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Counterexample intersections of 2 compacts is compact ?

  1. Jan 6, 2012 #1
    Counterexample "intersections of 2 compacts is compact"?

    Hello,

    I'm looking for a counterexample to "If A and B are compact subsets of a topological space X, then [itex]A \cap B[/itex] is compact." It's not for homework.

    I found one online, but it talked about "double-pointed" things which I didn't understand... My knowledge is an introductory course in topology: the first four chapters of Munkres.

    I realize I have to look for a non-Hausdorff space, but the only one I know
    --if I remember correctly--is the cofinite topology (or the related Zariski topology, but I'm not too familiar with that one).
     
  2. jcsd
  3. Jan 6, 2012 #2
    Re: Counterexample "intersections of 2 compacts is compact"?

    Here's an example:

    Take the circle with a doubled point. I.e. the circle with a disjoint point for which the topology can't distinguish the point [edit: I should probably elaborate this- take the circle with a disjoint point added. The topology is just the same as that of a regular circle but where the open sets are just the same as that as the circle but where we include the "doubled point" whenever its partner is in the open set).

    Take the following compact sets: the circle minus one of the doubled points, and the circle minus the other point. Each is homeo to a circle, so compact, but their intersection is equal to a circle with a point missing, not compact.
     
  4. Jan 6, 2012 #3

    micromass

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    Re: Counterexample "intersections of 2 compacts is compact"?

    I'll formalize the above comment:

    The line with two origins can be defined as follows:

    Take in [itex]\mathbb{R}^2[/itex] the set

    [tex]Y=\{(x,y)\in \mathbb{R}^2~\vert~y=0\}\cup\{(x,y)\in \mathbb{R}^2~\vert~y=1\}[/tex]

    (so this is just the union of two lines). Define the following equivalence relation on Y:

    We set [itex](x,y)\sim (x,y)[/itex] and if [itex]x\neq 0[/itex], then we also set [itex](x,0)\sim (x,1)[/itex] (and of course [itex](x,1)\sim (x,0)[/itex]). So we define all the points above eachother except (0,0) and (0,1).

    The set [itex]Y/\sim[/itex] is the double-pointed line.
     
  5. Jan 6, 2012 #4
    Re: Counterexample "intersections of 2 compacts is compact"?

    Thank you both :)
     
  6. Jan 6, 2012 #5
    Re: Counterexample "intersections of 2 compacts is compact"?

    I wonder, is there also a counterexample that is T1?
     
  7. Jan 6, 2012 #6

    micromass

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    Re: Counterexample "intersections of 2 compacts is compact"?

    That counterexample IS T1.
     
  8. Jan 6, 2012 #7
    Re: Counterexample "intersections of 2 compacts is compact"?

    Haha, of course, I'm sorry. Thanks again.
     
  9. Jan 7, 2012 #8
    Re: Counterexample "intersections of 2 compacts is compact"?

    I like both mine and micromasses formulation of what "doubled points" are.

    In mine, all the open sets are the same but include only the special point "as a pair". It's sort of like an extra ghost point which can't be separated from its friend, they always come as a pair. If (if being strong here, this probably can't be made formal) it could be made into a metric space, then the new point would be distance zero from the old one [actually, maybe this can be made precise, it will be a pseudo-metric space I think i.e. distinct points are allowed to be distance zero from each other.]

    Micromasse's formulation gives the same topology but in this way you are viewing things possibly a little more geometrically. If you imagine sitting on his line, then everywhere away from the "doubled point" it just looks like the line. As you approach the "doubled point" (perhaps in a continuous path) then in some way a choice of either doubled point is immaterial - both in some way "limit" to the correct place on the line, and if you were heading down a continuous path on the line, then whenever you had one of your doubled points, you could pick the other one instead with no trouble of continuity.
     
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