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Math Help - Two questions about neighborhoods

  1. #1
    Member pberardi's Avatar
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    Two questions about neighborhoods

    I think I have the idea but I am having trouble finishing off these. Can I just post the questions and my attempt at the solutions hoping that someone can explain how to finish or if I am right? Thanks.


    #1 Question
    Let x be in R(real #s) and let E(epsilon) > 0. Prove that if y is in (x-E,x+E), then (x-E,x+E) is a neighborhood of y.

    #1 Solution attempt
    Given that y is in (x-E,x+E) we want to show by definition of neighborhood that there exists a positive number E s.t. (y-(x-E),y+(x+E)) is a subset of the interval (x-E,x+E). Since y is in (x-E, x+E), x-E<|y|<x+E. And y<x+E, y<-(x+E). At this point I think to myself that this is almost obvious. If y is inside the interval then of course the interval is a neighborhood of y. But then I think that then why did they ask the question. So I am a bit stuck.

    #2
    Let x, y be in R(real #s) with x != y. Prove that there is a neighborhood U and a neighborhood V s.t. the intersection of U and V is the empty set. i.e. U and V are disjoint sets.

    #2 Solution attempt
    Here I just pick E to be 1
    Therefore x-1<x<x+1 and y-1<y<y+1
    or: (x-1,x+1) is U and (y-1,y+1) is V
    so |x| < 1 and |y| < 1. I thought I had it because I thought the the only way this could be true is if x = y. But then what if x was 1/2 and y was -1/2, this could still hold. So I thought maybe I am not thinking right.

    Any help on these two?
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  2. #2
    Super Member Gamma's Avatar
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    1) I am not sure what your definition of neighborhood is, but when I learned it, it was just an open set that contains the point. By assumption y\in (x-\epsilon , x + \epsilon ) and this is a basis element for \mathbb{R} so it is open. It doesnt matter that it is not centered at y.

    2)If x\not = y then let \delta= |x-y|. Then consider U=B(x, \frac{\delta}{2}) and V=B(y, \frac{\delta}{2}). These are the neighborhoods you are looking for.

    NOTE: B(x,\delta)=\{ y\in \mathbb{R}| |x-y|< \delta \}
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  3. #3
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    Quote Originally Posted by pberardi View Post
    I think I have the idea but I am having trouble finishing off these. Can I just post the questions and my attempt at the solutions hoping that someone can explain how to finish or if I am right? Thanks.




    #2
    Let x, y be in R(real #s) with x != y. Prove that there is a neighborhood U and a neighborhood V s.t. the intersection of U and V is the empty set. i.e. U and V are disjoint sets.

    #2 Solution attempt
    Here I just pick E to be 1
    Therefore x-1<x<x+1 and y-1<y<y+1
    or: (x-1,x+1) is U and (y-1,y+1) is V
    so |x| < 1 and |y| < 1. I thought I had it because I thought the the only way this could be true is if x = y. But then what if x was 1/2 and y was -1/2, this could still hold. So I thought maybe I am not thinking right.

    Any help on these two?
    In those cases a geometrical picture always help.So take two different points x,y on the real No line .The distance between them is |x-y|.

    SO if you define two neighborhoods with centers x and y respectively and radius |x-y|/3 ,then those two neighborhoods have no common points.

    An analytical proof goes as follows:

    Suppose z belongs to the U(x,|x-y|/3) ,a neighborhood round x,THEN ,|z-x|<|x-y|/3................................................. ...................................1

    Now we must show that z does not belong to the V(y,|x-y|/3) ,a neighborhood round y
    Assume it does.then:

    |z-y|<|x-y|/3................................................. .........................2

    Add (1) and (2) and we get:


    |x-y|<|x-y|/3 a contradiction
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