How do I prove that a subset of a metric space is open?

Question

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"?

Answer 1

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)

Answer 2

Let $A = \{y: r < d(x,y) < s\}$. We can see this set as $S_s(x) \cap O_r(x)$, where $O_r(x)=\{y: d(x,y) > r\}$.

So show first that all sets of the form $S_r(x)$ are open, as are all sets of the form $O_r(x)$, $x \in X, r>0$. The asked for set is then a finite intersection of opens sets, hence open too.

To see that a set like $S_r(x)$ is open, pick $y \in S_r(x)$ so that $d(x,y) < r$. Define $t=r-d(x,y) > 0$. Then $S_t(y) \subseteq S_r(x)$: $z \in S_t(y)$ implies $d(y,z) < t$ but then $$d(z,x) \le d(z,y) + d(y,x) < t + d(x,y) = (r-d(x,y))+d(x,y) = r$$ and so $z \in S_r(x)$, showing the inclusion, and as $y$ was arbitary, the criterion for being open is satisfied.

For $O_r(x)$ we just use the same proof idea with $t=d(x,y)-r$ etc.

Answer 3

For the fixed $x$ define $f:M\to\mathbb{R^+}$ as $f(y):=D(y,x)$. Then $f$ is continous and so the reverse image of the open interval $(r,s)\subset \mathbb{R^+}$ must be open as well. But this open set is $ f^{-1}( \quad (r,s) \quad )=\{y: r < D(x,y) < s\} $. So $ \{y: r < D(x,y) < s\} $ is open.