I know that C is closed if every point in C is a limit point of C. Intuitively this makes sense because for any neighborhood around an arbitrary p0 in C, there must be a point q in C not equal to p0 that is of distance d(p,q)<z+r, so p0 is a limit point of C. This is true for any point in C, but I'm having a hard time expressing it mathematically. I always get confused about this, but should I mention that the rationals are dense in the reals to prove this?
so that's the definition you are using for closed? that sounds like the definition for "complete". usually we say a set is closed if its complement is open, or if it is equal to its closure or something like that.
anyway, fine, all you have to do, is pick some point in C and show that you will always be able to construct a sequence (in C) converging to it.
Hint: consider the sets of the form
where , and
I would therefore also go for the proof that the complement of C is open. Therefore just show that there is a neighbourhood N(x,a) for any x in the complement of C where the radius a > 0 such that N only contains points outside N.