Let X be a metric space and fix p in X. Then define C={x: d(p,x)<=z} where z>0 is fixed in the positive reals.
Could someone show me how to prove C is closed?
Any help is greatly appreciated.
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?
we are talking about an arbitrary metric space here, you can't assume you're dealing with the reals.
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 $\displaystyle \dsiplaystyle \{ y ~:~ d(y,c) \le z/n \}$
where $\displaystyle \dsiplaystyle c \in C$, and $\displaystyle \dsiplaystyle n \in \mathbb N$
Here is another approach. Do you understand that the union of balls is an open set?
Do you understand that if $\displaystyle d(p,q)>r$ then $\displaystyle B(q;d(p,q)-r)$ contains no point of your set?
So the union of those balls is open; the complement is open.
Thus the set is closed.
This wouldn't prove that C is closed? Even after showing this, there could be other points outside C where I could conduct such a sequence, couldn't it? Therefore I would have to prove additionally that the complement of C is open what is already sufficient to show that C is closed.
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.
I agree with you. which is why i commented on this weird definition, i didn't think it worked to show the set is closed. but i decided to gave the OP what he/she asked for. what i did would accomplish what the OP asked for. were it up to me, i would have done this problem by showing the compliment is open (Plato's approach would work beautifully). i think there is perhaps a mistake in definitions here. it's not only the fact that you can do what i did for points outside of C that is troubling, it's the fact that you can do it for some point in C even if C were open.