Open ball

Definition

For a metric space $$(X,d)$$ 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 $$B_r(a)$$

Proof that an open ball is open

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

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

But by the Triangle inequality part of the metric $$d(y,p)\le d(y,x)+d(x,p)<\epsilon$$

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.

See Also

• Open set