Open ball

Definition

$(X,d)$ an "open ball" of radius

$r$ centred at

$a$ is the set

$\{x\in X|d(a,x)\lt r\}$ , it can be denoted several ways. I frequently encounter

$B_r(a)=B(a;r)=\{x\in X|d(a,x)\lt r\}$ and use

$B_r(a)$

Take the open ball

$B_\epsilon(p)$ .

Let

$x\in B_\epsilon(p)$ be arbitrary

Choose

$r=\epsilon-d(x,p)$ - then as

$x\in B_\epsilon(p)\iff d(x,p)<\epsilon$ we see

$r>0$

We now need to show that

$B_r(x)\subset B_\epsilon(p)$ using the Implies and subset relation we see:

$B_r(x)\subset B_\epsilon(p)$

$\iff y\in B_r(x)\implies y\in B_\epsilon(p)$

So let

$y\in B_r(x)$ be arbitrary, then:

$y\in B_r(x)\iff d(y,x)< r=\epsilon-d(x,p)$ so

$d(y,x)<\epsilon-d(x,p)$

$d(y,x)<\epsilon-d(x,p)\iff d(y,x)+d(x,p)<\epsilon$

$d(y,p)\le d(y,x)+d(x,p)<\epsilon$

So

$d(y,p)<\epsilon\iff y\in B_\epsilon(p)$

We have shown that

$y\in B_r(x)\implies y\in B_\epsilon(p)\iff B_r(x)\subset B_\epsilon(p)$ , since

$x\in B_\epsilon(p)$ was arbitrary, we have shown that

$B_\epsilon(p)$ is a neighbourhood to all of its points, thus is open.