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Prove if S is Open and Closed it must be Rn

  1. Oct 1, 2009 #1
    The main question:
    Let S be a subset in Rn which is both open and closed. If S is non-empty, prove that S= Rn. I am allowed to assume Rn is convex.

    Things I've considered and worked with:
    The compliment of Rn is an empty set which has no boundaries and therefore neither does Rn. Therefore there exists NO points P such that the delta neighborhood of P is located within Rn and its compliment. This means it is open, because any d-nbhd within Rn is a subset of Rn, and it is closed because all of the boundaries of Rn are within the set. Therefore Rn is both open and closed.

    Thats where I am stuck. And also, I am not sure If all of that is valid logic.

    The second question I had, as I have two questions remaining on my assignment after working on it for 10 hours today and yesterday another 10, and becoming quite desperate for help is:

    "Let f(x,y) be defined on the square -1 <= x <= 1, -1<=y<=1 as follows: f(x,y) = 1 for all (x,y) where Abs(x^3) <y < x^2 and f(x,y) = 0 otherwise. Show that f(x,y) approaches 0 as (x,y) approach zero on any straight line through the origin. Determine if the lim (x,y) -> 0 exists."
     
  2. jcsd
  3. Oct 2, 2009 #2

    Dick

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    I'm not buying your proof. It's way too vague. It's great you can assume R^n is convex, because it is. And it's not hard to prove. But if you assume it is, then if S is nonempty then there is a point x in S. If S is not equal to R^n then there is a point in y in complement(S). Define the continuous map f:R->R^n by f(t)=t*x+(1-t)*y. The set of all t such that f(t) is an element of S is a closed and open set of R, right? Why aren't there any such nontrivial subsets of R? In other words, why is R connected?
     
    Last edited: Oct 2, 2009
  4. Oct 2, 2009 #3
    This is my first assignment ever with topological proofs, so bear with me.

    first, what is the map you defined , and why did you define it? Is that to say that any subset within S must also be open and closed?

    secondly, I've never encountered the idea of being "connected" before. I took the time to research it a bit, but don't understand the relevance (simply because I lack the knowledge, not to say it IS irrelevant)

    I'm now working with this idea - If S does not Equal R^n, then there exists a point y in R^n - S. This implies the compliment also has boundaries. If the compliment has boundaries, then at least one boundary point must either belong to S or compliment(S) but not both. By definition, a set is open if and only if it's compliment is closed so by this fact, either S is not closed, which contradicts the original information, or S is not open, by virtue of the fact that it's compliment is not closed.
     
  5. Oct 2, 2009 #4

    Dick

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    When you say "This implies the compliment also has boundaries." nothing implies any such thing. Saying it does is basically assuming what you want to prove. A set doesn't HAVE to have boundary. I would suggest you start with the case n=1, i.e. R^n=R. And here's an instructive example. Let A=R-{0} (R with the origin removed) with the usual topology. Let S={x:x>0}. Then A-S={x:x<0}. S and A-S are open, clearly, but they are also CLOSED in A. Not closed in R, clearly, but they are closed in A. Can you show this? Now why can't R be split up like I split up A? This is the gist of what connectedness is about.
     
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