Open ball
Contents
Definition
Given a metric space [ilmath](X,d)[/ilmath] the open ball centred at [ilmath]x_0\in X[/ilmath] of radius [ilmath]r>0[/ilmath], denoted [ilmath]B_r(x_0)[/ilmath] (however many notations are used, see below), is given by^{[1]}^{[2]}:
 [math]B_r(x_0):=\{x\in X\vert\ d(x,x_0)<r\}[/math]  that is all the points of [ilmath]X[/ilmath] that are a distance (given by [ilmath]d[/ilmath]) strictly less than [ilmath]r[/ilmath] from [ilmath]x_0[/ilmath]
The open ball must be proved to be open, it is not true by definition. See below
Notations
Here the notations denote an open ball of radius [ilmath]r[/ilmath] centred at [ilmath]x[/ilmath] (in a metric space [ilmath](X,d)[/ilmath], this table is supposed to be complete, so preferred notations are marked from the others
#  Notation  Usage  Comments 

1  [ilmath]B_r(x)[/ilmath]  preferred  Use if the metric is implicit.

2  [ilmath]B_{r,d}(x)[/ilmath]  preferred  Preferred to [ilmath]1[/ilmath] if the metric needs to be explicitly stated. 
3  [ilmath]B(x;r)[/ilmath]^{[1]}^{[2]}  Very common in books. 
TODO: Ensure all these notations have references
Reasoning for preferred notations
The subset notation stops it from looking (too much) like a function. The notation [ilmath]B_r(x)[/ilmath] makes it very clear that that there are a whole family of balls for each [ilmath]x\in X[/ilmath]. I've seen the use of semicolons abused in functions (where they are used to let it take multiple parameters, for example [ilmath]B(x,y;r)[/ilmath] say, the semicolon distinguishes the radius from the second argument ([ilmath]y[/ilmath] in this example)).
It also reads very easily.
I will however say that the notation [ilmath]B(x;r)[/ilmath] is easier if you want to explicitly mention a metric, eg [ilmath]B_d(x;r)[/ilmath]. However this is quite a rare occurrence.
Notes about openness
Recall the definition of a topological space
Topological space
A topological space is a set [math]X[/math] coupled with a "topology", [ilmath]\mathcal{J} [/ilmath] on [math]X[/math]. We denote this by the ordered pair [ilmath](X,\mathcal{J})[/ilmath].
 A topology, [ilmath]\mathcal{J} [/ilmath] is a collection of subsets of [ilmath]X[/ilmath], [math]\mathcal{J}\subseteq\mathcal{P}(X)[/math] with the following properties^{[3]}^{[4]}^{[2]}:
 Both [math]\emptyset,X\in\mathcal{J}[/math]
 For the collection [math]\{U_\alpha\}_{\alpha\in I}\subseteq\mathcal{J}[/math] where [math]I[/math] is any indexing set, [math]\cup_{\alpha\in I}U_\alpha\in\mathcal{J}[/math]  that is it is closed under union (infinite, finite, whatever  "closed under arbitrary union")
 For the collection [math]\{U_i\}^n_{i=1}\subseteq\mathcal{J}[/math] (any finite collection of members of the topology) that [math]\cap^n_{i=1}U_i\in\mathcal{J}[/math]
 We call the elements of [ilmath]\mathcal{J} [/ilmath] "open sets", that is [ilmath]\forall S\in\mathcal{J}[S\text{ is an open set}] [/ilmath], each [ilmath]S[/ilmath] is exactly what we call an 'open set'
As mentioned above we write the topological space as [math](X,\mathcal{J})[/math]; or just [math]X[/math] if the topology on [math]X[/math] is obvious from the context.
It can be shown that every metric space gives rise to a topological space (see topology induced by a metric) and that in this topology the open balls are open.
Proof that open balls are open
Let [ilmath](X,d)[/ilmath] be a metric space, consider the ball [ilmath]B_r(x_0)[/ilmath]
Theorem: The ball [ilmath]B_r(x_0)[/ilmath] is an open set
TODO: This is easy to do, it's actually done on this page (in the old version)
See also
References
 ↑ ^{1.0} ^{1.1} Introduction to Topology  Theodore W. Gamelin & Robert Everist Greene
 ↑ ^{2.0} ^{2.1} ^{2.2} Introduction to Topology  Bert Mendelson
 ↑ Topology  James R. Munkres
 ↑ Introduction to Topological Manifolds  John M. Lee
Old page
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 Xd(a,x)\lt r\}[/math], it can be denoted several ways. I frequently encounter
[math]B_r(a)=B(a;r)=\{x\in Xd(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=\epsilond(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=\epsilond(x,p)[/math] so [math]d(y,x)<\epsilond(x,p)[/math]
[math]d(y,x)<\epsilond(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.