So I did this:Quote:

Prove that a subset of a metric space is open iff it is a union of open balls.

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?