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Math Help - Discussion following on from another thread.

  1. #1
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    Discussion following on from another thread.

    Quote Originally Posted by alice8675309 View Post
    What would the formal proof look like for this problem?

    Prove that the interior of an intersection is the intersection of the interiors:

    (A∩B) =A∩B
    A boundary point p of B can not be an interior point of A\cap B because every neighborhood contains points not common to A and B because there are points not in B.* Same for A.

    *Every neighborhood of a boundary point of B contains points in B and points not in B.

    EDIT: The above argument is wrong because if p is a boundary point of B the points of a neighborhood of p not in B wouldn't show up in A\cap B, and a neighborhood of p would exist containing only points common to A and B, which would make (A∩B) =A∩B wrong.



    [b[Moderator edit:[/b] These posts have been moved from another thread in order to maintain a cleaner thread elsewhere. The orignal thread is here: http://www.mathhelpforum.com/math-he...tml#post618242
    Last edited by mr fantastic; February 15th 2011 at 01:32 PM.
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  2. #2
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    Start off with set B in a metric space E. Let p be a boundary point of B. Then a neighborhood (ball) of p contains points of B and points of E. Now add the set A and let p fall in the interior of A. The points in the ball which were originally in E are now points of A and the points in the ball which were originally in B are now in A\cap B.

    So an accumulation point of B is not an interior point of A\cap B, and interior points of A\cap B can contain only interior points of A and B
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  3. #3
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    Reply to Hartlw:

    Why even talk about boundary points in this thread? The most direct, constructive, intuitive proof of the problem in the OP has nothing to do with boundary points: it never needs to mention boundary points.

    So an accumulation point of B is not an interior point of A\cap B...
    This is manifestly false. Let A=(0,2), B=(1,3), and let x=1.5.

    Clearly, x is both an accumulation point of B and an interior point of A\cap B=(1,2).

    If you meant to say that a boundary point of B is not an interior point of A\cap B, as your two posts seem to point to, again, I would ask, "Why bother saying that?"

    I am always very reticent to post in the Analysis forum, because, while I've had many courses in analysis, I find many of the questions to be beyond me at present. I merely point this out for your reference.
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    Quote Originally Posted by Ackbeet View Post
    Reply to Hartlw:

    Why even talk about boundary points in this thread? The most direct, constructive, intuitive proof of the problem in the OP has nothing to do with boundary points: it never needs to mention boundary points.



    This is manifestly false. Let A=(0,2), B=(1,3), and let x=1.5.

    Clearly, x is both an accumulation point of B and an interior point of A\cap B=(1,2).

    If you meant to say that a boundary point of B is not an interior point of A\cap B, as your two posts seem to point to, again, I would ask, "Why bother saying that?"

    I am always very reticent to post in the Analysis forum, because, while I've had many courses in analysis, I find many of the questions to be beyond me at present. I merely point this out for your reference.
    Jump in by all means, that's how we all learn.

    I assume a set consists of interior points and boundary points. The sets you give are open (do not contain any boundary points and consist only of interior points), so they automaticaly satisify the theorem.
    You might consider (0,2) and [1,3). Then 1 is a (the) boundary point of B but is not an interior point of the intersection. This was beautifully explained in my previous post, which is now slowly being buried. sic transit gloria
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  5. #5
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    Quote Originally Posted by alice8675309 View Post
    What would the formal proof look like for this problem?

    Prove that the interior of an intersection is the intersection of the interiors:

    (A∩B) =A∩B
    OK, you might like this better.

    A0 and B0 are open sets so there intersection is open (consists only of interior points)

    (and I didn't even mention boundary points)
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    What I'm saying, Hartlw, is that you're in danger of violating Rule # 13 here.

    Your introduction of boundary points is irrelevant, and only confuses the issue. In addition, a fairly straight-forward solution (outline) has already been posted. The OP'er appears to have understood the problem. Therefore, if your posts are not clarifying anything, and you're not asking or pointing out anything new, why post?

