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Thread: Question in topology

  1. #1 Question in topology 
    Forum Freshman
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    Mar 2012
    I have only a superficial acquaintance with topology, but I need to make some use of it in connection with an investigation in another field. I am perplexed by the following line of reasoning, which seems to lead to an erroneous conclusion, but I cannot see the flaw, and would appreciate emlightenment.

    Suppose that I have a space X with two sets, P and M, which are disjoint but not separate. That is, they have no points in common but can contain limit points of the other. Let A be a subset of P, and let it be defined as the collection of all points of P that are limit points of M. Now consider the boundary points of A that are members of P and call this set of points B.

    1. If x P, and if any neighborhood U X that contains x also contains some m M then x A by definition.

    2. Conversely, for any a A, any neighborhood U X of a contains some m M because every point of A is a limit point of M.

    3. Assume x B, the boundary of A in P. Then any neighborhood S of x contains some other point a A by definition of boundary.

    4. Choose some neighborhood U of a such that U S.

    5. Then (m M) U S. But S is an arbitrary neighborhood of x. Hence x is a limit point of M.

    6. Then by (1), x B is a point of A and hence A contains its boundary in P and is closed in P.
    (Have I made an error to this point?)

    7. Since B is the boundary of A in P, the nighborhood S of x B also contains some y exterior(A) P.

    8. But x is a limit point of M from (5) and hence every neighborhood S contains some m M and some y exterior(A) \ P.

    9. Hence some y is a limit point of M and therefore a member of A by definition.

    Clearly the conclusion that points of the exterior of A are contained in A is absurd. I think that step 5 is legitimate since S converges to x and always contains a point of M. I am less sure about steps 8-9 since as S converges to x, a different y is in S at each step. Still, in the limit there is always some y S and some m S. Where have I gone wrong?

    I have had a further thought on this. Suppose that X is T2. Then although any S containing x will contain some y and some m, we are guaranteed that we can find some neighborhood containing that y that does not contain x or that m, and there is no guarantee that that neighborhood will contain any other m. Hence if X is T2, 8-9 fails. I believe 5 survives however. Anyone have any thoughts in support or contradiction of this?

    Last edited by Ernie; March 25th, 2012 at 11:10 PM.
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  3. #2  
    Forum Professor river_rat's Avatar
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    Jun 2006
    South Africa
    Hi Ernie

    I'm confused by your step 5, do you mind walking me through your implication there.

    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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  4. #3  
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    Mar 2012
    Thanks for your interest, River_rat

    The goal of this step is to decide if the subset A of P is open or closed in P. The argument relies on knowing that A was defined as the set of all points of P that are limit points of M. So, the thinking goes that if we have a boundary for A in P, then by definition a point x of the boundary has some other point, a, of A in any of its neighborhoods. Then by the definition of A, any such point, a, is a limit point of M and so has some point, m, of M in each of its neighborhoods. Since in both cases the neighborhoods are arbitrary we are free to choose the neighborhood of a containing m to be contained in the neighborhood of x containing a. Hence m is contained in the neighborhood of x that contains a. Since these are arbitrary neighborhoods we can say that any neighborhood of x must contain a point of M. Thus x is a limit point of M. Since A is the set of all limit points of M in P, x must be a member of A. Since we initially chose x to be an arbitrary point in the boundary of A in P, A must contain its boundary in P and thus be closed in P.


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