Latest revision as of 02:00, 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

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

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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