1. ## Open Ball

Prove that a subset of a metric space is open iff it is a union of open balls.
So I did this:

Suppose a subset of a metric space is open.
Let $\displaystyle (X,d)$ be a metric space.
Let $\displaystyle U \subset X$
Then there exists some $\displaystyle \varepsilon >0$ such that $\displaystyle B_{\varepsilon}(x;d) \subset U$.

But $\displaystyle x$ is an arbitrary element of $\displaystyle U$.
Therefore centre a ball around every point in $\displaystyle U$.
$\displaystyle \cup_{x \in U} B_{\varepsilon}(x;d)=U$.

(Since the centre of each of these balls is a point in $\displaystyle U$, the union of them must be the entire set $\displaystyle U$.)

Conversely, suppose a subset $\displaystyle U$ of a metric space $\displaystyle X$ is $\displaystyle \cup_{x \in U} B_{\varepsilon}(x;d)=U$.

Then by definition $\displaystyle U$ is open.

Is this right?

2. Actually it is not correct. The idea is right but not the details.
If $\displaystyle U$is an open set then $\displaystyle \left( {\forall x \in U} \right)\left( {\exists \varepsilon _x > 0} \right)\left[ {B\left( {x;\varepsilon _x } \right) \subseteq U} \right]$.
Now you can say that $\displaystyle U = \bigcup\limits_{x \in U} {B\left( {x;\varepsilon _x } \right)}$.

The other way is trivial.

3. Is there any significance to the fact that, in the definition you've provided, you've put the part about the ball in square brackets and not "()"?

I know it's a bit trivial, but it looks like a useful thing to know.

4. Originally Posted by Showcase_22
Is there any significance to the fact that, in the definition you've provided, you've put the part about the ball in square brackets and not "()"? I know it's a bit trivial, but it looks like a useful thing to know.
The point I made is that you used a fixed $\displaystyle \varepsilon$.
That is incorrect. Each $\displaystyle x\in U$ corresponds to its own $\displaystyle \varepsilon_x$