1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Question about infimums and closed sets

  1. Feb 14, 2013 #1
    1. The problem statement, all variables and given/known data
    So this question arose out of a question about showing that a set χ is dense in γ a B* space with norm ||.||, but I think I can safely jump to where my question arises. I think I was able to solve the problem in another way, but one approach I tried came to this crux and I wasn't sure if I was correct.

    Suppose that χ is closed and that we have both that inf x ε χ ||y-x|| = c for all y in γ and that ||y-x|| > C for some y in γ and for all x in χ.

    My question is if this is a contradiction. It seems like that the above would imply that the infimum is something that is not an element of χ, but then by definition of infimum there must be elements in χ such that for any ε > 0 ||y-x|| < ε + C. Then I want to say it seems like this infimum is a limit point of χ which contradicts χ being closed. However, I am concerned that this is incorrect and that really the infimum is a limit point of y - χ not χ.

    I'm not sure if this was clear or not. I guess my concern is that there existing elements x of χ such that ||y-x|| is arbitrarily close to ||y - infimum|| does not necessarily mean that there elements of χ that approach the infimum. In fact, when I put it like that, I think that perhaps it is not a contradiction. I think I just don't have a clear idea of infimum's of more complex sets. Thanks for your time!

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Feb 14, 2013 #2


    User Avatar
    Science Advisor
    Homework Helper

    Draw an example in the xy plane. Take X={(x,y):y>=1+1/x for x>0} and Y={(x,y):y<=0}. Both X and Y are closed, right? The infimum of ||a-b|| for a in X and b in Y is 1. What happens if you try to find pairs of points whose distance is arbitrarily close to 1?
    Last edited: Feb 14, 2013
  4. Feb 14, 2013 #3
    One thing is that ||y-x|| > C for a fixed y and any x and I think that in the counter example you gave for any fixed y, the infimum of ||y-x|| would be equal to ||y-r|| where r is actually in X. Does the y being fixed change anything?
  5. Feb 14, 2013 #4


    User Avatar
    Science Advisor
    Homework Helper

    Maybe I'm not understanding what the question is, but then try this one. Take the space of all points in the plane with rational coordinates. Take X={(x,y):x^2+y^2<=1} (x,y rational of course). Take Y={(1,1)}. Then the infimum is sqrt(2)-1. But there is no point in X that has that minimum value. X is closed, but it isn't compact.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Question about infimums and closed sets
  1. Infimum of a set. (Replies: 3)