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The illusory generality of metric spaces

  1. Feb 13, 2013 #1
    So I was flipping through Lang's Undergraduate Analysis and noticed the absence of the important concept of metric spaces. I checked the index and was referred to problem 2, Chapter 6 Section 2. There he defines a metric space, what a bounded metric is, and gives a few straightforward problems about specific bounded metrics. This is not very interesting, but the next problem is.

    In fact, I'll just copy the problem here for your convenience. (FYI, I have 'solved' the problem so I'm not asking for help.)

    Problem 3:
    Let S be a metric space. For each x in S, define a function [itex] f_x : S \rightarrow \mathbb{R} [/itex] by the formula [itex] f_x(y) = d(x,y) [/itex].
    (a) Given two points x, a in S, show that [itex] f_x - f_a [/itex] is a bounded function on S.
    (b) Show that [itex] d(x,y) = \left|\left|f_x - f_y \right|\right| [/itex]
    (c) Fix an element a of S. Let [itex] g_x = f_x - f_a [/itex]. Show that the bounded map [itex] x \mapsto g_x [/itex] is a distance preserving embedding (i.e., injection) of S into the normed vector space of bounded functions on S with the sup norm. [If the metric on S was orginally bounded, you can use f_x instead of g_x.] This exercise shows that the generality of metric spaces is illusory. In applications, metric spaces arise naturally as subsets of normed vector spaces.

    So there it is. I don't think metric spaces are ever mentioned again in the book, although admittedly I have not carefully read the entire book looking for their mention. Lang develops the rest of the standard topics in undergraduate analysis in the context of normed linear spaces (and of course, after the notion of completeness has been expounded, most of the analysis takes place in Banach spaces, although he does not use that word, for whatever reason.)

    Now like most of you, I first learned about continuity, completeness, compactness, etc., in the context in metric spaces. Then, upon reaching normed linear spaces, which are obviously also metric spaces, we could just 'transfer' over all the results from metric space theory. After reaching normed linear spaces, metric spaces are really never mentioned again, except maybe to quote a result from their theory.

    So why the focus in standard undergraduate analysis textbooks on metric spaces when (a) as Lang, shows, their generality is illusory and (b) we are really after normed linear space theory?
  2. jcsd
  3. Feb 13, 2013 #2


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    Yes, I've read that section of Lang's book too. It's an interesting intellectual exercise, but I'm not sure I see the point. Great, so you can embed a metric space S into a vector space V. But what good does it do me? The only reason to do this would be if I want a way to add the elements of S, or multiply them by scalars. However, if we call the embedding S', there's no reason to expect S' to be a subspace of V, so if I work in S', what stops me from adding two elements together and inadvertently leaving S'?

    And if I'm not going to add elements, why do I need to be embedded in a vector space in the first place? It just adds unnecessary baggage and potential confusion. Furthermore, it makes it harder to see which properties of a space depend only upon distance and not upon norm.

    Moreover, what if S is already a vector space and I want to give it, for example, the discrete metric:
    $$d(x,y) = \begin{cases}
    1 & \textrm{ if } x \neq y \\
    0 & \textrm{ if } x = y\\
    As I mentioned in a recent post ( https://www.physicsforums.com/showpost.php?p=4268270&postcount=2 ), there is no way to define a norm on S which induces this metric. So Serge Lang's construction will end up embedding S into some normed vector space V. But what's the metric on V? It surely isn't the discrete metric, because that can't be induced by a norm. So great, now I'm working in a normed vector space, but it doesn't have the topology I want.

    These criticisms aside, there's nothing wrong with Lang using normed vector spaces throughout the book, as every space he deals with is already inherently a vector space, so he doesn't need to actually perform an embedding into one.
    Last edited: Feb 13, 2013
  4. Feb 13, 2013 #3


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    For a more useful example of a metric space with no obvious "natural" embedding into a vector space, suppose ##S## is the surface of a sphere, with the distance between two points being defined as the (shortest) distance along a great circle containing both points.

    As it stands, ##S## is not a vector space. The only obvious vector space into which we might embed ##S## is ##\mathbb{R}^3##. But there is no norm we can define on ##\mathbb{R}^3## which will induce the great-circle distance on ##S##. Therefore Lang's construction must embed us in some other vector space, who knows what it looks like or what metric it has outside its restriction to ##S##, just so he can write ##||x-y||## instead of ##d(x,y)##.
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