I know that a subset $U$ of a metric space $M$ with corresponding metric $D(x,y)$ is called open in $M$ if for every $x\in U$ there is an open ball $S_r(x)$ (the set of all $y\in M | D(x,y) < r$) contained in $U$.
If the set is continuous I find it intuitive that every element $x\in U$ is surrounded by an open ball and that $U$ therefore is open. We "never reach the boundary". But how am I supposed to prove this?
Example:
Let $M$ be a metric space with given metric $D(x,y)$, and let $x\in M.$ I'm aksed to prove that the set of all $y\in M$ satisfying $r<D(x,y)<s$ is open.
As I said, I find it intuitive that this is the case. But how do I prove it? And what happens if the metric space is not "continuous"?
Best Answer
First of all, you need to be more accurate when talking about such things. There is no definition for a "continuous" space. If I understand correctly, you are confused with the notion of discreteness, e.g. the space being a finite set. My advice would be this: before trying to get intuition in the game, stick to the definitions and strict mathematical proofs.
Your current problem is easily tackled: let $E=\{y\in M: r<d(x,y)<s\}$. Take $y\in E$ and set $t=\min\{d(x,y)-r, s-d(x,y)\}>0$; consider the open ball $B(y,t)$, your task is to show that $B(y,t)\subset E$. Indeed, if $z\in B(y,t)$, then $d(x,z)\leq d(x,y)+d(y,z)<t+d(x,y)\leq s-d(x,y)+d(x,y)=s$ and $d(x,z)\geq d(x,y)-d(y,z)\geq d(x,y) - (d(x,y)-r)=r$, hence $z\in E$.
How did we find this radius $t$? We used our geometric intuition from $\mathbb{R}^2$; as you can see, it worked. In time you will understand when things from euclidean spaces cannot be generalized to arbitrary metric spaces but it always seemed like a good start to me.
Edit: In order to see what I mean by my advice, look at this example: suppose that our space is $X=\{1,2\}$ and our metric is $d_0$, with $d_0(x,y)=1$ if $x\neq y$ and $d_0(x,y)=0$ if $x=y$. This space is as "discontinuous" (to use your words) as it gets. But none of these affect the proof above: for example, for $x=1$, $r=0.5$ and $s=3$, our set $E$ is $\{2\}$. So what? $\{2\}$ is open in this topology! (verify it using definitions, it is easy)