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Help finishing connectedness proof.

  1. May 10, 2012 #1
    Been having trouble with connectedness. Tried out a problem to practice and not sure how to finish it.

    The problem is this:


    Let A and B be closed subsets of the space X. If A U B and A n B are connected, so also are A and B. Show by counter-example that both A and B must be closed, i.e., if one of A and B is not closed then in general the thm fails.


    My proof went like this:

    Let C = A ᴜ B. Suppose that U, V give a separation of C, where U and V are open in C.
    Since A is connected, we must have A ᴜ U or A ᴜ V (otherwise, A ∩ U and A ∩ V give a separation of A).
    Without loss of generality, assume A ᴜ U. Since A ∩ B = Ø and B is connected,
    we must have B ᴜ U. This implies that C ᴜ U, and therefore, V = Ø.
    This contradicts that U, V is a separation of C. Hence the sets A or B must be closed,


    The response I got back was this:

    To prove your result, assume (to get a contradiction) that one of the closed sets, A or B is not connected. WOLOG, we may assume it is A. Then there are open sets V and W in X s.t. A= (A ∩V) U (A∩ W) and = (A ∩V) ∩ (A∩ W)=ᴓ.
    Now, you should be able to use V and W to build a separation of either AUB or A∩B. (Try “unioning“ B to both sides of the above equations, and it may give you an idea of what to use for a separation.)


    I thought I was doing it right, but by my professor's response, I'm not sure. Could anyone tell me what I'm doing wrong and what he means by "unioning" B to both sides?

    Any feedback would be greatly appreciated.
     
  2. jcsd
  3. May 10, 2012 #2

    You already repeated this several times to be a mere mistake: you write "we must have [itex]\,A\cup U\,\,or\,\,A\cup V\,[/itex]", and then again " assume [itex]\,A\cup U\,[/itex]"

    WHAT are we assuming, anyway?? My guess is that you actually mean "assume [itex]\,A\cup U=A\,[/itex]" , but then what do you expect from this?

    I think you first must be clear above in order to meet your instructor's criticism.




    DonAntonio
     
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