# Open ball

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## 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]:

• $B_r(x_0):=\{x\in X\vert\ d(x,x_0)<r\}$ - 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

1 [ilmath]B_r(x)[/ilmath] preferred Use if the metric is implicit.
• Very common in lecture notes, less common in books (weirdly)
• Very easy to read, eg [ilmath]B_\delta(x_0)[/ilmath] becomes "the ball of radius delta at [ilmath]x_0[/ilmath]..."
• The subscript radius feels very familiar in proofs, eg [ilmath]B_{\delta_1}(x)[/ilmath], [ilmath]B_{\delta_2}(x)[/ilmath] is very easy to see as "a ball of radius [ilmath]\delta_1[/ilmath] and another of [ilmath]\delta_2[/ilmath]", rather than the other "functional" notations that look like a function that returns an open ball.
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 semi-colons abused in functions (where they are used to let it take multiple parameters, for example [ilmath]B(x,y;r)[/ilmath] say, the semi-colon distinguishes the radius from the second argument ([ilmath]y[/ilmath] in this example)).

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.

Recall the definition of a topological space

### Topological space

A topological space is a set $X$ coupled with a "topology", [ilmath]\mathcal{J} [/ilmath] on $X$. 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], $\mathcal{J}\subseteq\mathcal{P}(X)$ with the following properties[3][4][2]:
1. Both $\emptyset,X\in\mathcal{J}$
2. For the collection $\{U_\alpha\}_{\alpha\in I}\subseteq\mathcal{J}$ where $I$ is any indexing set, $\cup_{\alpha\in I}U_\alpha\in\mathcal{J}$ - that is it is closed under union (infinite, finite, whatever - "closed under arbitrary union")
3. For the collection $\{U_i\}^n_{i=1}\subseteq\mathcal{J}$ (any finite collection of members of the topology) that $\cap^n_{i=1}U_i\in\mathcal{J}$
• 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 $(X,\mathcal{J})$; or just $X$ if the topology on $X$ 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)

# Old page

## 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.