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Need help with a proof on closure

  1. Feb 29, 2012 #1
    Hi, my instructor left this as an exercise, but I got confused in the second part. Could you please help me?

    cl(A[itex]\cap[/itex]B)[itex]\subseteq[/itex]cl A [itex]\cap[/itex] cl B
    But the reverse is not true. Prove this and give a counterexample on the reverse statement.

    My attempt:

    If x[itex]\in[/itex] A[itex]\cap[/itex]B, then x[itex]\in[/itex] cl(A[itex]\cap[/itex]B)
    x[itex]\in[/itex] A and x[itex]\in[/itex] B [itex]\Rightarrow[/itex] x[itex]\in[/itex] cl(A) and x[itex]\in[/itex]cl(B). Hence,

    cl(A[itex]\cap[/itex]B)[itex]\subseteq[/itex]cl A [itex]\cap[/itex] cl B

    Does this proof have any flaws? It is an easy one, I guess, but I feel a bit confused. And I do not understand why the reverse is wrong. Can't we use the same method? Thanks for any help.
  2. jcsd
  3. Feb 29, 2012 #2
    I presume this is topology? Closure can mean a few different things.

    In your proof, you only showed that [itex]A\cap B\subset[/itex] the right-hand side.

    However, this is a good start. The LHS is a subset of the RHS, which is closed. Now take the closure of the LHS. Can we show that it doesn't get "bigger" than the RHS, keeping in mind that the RHS is closed.
  4. Feb 29, 2012 #3
    You seem to prove that if [itex]x\in A\cap B[/itex], that then [itex]x\in cl(A)\cap cl(B)[/itex]. This is certainly true, but it only proves that [itex]A\cap B\subseteq cl(A)\cap cl(B)[/itex].
    You need an extra argument to conclude that [itex]cl(A\cap B)\subseteq cl(A)\cap cl(B)[/itex].
  5. Mar 1, 2012 #4
    As noted, your "proof" is incomplete; you must also consider elements that are limit points of A intersect B.

    For the counterexample, consider the open intervals (0,1) and (1,2).
  6. Mar 1, 2012 #5


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    What is your definition of "closure"? There are several equivalent definitions and how you would prove this depends upon which you are using.

    For example, some texts use "cl(A) is A union all limit points of A" while others use "the smallest closed set containing A" and still others "the intersection of all closed sets containing A".
    Last edited by a moderator: Mar 1, 2012
  7. Mar 1, 2012 #6
    Thanks, micromass, but I don't understand why I need an extra argument. Is it for the case that A and B are not both open or closed? Or neither open nor closed?

    Spherics, I had found this myself while thinking before sleep :) cl(0,1)=[0,1] and cl(1,2)= [1,2].
    Their intersection is 1, but the closure of their intersection is 0. Therefore, the reverse is wrong. This is how to do it, right?

    HallsofIvy, I tried to use the definition of closure as the one ''The smallest closed set containing A'' which is IntA[itex]\cap[/itex] Bd(A).

    Thanks for your time and effort, and I would be greatful for further help :)
  8. Mar 1, 2012 #7
    The definition you wrote is always empty. You probably mean union, not intersection.
  9. Mar 2, 2012 #8
    That was a mistake, sorry. I meant IntA[itex]\cup[/itex]IntB.
  10. Mar 2, 2012 #9
    Whether or not an extra argument is needed is a matter of context, really. In your particular course, it may be that your professor is okay with "(A[itex]\subset[/itex]B) and (B is closed) implies cl(A)[itex]\subset[/itex]B". That is, if you're inside a closed set, and you take a closure, you won't move outside that closed set.

    You might have skipped this step in your logic, or you might've been aware and just not mentioned it.
  11. Mar 2, 2012 #10
    Almost, but not quite. To contradict the statement

    cl(A[itex]\cap[/itex]B)[itex]\supseteq[/itex][cl A [itex]\cap[/itex] cl B]

    you must exhibit an element of [cl A [itex]\cap[/itex] cl B] that is not in cl(A[itex]\cap[/itex]B). As you noted, if A=(0,1) and B=(1,2) such an element is 1, for

    1[itex]\in[/itex][cl A [itex]\cap[/itex] cl B] but 1[itex]\notin[/itex]cl(A[itex]\cap[/itex]B)=[itex]\phi[/itex]
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