    Your Post # 24, while true, does not prove the problem in the OP, because you haven't shown the two sides of the equals sign are equal. You've only shown that both sides are open sets.
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    Quote Originally Posted by Ackbeet View Post
    What I'm saying, Hartlw, is that you're in danger of violating Rule # 13 here.

    Your introduction of boundary points is irrelevant, and only confuses the issue. In addition, a fairly straight-forward solution (outline) has already been posted. The OP'er appears to have understood the problem. Therefore, if your posts are not clarifying anything, and you're not asking or pointing out anything new, why post?

    Your Post # 24, while true, does not prove the problem in the OP, because you haven't shown the two sides of the equals sign are equal. You've only shown that both sides are open sets.
    Astute observation Ackbeet. However, I don't believe it is for you to say a proof can't involve boundary points.

    OK, I'll take a shot at a formal proof.
    Let Ap, and Bp be boundary points of A and B belonging to A and B. Then:
    A=AoUAp
    B=BoUBp

    A^B = (AoUAp)^(BoUBp)
    If you expand using DeMorgans laws* you get:

    A^B = (Ao^Bo)U(Ap^Bp)U(Ao^Bp)U(Ap^Bp)

    The only interior (open) points are (Ao^Bo), which proves the theorem.

    I would gladly see another proof, with or without boundary points.

    I find your referring me to Rule # 13 here quite intimidating. I do not instantly see perfect answers, I zero in on them, and I would hope the process generally applied would be enlightening to everyone. I am reminded of a math student who said the way they solved math problems was to get together and wack away at it.

    In any event, I note you are applying the rule only to me.
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  8. #8
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    Hartlw: It seems to me that you are essentially assuming the result in order to prove it - how do you know that the only interior points of (A^o \cap B^o)\bigcup(A^p \cap B^p)\bigcup (A^o \cap B^p) \bigcup (A^p \cap B^p) are A^o \cap B^o? This needs further justification.
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    Quote Originally Posted by Defunkt View Post
    Hartlw: It seems to me that you are essentially assuming the result in order to prove it - how do you know that the only interior points of (A^o \cap B^o)\bigcup(A^p \cap B^p)\bigcup (A^o \cap B^p) \bigcup (A^p \cap B^p) are A^o \cap B^o? This needs further justification.
    It was proved by me in a previous post:

    "Start off with set B in a metric space E. Let p be a boundary point of B belonging to B. Then a neighborhood (ball) of p contains points of B and points of E. Now add the set A and let p fall in the interior of A. The points in the ball which were originally in E are now points of A and the points in the ball which were originally in B are now in .

    So an accumulation point of B is not an interior point of , and interior points of can contain only interior points of A and B"

    which really should have ended the thread.

    EDIT I love the Tex typography. Wish I could do that. I know, I know, the tutorial. I'm just slow.
    Last edited by Hartlw; February 15th 2011 at 09:27 AM. Reason: underline added
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  10. #10
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    Sorry, I really don't understand any of that. Could you possibly rephrase this in a more rigorous way?
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    Quote Originally Posted by Defunkt View Post
    Sorry, I really don't understand any of that. Could you possibly rephrase this in a more rigorous way?
    The intersection of an interior point with a boundary point is not an interior point (open).
    Last edited by mr fantastic; February 15th 2011 at 01:53 PM. Reason: Deleted insulting comment.
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  12. #12
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    Quote Originally Posted by Defunkt View Post
    Sorry, I really don't understand any of that. Could you possibly rephrase this in a more rigorous way?
    A^B consists only of points common to A and B. Since boundary points of B include points not in B they can't be in A^B. Same for A. So A^B consists only of interior points, and the OP is proved.

    Please note that most set-theoretic fundamental proofs are verbal, based on axioms and definitions. DeMorgans Laws for example. Consult any analysis text
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