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		+	{{Refactor notice}}
		+	==Definition==
		+	Given a [[metric space]] {{M|(X,d)}} the ''open ball centred at {{M|x_0\in X}} of radius {{M|r>0}}'', denoted {{M|B_r(x_0)}} (however many notations are used, see below), is given by{{rITTGG}}{{rITTBM}}:
		+	* {{MM|1=B_r(x_0):=\{x\in X\vert\ d(x,x_0)<r\} }} - that is all the points of {{M|X}} that are a distance (given by {{M|d}}) strictly less than {{M|r}} from {{M|x_0}}
		+	{{Note|1=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 {{M|r}} centred at {{M|x}} (in a [[metric space]] {{M|(X,d)}}, this table is supposed to be complete, so preferred notations are marked from the others
		+	{| class="wikitable" border="1"
		+	|-
		+	! #
		+	! Notation
		+	! Usage
		+	! Comments
		+	|-
		+	! 1
		+	| {{M|B_r(x)}}
		+	| '''preferred'''
		+	| Use if the metric is implicit.
		+	* Very common in lecture notes, less common in books (weirdly)
		+	* Very easy to read, eg {{M|B_\delta(x_0)}} becomes "the ball of radius delta at {{M|x_0}}..."
		+	* The subscript radius feels very familiar in proofs, eg {{M|B_{\delta_1}(x)}}, {{M|B_{\delta_2}(x)}} is very easy to see as "a ball of radius {{M|\delta_1}} and another of {{M|\delta_2}}", rather than the other "functional" notations that look like a function that returns an open ball.
		+	|-
		+	! 2
		+	| {{M|B_{r,d}(x)}}
		+	| '''preferred'''
		+	| Preferred to {{M|1}} if the metric needs to be explicitly stated.
		+	|-
		+	! 3
		+	| {{M|B(x;r)}}<ref name="ITTGG"/><ref name="ITTBM"/>
		+	|
		+	| 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 {{M|B_r(x)}} makes it very clear that that there are a whole family of balls for each {{M|x\in X}}. I've seen the use of semi-colons abused in functions (where they are used to let it take multiple parameters, for example {{M|B(x,y;r)}} say, the semi-colon distinguishes the radius from the second argument ({{M|y}} in this example)).
		+	It also reads very easily.
		+
		+	I will however say that the notation {{M|B(x;r)}} is easier if you want to explicitly mention a metric, eg {{M|B_d(x;r)}}. However this is quite a rare occurrence.
		+
		+
		+	==Notes about open-ness==
		+	Recall the definition of a [[topological space]]
		+	===[[Topological space]]===
		+	{{:Topological space/Definition}}
		+	<div style="width:30%;"><hr/></div>
		+	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 {{M|(X,d)}} be a [[metric space]], consider the ball {{M|B_r(x_0)}}
		+	{{Begin Theorem}}
		+	Theorem: The ball {{M|B_r(x_0)}} is an [[open set]]
		+	{{Begin Proof}}
		+	{{Todo|This is easy to do, it's actually done on this page (in the old version)
		+	{{End Proof}}
		+	{{End Theorem}}
		+	==See also==
		+	* [[Neighbourhood]]
		+	* [[Open set]]
		+	* [[Metric space]]
		+	* [[Closed ball]]
		+	* [[Closed set]]
		+	* [[Interior]]
		+	* [[Interior point]]
		+
		+	==References==
		+	<references/>
		+
		+	{{Definition|Metric Space|Topology|Functional Analysis}}
		+
		+	=Old page=
	==Definition==		==Definition==
	For a [[Metric space|metric space]] <math>(X,d)</math> an "open ball" of radius <math>r</math> centred at <math>a</math> is the set		For a [[Metric space|metric space]] <math>(X,d)</math> an "open ball" of radius <math>r</math> centred at <math>a</math> is the set

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Revision as of 01:59, 29 November 2015

Definition

Given a metric space $(X,d)$ the open ball centred at $x_0\in X$ of radius $r>0$, denoted $B_r(x_0)$ (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 $X$ that are a distance (given by $d$) strictly less than $r$ from $x_0$

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 $r$ centred at $x$ (in a metric space $(X,d)$, this table is supposed to be complete, so preferred notations are marked from the others

#	Notation	Usage	Comments
1	$B_r(x)$	preferred	Use if the metric is implicit. Very common in lecture notes, less common in books (weirdly) Very easy to read, eg $B_\delta(x_0)$ becomes "the ball of radius delta at $x_0$..." The subscript radius feels very familiar in proofs, eg $B_{\delta_1}(x)$, $B_{\delta_2}(x)$ is very easy to see as "a ball of radius $\delta_1$ and another of $\delta_2$", rather than the other "functional" notations that look like a function that returns an open ball.
2	$B_{r,d}(x)$	preferred	Preferred to $1$ if the metric needs to be explicitly stated.
3	$B(x;r)$[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 $B_r(x)$ makes it very clear that that there are a whole family of balls for each $x\in X$. I've seen the use of semi-colons abused in functions (where they are used to let it take multiple parameters, for example $B(x,y;r)$ say, the semi-colon distinguishes the radius from the second argument ($y$ in this example)).

It also reads very easily.

I will however say that the notation $B(x;r)$ is easier if you want to explicitly mention a metric, eg $B_d(x;r)$. However this is quite a rare occurrence.

Notes about open-ness

Recall the definition of a topological space

Topological space

A topological space is a set

$$X$$ coupled with a "topology", $\mathcal{J}$ on $$X$$ . We denote this by the ordered pair $(X,\mathcal{J})$.

• A topology, $\mathcal{J}$ is a collection of subsets of $X$, $$\mathcal{J}\subseteq\mathcal{P}(X)$$ 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 $\mathcal{J}$ "open sets", that is $\forall S\in\mathcal{J}[S\text{ is an open set}]$, each $S$ 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 $(X,d)$ be a metric space, consider the ball $B_r(x_0)$

See also

• Neighbourhood
• Open set
• Metric space
• Closed ball
• Closed set
• Interior
• Interior point

References

1. ↑ ^{Jump up to: 1.0 1.1} Introduction to Topology - Theodore W. Gamelin & Robert Everist Greene
2. ↑ ^{Jump up to: 2.0 2.1 2.2} Introduction to Topology - Bert Mendelson
3. Jump up ↑ Topology - James R. Munkres
4. Jump up ↑ Introduction to Topological Manifolds - John M. Lee

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.

See Also

• Open set