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Finding the Induced Metric

  1. May 31, 2015 #1
    1. The problem statement, all variables and given/known data
    If ##A## is the subspace of ##l^\infty## consisting of all sequences of zeros and ones,
    what is the induced metric on ##A##?

    2. Relevant equations


    3. The attempt at a solution
    The metric imposed on ##l^\infty## is ##d(x,y) = \underset{i \in \mathbb{N}}{\sup} |x_i - y_i|##. I suspect that the induced (or, as a I call it, the reduced) metric is ##d(x,x) = 0## and ##d(x,y) = 1##. However, I am having difficulty showing this. Here is what I came up with:

    Let ##I_{x,0} = \{i : x_i = 0\}## and ##I_{x,1} = \{i : x_i = 1\}##, and similarly define ##I_{y,0}## and ##I_{y,1}##. Hence, ##I_{x,1} \cap I_{y,1} \ne \emptyset## implies that there exists an ##i## such that ##x_i = 1## and ##y_ i = 1##; furthermore, ##x_i - y_i = 0##, which means ##\underset{i \in \mathbb{N}}{\sup} |x_i - y_i| = 0##...right?

    This, however, does not appear to be a very elegant solution, as there will be many cases to deal with.
     
  2. jcsd
  3. May 31, 2015 #2

    pasmith

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    If [itex]x_i[/itex] and [itex]y_i[/itex] take values in [itex]\{0,1\}[/itex] then the only possible values of [itex]|x_i - y_i|[/itex] are 0 if [itex]x_i = y_i [/itex] and 1 if [itex]x_i \neq y_i[/itex].

    If [itex]x = y[/itex] then by definition [itex]x_i = y_i[/itex] for all [itex]i \in \mathbb{N}[/itex].
    If [itex]x \neq y[/itex] then by definition there exists an [itex]i \in \mathbb{N}[/itex] such that [itex]x_i \neq y_i[/itex].
     
  4. Jun 3, 2015 #3

    micromass

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    This metric has a special name that you might have seen. Do you know what it is?
     
  5. Jun 3, 2015 #4
    Yes, I do indeed know: it is the discrete metric!
     
  6. Jun 3, 2015 #5

    micromass

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    Exactly! Well done.
     
  7. Jun 3, 2015 #6

    micromass

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    Some further information on why this problem is important: A metric space is separable if (by definition) it has a countable dense subset. Separability is an important condition in functional analysis and (among others) it ensures that a Hilbert space has a countable orthonormal basis. Now you can show that a discrete metric space is separable iff it is countable. So your ##A## is not separable. This implies immediately that ##\ell^\infty## is not separable since subspaces of separable spaces are separable (this is not that easy to show). ##\ell^\infty## is one of the most important spaces that is nonseparable. The other ##\ell^p## spaces are separable.
     
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