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Rough and ready compactness

  1. Nov 1, 2007 #1
    Rough and ready "compactness"

    I'm not going to take real analysis until next fall, but I don't know how I'll survive complex without knowing what "compactness" is. I've been working through a real analysis book with some success, but can anyone help me with web resources, or intuitive descriptions, of this concept? You could even suggest a good book.

    So, far I know that it is a property of regions. It's closely related to uniform continuity. On a closed compact set the same delta will work for all vales of Z. But why must it be closed? ... I'm finding this idea really hard to learn on my own.
     
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  3. Nov 1, 2007 #2
    There are 2 equivalent definitions of compactness. One is that given a sequence in the set, you can always find a subsequence that converges to a value in the set. The other is that given any open cover of the set (a collection of open sets that contain every point of the set), you can find a finite sub cover (you can pick out a finite amount of those sets that cover the entire set). The first definition is valid in any metric space, but not in a topological space in general (in which sequences are not defined). The second is valid in any topological space

    The first definition directly leads to closure of the set (the second does too if you use that the complement of an open set is a closed set, but I haven't seen the proof, so I don't know how). Simply take any sequence in your compact set that converges in the metric space. It has a subsequence that converges to some value in your set, say x. However, since the sequence is convergent, it must converge to x (since all subsequences converge to the same value), which we just showed is in the compact set. Thus, the set is closed.

    Also, compact sets must be bounded because otherwise you can construct unbounded sequences that have no convergent subsequence
     
  4. Nov 2, 2007 #3

    Hurkyl

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    Not true for arbitrary spaces. For example, any finite topological space is compact, but it need not be true that every finite subset of a given space is closed.
     
  5. Nov 2, 2007 #4

    Hurkyl

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    Compactness is a type of finiteness condition. If you have some class of 'interesting' open sets, then you only need finitely many to cover any given compact set.

    Exercise: By considering the unit disks of R², prove that compact subsets of R² have finite area.


    Compact sets also have the wonderful property that the image of a compact set is compact.

    Exercise: Let X be a compact space. Let f be a real-valued function on X. Prove that f has a maximum value. (Hint: use the fact that in R, compactness is equivalent to being closed and bounded)
     
    Last edited: Nov 2, 2007
  6. Nov 2, 2007 #5
    Hmm... I guess that sort of makes sense, but could you give me an example? I've never seen any examples of topological spaces that do not result from metrics.

    Also, you meant "the continuous image of a compact set is compact", correct?
     
  7. Nov 2, 2007 #6
    Nevermind, I think I came up with one:

    If we take our topological space to be R and defined open sets as any set that contains only a countable number of elements (and closed sets to be any sets that has an open complement)

    Then every open set is compact and no closed set is compact.

    This works, right? Or is there something preventing me from using these definitions?
     
  8. Nov 2, 2007 #7
    Nope... that's not a topology. If all sets that contain a countable number of elements is open, then all 1-point sets are open, and all unions of 1-point sets (i.e. every subset of R) must be open. This results in the discrete topology on R. Every subset is closed and open, but only finite sets are compact.
     
  9. Nov 2, 2007 #8

    matt grime

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    These are not equivalent definitions. Convergence makes sense in any space, not just metric spaces. What you've written down are the definitions of sequential compactness and compactness. These will agree in a complete metric space, but in general are different notions.
     
  10. Nov 2, 2007 #9
    Yeah, I realized that what I did was quite stupid. I was a little unclear on the axioms of topology, and I was thinking that I could have sets that contained an uncountable number of elements be not open, but then R is not open and uncountable unions of open sets are not necessarily open, so this doesn't at all work.

    I realize this now. I've only had some metric space topology though and no general topology, and I didn't realize that in general covering compactness does not imply closure.

    Anyway, the original poster seems to be asking about metric space topology, so at least the statement about sequential compactness is valid.

    Futurebird: Is this a beginning real analysis course or will the students be expected to already have exposure to most of this? If it is a beginning real analysis course, you're probably already ahead of where you will be expected to be for the first 2 months or so. Also, by next fall do you mean in 10 months? Or are you in the southern hemisphere? Either way, just keep reading. You should be fine.
     
