Difference between revisions of "Open ball"

From Maths
Jump to: navigation, search
m
m (Proof that open balls are open: fixed typo)
 
Line 51: Line 51:
 
Theorem: The ball {{M|B_r(x_0)}} is an [[open set]]
 
Theorem: The ball {{M|B_r(x_0)}} is an [[open set]]
 
{{Begin Proof}}
 
{{Begin Proof}}
{{Todo|This is easy to do, it's actually done on this page (in the old version)
+
{{Todo|This is easy to do, it's actually done on this page (in the old version)}}
 
{{End Proof}}
 
{{End Proof}}
 
{{End Theorem}}
 
{{End Theorem}}
 +
 
==See also==
 
==See also==
 
* [[Neighbourhood]]
 
* [[Neighbourhood]]

Latest revision as of 02:00, 29 November 2015

This page is currently being refactored (along with many others)
Please note that this does not mean the content is unreliable. It just means the page doesn't conform to the style of the site (usually due to age) or a better way of presenting the information has been discovered.

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.
  • 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)).

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 open-ness

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]:
  1. Both [math]\emptyset,X\in\mathcal{J}[/math]
  2. 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")
  3. 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. 1.0 1.1 Introduction to Topology - Theodore W. Gamelin & Robert Everist Greene
  2. 2.0 2.1 2.2 Introduction to Topology - Bert Mendelson
  3. Topology - James R. Munkres
  4. 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 X|d(a,x)\lt r\}[/math], it can be denoted several ways. I frequently encounter

[math]B_r(a)=B(a;r)=\{x\in X|d(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=\epsilon-d(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=\epsilon-d(x,p)[/math] so [math]d(y,x)<\epsilon-d(x,p)[/math]

[math]d(y,x)<\epsilon-d(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.


See Also