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Metric Spaces

  1. Nov 23, 2013 #1
    Technically, this is not a homework question, since I solely seek an answer for self-indulgence.

    1. The problem statement, all variables and given/known data

    Example 1.1.4. Suppose f and g are functions in a space X = {f : [0, 1] → R}. Does
    d(f, g) =max|f − g| define a metric?

    2. Relevant equations

    (1) d(x, y) ≥ 0 for all x, y ∈ X
    (2) d(x, y) = 0 if and only if x = y
    (3) d(x, y)=d(y, x)
    (4) d(x, z) ≤ d(x, y) + d(y, z)

    3. The attempt at a solution


    So, from my understanding: for d(f, g) to define a metric on X, it has to satisfy all the given properties of a metric.
    Well, my question is not necessarily whether d(f, g) defines a metric (though I wouldn't mind a proof of it); I was wondering if property (2) is satisfied.
    Because in my pursuit of an understanding in topology, I stumbled across a compilation of notes, in which the note-taker mentions that the second property is not satisfied.
    The reasoning is: that, "by considering two arbitrary functions at any point within the interval [0, 1]. If |f(x) − g(x)| = 0, this does
    not imply that f = g because f and g could intersect at one, and only one, point."
    However, I was wondering if that could also be said about d(f, g) =max|f − g|, which is the function being originally considered; since if d(f, g) = 0, then max|f − g|= 0, which means that for all points in [0,1], 0 ≤|f − g| ≤ max|f − g| = 0, or |f − g|= 0; which would further imply f = g.
     
  2. jcsd
  3. Nov 24, 2013 #2
    If f and g are bounded, then that's a metric.
    I know that metric by the name "supremum distance" (It has to be a supremum, not maximum, because the maximum doesn't always exists).
     
  4. Nov 24, 2013 #3
    Oh ok.
    Would it be safe to say, then, that d(f, g) =max|f − g| does not define a metric on X for this particular case, because X is a set of functions that map [0,1] to R, and R is unbounded?

    So one might be able to prove that "if d(f, g) =max|f − g| = 0, then f = g", but not the converse; that is "if f = g, then max|f − g|= 0", since max|f − g| might not even exist.
     
  5. Nov 24, 2013 #4

    jbunniii

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    Yes, that's correct. Here is an example where the functions are bounded but the max still doesn't exist. Let
    $$f(x) = \begin{cases}
    x & \text{ if } 0 \leq x < 1 \\
    0 & \text{ if } x = 1 \\
    \end{cases}$$
    and let ##g(x) = 0## for all ##x \in [0,1]##. Then ##|f - g| = f## has no maximum value. Since ##d(f,g)## is not even defined for every choice of ##f## and ##g##, it certainly can't be a metric.
     
  6. Nov 24, 2013 #5
    Sorry, I wasn't clear enough the condition 2) always work, if if f = g, then max|f − g|= 0, and it will exists.
    The student who toke the notes was wrong.

    I was only saying what amends are needed for this to be a metric.

    The point that fails, is that this isn't always a function from (X,X) to R, existence fails sometimes.

    But once existence is satisfied, the other 4 are always satisfied.
     
  7. Nov 24, 2013 #6

    PeroK

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    If you restrict the set X to continuous functions, then max will always be defined. And you will have a metric.
     
  8. Nov 24, 2013 #7
    Thank you guys.
     
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