1. Not finding help here? Sign up for a free 30min 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!

Showing that R is not compact

  1. Oct 14, 2011 #1
    1. The problem statement, all variables and given/known data
    I'm trying to prove that R with the usual topology is not compact.

    2. Relevant equations
    3. The attempt at a solution
    According to the solutions, there are two "simple" counterexamples of open coverings that do not contain finite subcoverings: (-n, n) and (n, n+2). Of course finding just one counter-example is sufficient to show that R is not compact. However, it is not clear to me why those are not compact (and something like (0, n) would be compact?).

    Couldn't I also use a counter-example where each subcovering of R covers exactly one element of R (i.e. the intersection of the subcovering and R has one element), then observe that R has infinitely many elements and therefore must have infinitely many subcoverings? Having one open covering of A with infinitely many subcoverings would then violate compactness.

    edit: I realized why the latter doesn't work - not all elements of R are open sets. I'm still not clear why the subcoverings in the solutions are not compact, though.
     
    Last edited: Oct 14, 2011
  2. jcsd
  3. Oct 14, 2011 #2
    An open set like (0,n) is not compact, consider the family {C_n}={(1/n,+infinity) for n in N\{0}}. (0,n) is contained in the union, so {C_n} is a covering of (0,n). Assume it has a finite subcover. Then a finite union of C_i will cover (0,n). It's finite, so there is a largest
    value for i, however 1/i > 0, so your union fails to completely cover (0,n). Reading your last counter-example, I think you may need to reread the definitions carefully; with compactness it doesn't matter how many subcoverings you have, but that at least one of those subcoverings has a finite number of elements. Covers can be infinite, but if it is compact you can take a finite number of elements from your cover and make a subcover. This in turn is also a cover, but it is already finite so you don't need to look further. You have to show this for any covering of the set in question.
     
  4. Oct 14, 2011 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    It is true of any metric space that a compact set is both closed and bounded. R is not, of course, bounded so cannot be a compact set.

    If you want a proof directly from the definition, consider the open cover [itex]\{(n- 3/4, n+ 3/4)\} where n can be any integer. Any real number is within distance 1/2 of an integer so every real number is contained in at least one such set. And because any reall number is contained in at most two of them, it is impossible to have every real number in any finite subcollection.
     
  5. Oct 14, 2011 #4

    Deveno

    User Avatar
    Science Advisor

    one of the more general ways to define compact is:

    every open cover admits a finite subcover.

    the "standard" counter-example for R is simplicity itself: the cover {(-n,n):n in N}.

    clearly any real number x is finite, so it lies in some interval (-k,k).

    now, suppose some finite subcover, also covered R.

    since our subcover is finite, there is some N for which (-N,N) contains every other element of our cover.

    but if x in R is larger than N, x is not in (-N,N), and is therefore not in our subcover.

    why does this happen? because R itself is unbounded, and therefore requires an infinite number of bounded sets to cover it. but we can find a cover of bounded open sets, which means R cannot be compact.

    now, let's see what happen when we take the set [0,n], where n is a positive integer.

    we have an open cover, we don't even know what it is (there are a gazillion possible open covers). at least one of these open sets contains n. we'll keep that one (call it U1).

    well, if U1 contains all of [0,n], whoopee! we're done. but that would be too easy, right?

    but...if not, then the set A1 = {x in [0,n]: x is not in U1} is non-empty. this set is bounded below, and as such, has an inf. so let's call inf(A1), α1.

    the next question is, is α1 in [0,n]? well, U1 contains n (and U1 is open), so U1 contains a neighborhood of n, so there is some ε > 0 such that (n-ε,n+ε) is in U1. so α1 ≤ n-ε < n, so the only way α1 could NOT be in [0,n] is if α1 < 0. but 0 is a lower bound for [0,n], and is thus a lower bound for the subset A1. if α1 < 0, this contradicts the meaning of inf(A1).

    so, yep, α1 is in [0,n]. since it is in [0,n], some open set of our open cover contains it. we'll call this set U2.

    repeat: define A2 = {x in [0,n]: x is not in U1 U U2}. let α2 = inf(A2). it is easy to see that α2 < α1 (because U2 is open, and therefore contains a neighborhood of α1).

    this gives us a sequence U1,U2,U3,..... with a sequence of inf's:

    n > α1 > α2 > α3 >.......

    with each αk in [0,n].

    the question is: could this sequence perhaps be infinite? again, if it was finite, there's our finite subcover, and woo-hoo! problem solved!

    but, maybe not....maybe it takes us an infinite number of U's to finish. what to do?

    so let's try an new set: B = {x in [0,n]: [x,n] is covered by a finite number of sets in our open cover}.

    as we saw above, all of our αk are in B, so B is non-empty. let's set β = inf(B).

    now 0 is a lower bound for B, so β ≥ 0. also β ≤ αk for each k, so β is in [0,n]. now some open set (say V) in our cover contains B, and therefore contains (β-δ,β+δ) for some δ>0.

    since β+δ is in B (otherwise β is not a greatest lower bound), [β+δ,n] is covered by some finite collection in our open cover, say V1,...,Vm. and since V contains β, V U V1 U.....U Vm contains [β,n]. this shows that β is in B.

    now, suppose β > 0. since V contains (β-δ,β+δ), this means that there is some β-δ < x' < β, with [x',n] covered by V U V1 U ...U Vm. but this contradicts the fact that β = inf(B). the only way to avoid this contradiction, is for β = 0, so that we can find no such x'.

    but then...ohmigosh, we've found a finite subcover, from our arbitrary cover. so [0,n] must be compact.

    now, we didn't really use the fact that n was an integer, so it could have been ANY positive real number. so [0,b] is compact for all real b. an entirely analogous argument, shows that [c,0] is also compact. as a bonus, any sub-interval of [a,b] of [0,b] is compact (we can even use the same subcover, throwing out ones we don't need).

    similarly any sub-interval [c,d] of [c,0] is compact. for any interval [r,s], we either have:

    r ≥ 0 (compact)
    r < 0, s ≤ 0 (compact)
    r < 0, s > 0, [r,s] = [r,0] U [0,s] (compact, one finite cover for "each side").


    **********

    no, you can't pick a cover of R where each open set covers just "one" element of R. open intersect open is always open, a singleton set (in the usual topology) is closed.
     
  6. Oct 16, 2011 #5
    Thanks for your responses - and thanks especially Deveno for taking the time to write out such a detailed explanation. Got me a lot closer to understanding the topic, I think. :)
     
  7. Oct 16, 2011 #6

    Bacle2

    User Avatar
    Science Advisor

    Look up the definition of compactness for metric spaces, and consider the sequence:

    an=n , and use it to show R is not compact.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook