# Open Ball

• Jan 26th 2010, 01:40 PM
Showcase_22
Open Ball
Quote:

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

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

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

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

Then by definition $U$ is open.

Is this right?
• Jan 26th 2010, 01:51 PM
Plato
Actually it is not correct. The idea is right but not the details.
If $U$is an open set then $\left( {\forall x \in U} \right)\left( {\exists \varepsilon _x > 0} \right)\left[ {B\left( {x;\varepsilon _x } \right) \subseteq U} \right]$.
Now you can say that $U = \bigcup\limits_{x \in U} {B\left( {x;\varepsilon _x } \right)}$.

The other way is trivial.
• Jan 26th 2010, 02:21 PM
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.
• Jan 26th 2010, 02:53 PM
Plato
Quote:

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 $\varepsilon$.
That is incorrect. Each $x\in U$ corresponds to its own $