1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Question about compactness

  1. Dec 7, 2012 #1
    1. So I worked on this problem, only to find that the solutions claim a fairly simple answer. The question is as follows

    Let X be a metric space with metric d; let A [itex]\subset[/itex] X be nonempty. Show that if A is compact, d(x,A)=d(x,a) for some a[itex]\in[/itex]A.

    2. Relevant equations

    3. The attempt at a solution
    So the solutions claimed an easy fix. The function is continuous in both variables so a continuous image of a compact set is compact and so on the set {x} × A, it reaches a minimum.

    My solution was more convoluted, because I did not immediately see that the distance function was continuous in the second variable. This is how I did it.

    Consider the collection, ℂ, of all sets ℂε, ε>d(x,A) , such that a[itex]\in[/itex]ℂε if and only if d(x,a)<ε.

    Then, for any arbitrary finite collection {Cεi}, we can order them by set inclusion and we see that their intersection is nonempty, for if it were empty, then this would imply that d(x,A)≠inf{d(x,a)|a[itex]\in[/itex]A}.

    Then, by the finite intersection property of a compact set, the intersection of all of these sets must be nonempty, so let p be that element in the intersection. Then I claim that d(x,p)≤d(x,a) for all a[itex]\in[/itex]A, for if not then d(x,p)>d(x,a) for some a, so then we can fit an ε between d(x,p) and d(x,a) (by the order properties of ℝ). Then, the infinite intersection will not contain p, a contradiction.

    So, then d(x,p)≤d(x,a) for all a[itex]\in[/itex]A. So, by the definition, d(x,A)≤d(x,p) but similarly, d(x,p)≤d(x,A) since it is a lower bound of the set. Therefor d(x,A)=d(x,p).

    There are a few places where I was lax on the rigor (like claiming my formation of the collection is actually ok), but let me know if it looks right.
  2. jcsd
  3. Dec 7, 2012 #2


    User Avatar
    Homework Helper

    I'm going to assume that X is complete, because otherwise the proposition that a continuous function on a compact set attains its bounds does not hold.

    By definition d(x,y) = d(y,x) for all x and y. Thus if f(x) = d(x,a) for some fixed a is continuous, we have immediately that f(x) = d(a,x) is continuous.

    So [itex]C_\epsilon = \{a \in A: d(x,a) < \epsilon \}[/itex] and [itex]\mathcal{C}
    = \{C_\epsilon: \epsilon > d(x,A)\}[/itex].

    It's obvious that, for a finite collection,
    [tex]\bigcap_{i} C_{\epsilon_i} = C_{\min \epsilon_i}
    = \{a \in A : d(x,a) < \min \epsilon_i \}.[/tex]
    But since [itex]\min \epsilon_i > d(x,A),[/itex] there must by definition of infimum exist some [itex]y \in A[/itex] such that [itex]d(x,A) < d(x,y) < \min \epsilon_i [/itex], and so the intersection is not empty. This is so whether or not A is compact, provided the infimum exists (which it must do, because 0 is always a lower bound for d(x,y)).

    The remainder of your proof is suspect since you've never actually used the assumption that A is compact (other than to invoke a proposition about intersections of compact sets, which as you are applying it says that if the [itex]C_{\epsilon_i}[/itex] are compact then their intersection is compact; unfortunately the [itex]C_{\epsilon_i}[/itex] are open and so not compact).

    I think what you need to do is take a countable intersection, so that you may rightly conclude that
    [tex]\bigcap_{i} C_{\epsilon_i} = \{a \in A : d(x,a) = d(x,A)\}.[/tex]
    Then you need to use compactness of A to show that this intersection is not empty: for example, by constructing a Cauchy sequence [itex](a_i)[/itex] whose limit is in the intersection (X is complete, so by definition every Cauchy sequence converges, and A is closed, so the limit of the sequence must be in A). But that's essentially the proof that a continuous function on a compact set is bounded and attains its bounds.
    Last edited: Dec 7, 2012
  4. Dec 7, 2012 #3
    Well the solutions claim that d is a continuous function into ℝ, a set in the order topology, and so for fixed x, d(x,A) attains a minimum in ℝ (I guess by extreme value theorem). Is this not correct?

    Also, there is no mention of completeness, and so I'm guessing the continuous function proof is the best way.

    Anyway, I see now where I messed up in my construction, thanks very much.
  5. Dec 7, 2012 #4


    User Avatar
    Homework Helper

    Yes. I had overlooked that a continuous function from a compact space to [itex]\mathbb{R}[/itex] is bounded and attains its bounds, whether or not that space is complete.
  6. Dec 7, 2012 #5
    Not to worry, you've helped me plenty.

    Out of curiosity then, is it true that a continuous function from a compact subset of a complete metric space attains a minimum?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook