Open ball
Definition
For a metric space [math](X,d)[/math] an "open ball" of radius [math]r[/math] centred at [math]a[/math] is the set [math]\{x\in X|d(a,x)\lt r\}[/math], it can be denoted several ways. I frequently encounter
[math]B_r(a)=B(a;r)=\{x\in X|d(a,x)\lt r\}[/math] and use [math]B_r(a)[/math]
Proof that an open ball is open
Take the open ball [math]B_\epsilon(p)[/math].
Let [math]x\in B_\epsilon(p)[/math] be arbitrary
Choose [math]r=\epsilon-d(x,p)[/math] - then as [math]x\in B_\epsilon(p)\iff d(x,p)<\epsilon[/math] we see [math]r>0[/math]
We now need to show that [math]B_r(x)\subset B_\epsilon(p)[/math] using the Implies and subset relation we see:
[math]B_r(x)\subset B_\epsilon(p)[/math][math]\iff y\in B_r(x)\implies y\in B_\epsilon(p)[/math]
So let [math]y\in B_r(x)[/math] be arbitrary, then:
[math]y\in B_r(x)\iff d(y,x)< r=\epsilon-d(x,p)[/math] so [math]d(y,x)<\epsilon-d(x,p)[/math]
[math]d(y,x)<\epsilon-d(x,p)\iff d(y,x)+d(x,p)<\epsilon[/math]
But by the Triangle inequality part of the metric [math]d(y,p)\le d(y,x)+d(x,p)<\epsilon[/math]
So [math]d(y,p)<\epsilon\iff y\in B_\epsilon(p)[/math]
We have shown that [math]y\in B_r(x)\implies y\in B_\epsilon(p)\iff B_r(x)\subset B_\epsilon(p)[/math], since [math]x\in B_\epsilon(p)[/math] was arbitrary, we have shown that [math]B_\epsilon(p)[/math] is a neighbourhood to all of its points, thus is open.