    Last edited: Nov 2, 2007
  11. Nov 2, 2007 #10
    Yes they are, arbitrary unions of open sets are open.
     
  12. Nov 2, 2007 #11
    I believe he was referring to his "topology" (which wasn't a topology), in which it turned out that some uncountable unions of open sets were not open, and hence concluded that it was in fact not a topology by definition.
     
  13. Nov 2, 2007 #12

    It's an into course. But I'm in intro to complex analysis now and there are a few things that make no sense becuse they depened on compactness.
     
  14. Nov 2, 2007 #13
    I just need to know about compactness in the complex plane. That's a metric space, right? I guess so, since the triangle inequality works... I'm not ready for all of these topology ideas!

    For example, how is compactness related to uniform convergence?


    Compact is closed and bounded.
    What would a non-compact region in the complex plane be like?
    So, Open and bounded fails. Why? (in terms of convergence)


    Can you have compactness in the real number line? Or is it only in a plane and those scary non-metric things you guys were talking about?
     
  15. Nov 3, 2007 #14
    Yes, the complex plane is a metric space, with the standard metric being the euclidean metric; hence the topology in the complex plane is the same as in R^2. Addition and multiplication are different, but that's irrelevant in topology.

    Compact sets have lots of nice properties in metric spaces: they are cauchy complete, they are totally bounded (stronger than boundedness) and they are sequencially compact. Continuous functions on compact sets are automatically uniformly so. Images of compact sets under continuous functions achieve their max/mins.

    The big theorem for R^n is the heine borel theorem: a set in R^n (complex plane follows from R^2) is compact iff and it is bounded and closed. This answers your question for a noncompact region.

    I don't think compactness is related to uniform convergence in the case of complex valued functions. I could be wrong. However, there is a special result in the real case which is called Dini's theorem, if I remember correctly.

    EDIT: You probably meant uniformly continuous. Uniform convergence is something totally different. Uniformly continuous means for any epsilon there exists a delta that works for all x.

    Claim: Continuous functions on some compact set K is uniformly continuous on K.

    Proof. Let [itex]\epsilon>0[/itex]. For each a in K find [itex]\delta_a[/itex] that makes 'f continuous at a' (this makes no sense, hopefully you know what I mean). It is clear that

    [tex]{\bigcup_{a\in K}B_{\delta_a/2}(a)}[/tex] is an open cover of K.

    Since K compact there exist a_1,a_2,...,a_n in K such that

    [tex]{\bigcup_{i=1}^nB_{\delta_a_i/2}(a_i)}\supset K[/tex]

    Let [itex]\delta:=\frac{1}{2}\min_{i}\{\delta_a_i\}.[/itex]

    The rest is all just fact checking. Suppose [itex]|x-y|<\delta[/itex](x,y in K). Then there exists an i such that [itex]|x-a_i|<\delta_a_i/2[/itex] and so

    [tex]|y-a_i|\le|y-x|+|x-a_i|<\delta+\delta_a_i/2<\delta_a_i[/tex] and concluding

    [tex]|f(x)-f(y)|\le |f(x)-f(a_i)|+|f(y)-f(a_i)|<2\epsilon[/tex]

    Voila! I realized I forgot how this proof went while in the middle of writing it...not a good feeling. The 2epsilon is a product of me avoiding an extra line of text at the very beginning.
     
    Last edited: Nov 3, 2007
  16. Nov 3, 2007 #15

    matt grime

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    What is the definition of unbounded? Hence... Now, open - look at an open set and try to find a sequence that doesn't have a limit point in the set.

    The line is a metric space. It is a subspace of the plane so it is metric, is one way of looking at it.

    Topological spaces are not scary. But of course since no one defined them for you, they must look qute intimidating. But don't worry, they aren't really. In fact they're a lot nicer than metric spaces (no epsilon delta arguments, for example).
     
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