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Is (a,b) open in R^n

  1. Sep 26, 2007 #1
    1. The problem statement, all variables and given/known data

    We know that (a,b) is open in R. But is it open R^n?

    2. Relevant equations



    3. The attempt at a solution

    I don't think (a,b) is open in R^n even if it is open in R. Let's take for example n=2, then

    E = {(x,y) | x^2+y^2 < r^2} , where r^2 = |b|^2 + Y^2 , for all y belongs to R. However , we can see that even -b satisfies the above equation but -b is not part of (a,b). Hence, we have a point in R^2 which is not internal point of the segment (a,b). Hence, in R^2 (a,b) is not open.

    It looks like (a,b) is not closed either for the neighborhood (-a-e, -a+e), for e >0 for any y belongs to R for any limit point (-a,y) in R^2 has no intersection with (a,b).

    Any comments on the proof. Can this be extended to the n dimensional space?

    Thanks.
     
  2. jcsd
  3. Sep 26, 2007 #2

    HallsofIvy

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    What do you mean by (a,b)in Rn? In R, (a, b) means the interval from a to b on the number line, not including the endpoints a and b. But there is no one "number line" in Rn. I suspect you mean the straight line interval from (a, 0,0,..., 0) to (b, 0,0,...,0)- i.e. on the 'x' axis, but you should say that clearly. Or do you intend "a" and "b" to be, not numbers put points in Rn? Unfortunately, in that case, your "proof" below still makes no sense.

    Your example:
    makes no sense. Again, there is no "(x,y)" in Rn- points don't have only two coordinates. If, as I said above, you mean these as points, then "x^2+ y^2< r^2" makes no sense. You seem to be trying to construct circles (or n-spheres?) around the origin with varying radii. But the origin may be nowhere near your original segment.

    What you want to do is choose a point on your interval and construct a neighborhood (an n-sphere) with that point as center and show that, no matter how small the radius is, it will contain some points that are not in the interval. That will show that the point is not an interior point and the interval is not only not open but has empty interior.
     
  4. Sep 26, 2007 #3
    What I meant was what I copied from Rudin "Principle of mathematical analysis". It is quoted in the second chapter that "(a,b) is open in R but it is not open in R^2". Period.

    I only extended my problem to n dimensional. If my question does not makes sense to you, then I suppose you have a better way of interpreting the statement from Rudin's book

    That I think I should have clarified. My example was for R^2 , which is why I was constructing circles and not n-spheres.

    Thanks. Yeah, I was thinking in those lines myself.
    That was actually my first post but I am a bit surprised. Are people on this post usually a bit cynical?
     
    Last edited: Sep 26, 2007
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