. Of course I'm assuming you have a countably infinite index set but you can work around that if you don't want that stipulation.

    Hopefully with that idea of the supremum in mind you should be able to see why
    [tex]\bigcup_{\delta < \epsilon}{\prod_i{(x_i-\delta, x_i+\delta)}}} \ne \prod_i{\bigcup_{\delta<\epsilon}{(x_i-\delta,x_i+\delta)}}}[/tex]

    Hope that's some help,
  4. May 11, 2005 #3

    Perfect, snoble, thank you.
    Now that I see it, I can't believe what I was thinking.
    I think at some point, I was equating the open ball with the closed ball in the uniform tolology. As in, it included the values for which the supremum was reached. Obviously, that's not the definition of a basis element for the uniform metric topology. Silly me :redface:

    In any event, it's now painfully clear to me that the uniform topology is an entirely distinct animal from the product and box topologies. Now to try and wrap my head around it.

    Thanks again for pointing out my broken line of thought.
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