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Connected and compact space

  1. Aug 12, 2010 #1


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    Hello to all, here's another problem the answer of which I'd like to check.

    Let X be a non-empty infinite set, and U = {Φ} U {X\K : K is a finite subset of X} (Φ denotes the empty set) a topology on X. One needs to prove that the topological space (X, U) is connected and compact.

    Now, it seems quite obvious that the space isn't connected, since there don't exist two non-empty disjoint open sets which cover the space (for any two chosen finite sets, the open sets of the form X\K in the topology will have an infinite number of elements in common).

    Further on, to show that the space is compact, let C be an open cover for X. Then X = [tex]\cup[/tex] X\K = X\[tex]\cap[/tex]K (I didn't index the set operators for practical reasons). Now, it follows that the intersection of finite sets [tex]\cap[/tex]K is empty, and so it must contain at least two disjoint finite sets Ki and Kj. It we take these two sets, then (X\Ki)U(X\Kj) is a finite cover for X, so (X, U) is compact.
  2. jcsd
  3. Aug 12, 2010 #2
    The proof seems good to me.
    The co-finite topology is interesting & would yield a good example of a non-metrizable space for sufficiently 'big' X. The space is normal,too.
  4. Aug 12, 2010 #3


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    OK, thanks a lot.

    Could you please elaborate on this one a bit?

    A space is normal if: i) for every pair of distinct points, every one of them has a neighbourhood which doesn't contain the other one - this one seems obvious, yes. ii) for every two closed disjoint sets, there exist neighbourhoods for these sets which are disjoint, too. Now, the closed sets in this topology are the finite sets, right? (Since they have open complements) But if we take two such disjoint finite sets, it seems to me that every neighbourhood intersects with the other one. Or I'm missing something?
  5. Aug 12, 2010 #4


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    Yes, and another question I find interesting (no need to start another thread, I guess).

    Let (N, U) be a topological space the basis of which is given with B = {{1, 2, ..., n} : n is a natural number}. Is (N, U) compact?

    My first thoughts were that it's not compact, since if C is an open cover for N, and every element of C is finite, we can't cover N by a finite number of finite sets.

    But then, I tried to use a theorem which states that a topological space is compact iff every family of closed sets with the property of finite intersections (i.e., for every finite subset of their indexing sets their finite intersection is non-empty) has a non-empty intersection.

    So, the closed sets in our topology are of the form {n, ... : n is a natural number}, and their intersection is always non-empty, so (N, U) should be compact.

    Obviously I'm missing something here.
  6. Aug 12, 2010 #5
    The proof of compactness in your first post is wrong. For instance: {1,2}, {1,3}, {2,3} are finite sets with empty intersection, but no two of the sets are disjoint.

    It's still fixable, though. For instance, choose any X \ K0 in C. For each x in K0 there is a set X \ Kx in C containing x. Then {X \ K0, X \ Kx | x in K0} is a finite cover of X, because K0 is finite.

    Also, X is not normal. It's not even Hausdorff.

    Now, for your last post: Consider, for each n in N, the closed set {n, n+1, ...}. The intersection of all of these sets is clearly empty. In fact, the intersection of some number of such sets is nonempty if and only if that number is finite. This proves that N is not compact in this topology.
    Last edited: Aug 12, 2010
  7. Aug 13, 2010 #6
    Let a, b be two distinct points in X. Since X is infinite, we can choose a countable set C= {c1,c2,...} not containing either a or b.
    Then the open sets {a, c1,c3,..} and { b,c2,c4,...} are disjoint & separate a from b.
    The normality can be proved on similar lines.
  8. Aug 13, 2010 #7


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    adriank, thanks for the reply.

    Actually, I didn't express what I meant in a clear enough manner. If we take the intersection [tex]\cap[/tex]K, at some point, after intersecting some number of sets, and another one (the operation is associative and commutative), we must arrive at a combination of two families of intersected sets which give the empty set, so Ki and Kj were thought of as two families of intersected sets. Would it work this way?

    Oh, I didn't realize that an infinite intersection of such sets is empty, it doesn't seem quite intuitive to me.
  9. Aug 13, 2010 #8


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    Are you referring to the problem from the first post? How can these sets be open in this topology?

    By the way, it would be enough to show the space is normal, since its normality would imply it being Haussdorf. Although, if we show it not being normal, it can still be Haussdorf.

    If we take two distinct points a, b from X, I don't see any way to find open neighborhoods of these points which are disjoint. These neighborhoods would need to be of the form X\{a, ...} and X\{b, ...} (where the finite set {a, ...} mustn't contain b and vise versa), and they are certainly not disjoint, right?
  10. Aug 13, 2010 #9
    Those two sets aren't open, since their complements are infinite.

    At best this is a very unclear argument.
  11. Aug 14, 2010 #10


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    adriank, I have thought about what you wrote, and it makes perfect sense.

    By the way, I just realized that the biggest mistake in the attempt of a proof in the first post is that (even if it "worked"), I didn't actually find a finite sub-cover for the given cover, since the sets X\Ki and X\Kj are not necessarily contained in the first given open cover.
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