Question about infimums and closed sets

[Fractal20]

Homework Statement

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!

Homework Equations

The Attempt at a Solution

 

[Dick]

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?

 

[Fractal20]

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?

 

[Dick]

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?

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.