Difference between revisions of "Discrete metric and topology"
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==Metric space definition== | ==Metric space definition== | ||
{{:Discrete metric and topology/Metric space definition}} | {{:Discrete metric and topology/Metric space definition}} | ||
| − | ==Open balls== | + | ===Open balls=== |
The [[Open ball|open balls]] of {{M|X}} with the discrete topology are entirely {{M|X}} or a single point, that is: | The [[Open ball|open balls]] of {{M|X}} with the discrete topology are entirely {{M|X}} or a single point, that is: | ||
{{Begin Theorem}} | {{Begin Theorem}} | ||
| Line 13: | Line 13: | ||
This completes the proof | This completes the proof | ||
{{End Proof}}{{End Theorem}} | {{End Proof}}{{End Theorem}} | ||
| − | ==Open sets== | + | ===Open sets=== |
The [[Open set|open sets]] of {{M|(X,d_\text{discrete})}} consist of every subset of {{M|X}} (the [[Power set|power set]] of {{M|X}}) - this is how the [[Topology induced by a metric|topology induced by the metric]] may be denoted {{M|(X,\mathcal{P}(X))}} | The [[Open set|open sets]] of {{M|(X,d_\text{discrete})}} consist of every subset of {{M|X}} (the [[Power set|power set]] of {{M|X}}) - this is how the [[Topology induced by a metric|topology induced by the metric]] may be denoted {{M|(X,\mathcal{P}(X))}} | ||
{{Begin Theorem}} | {{Begin Theorem}} | ||
| Line 19: | Line 19: | ||
{{Begin Proof}} | {{Begin Proof}} | ||
: Let {{M|A}} be a subset of {{M|X}}, we will show that {{M|\forall x\in A\exists r>0[B_r(x)\subseteq A]}} | : Let {{M|A}} be a subset of {{M|X}}, we will show that {{M|\forall x\in A\exists r>0[B_r(x)\subseteq A]}} | ||
| − | {{ | + | :* Let {{M|x\in A}} be given |
| + | :** Choose {{M|1=r=\tfrac{1}{2} }} | ||
| + | :**: Now {{M|1=B_r(x)=\{x\}\subseteq A}} | ||
| + | :**:* This must be true as we know already that {{M|x\in A}} (to show this formally use the [[Implies and subset relation|implies-subset relation]]) | ||
| + | :** We have shown that given an {{M|x\in A}} we can find an open ball about {{M|x}} entirely contained within {{M|A}} | ||
| + | :* We have shown that for any {{M|x\in A}} we can find an open ball about {{M|x}} entirely contained within {{M|A}} | ||
| + | This completes the proof | ||
{{End Proof}}{{End Theorem}} | {{End Proof}}{{End Theorem}} | ||
| + | ==Discrete topology== | ||
| + | The ''discrete topology'' on {{M|X}} is the [[Topological space|topology]] that considers every subset to be open. We may write {{M|X}} imbued with the discrete topology as: | ||
| + | * {{M|(X,\mathcal{P}(X))}} where {{M|\mathcal{P} }} denotes [[Power set|power set]] | ||
| + | {{Todo|find reference - even though it is obvious as I show above that every subset is open}} | ||
==Notes== | ==Notes== | ||
<references group="Note"/> | <references group="Note"/> | ||
Revision as of 20:54, 23 July 2015
Contents
Metric space definition
Let [ilmath]X[/ilmath] be a set. The discrete[1] metric, or trivial metric[2] is the metric defined as follows:
- [math]d:X\times X\rightarrow \mathbb{R}_{\ge 0} [/math] with [math]d:(x,y)\mapsto\left\{\begin{array}{lr}0 & \text{if }x=y \\1 & \text{otherwise}\end{array}\right.[/math]
However any strictly positive value will do for the [ilmath]x\ne y[/ilmath] case. For example we could define [ilmath]d[/ilmath] as:
- [math]d:(x,y)\mapsto\left\{\begin{array}{lr}0 & \text{if }x=y \\v & \text{otherwise}\end{array}\right.[/math]
- Where [ilmath]v[/ilmath] is some arbitrary member of [ilmath]\mathbb{R}_{> 0} [/ilmath][Note 1] - traditionally (as mentioned) [ilmath]v=1[/ilmath] is used.
- Where [ilmath]v[/ilmath] is some arbitrary member of [ilmath]\mathbb{R}_{> 0} [/ilmath][Note 1] - traditionally (as mentioned) [ilmath]v=1[/ilmath] is used.
Note: however in proofs we shall always use the case [ilmath]v=1[/ilmath] for simplicity
Open balls
The open balls of [ilmath]X[/ilmath] with the discrete topology are entirely [ilmath]X[/ilmath] or a single point, that is:
- [math]B_r(x):=\{p\in X\vert\ d(x,p)<r\}=\left\{\begin{array}{lr}\{x\} & \text{for }r\le 1\\ X & \text{otherwise}\end{array}\right.[/math]
- By definition [math]B_r(x):=\{p\in X\vert\ d(x,p)<r\}[/math] note that for:
- [ilmath]r\le 1[/ilmath] we have [ilmath]B_r(x)=\{x\}[/ilmath] as
- [ilmath]d(x,p)< r \le 1[/ilmath] so [ilmath]d(x,p)<1[/ilmath] only when [ilmath]x=p[/ilmath], as if [ilmath]x\ne p[/ilmath] then [ilmath]d(x,p)=1\not<1[/ilmath] (proof by contrapositive)
- [ilmath]r> 1[/ilmath] we have [ilmath]B_r(X)=X[/ilmath] as
- [ilmath]d(x,y)\le 1[/ilmath] always, so we have [ilmath]\forall x,y\in X[d(x,y)\le 1< r][/ilmath] so [ilmath]\forall x,y\in X[d(x,y)< r][/ilmath] thus the ball contains every point in [ilmath]X[/ilmath]
- [ilmath]r\le 1[/ilmath] we have [ilmath]B_r(x)=\{x\}[/ilmath] as
This completes the proof
Open sets
The open sets of [ilmath](X,d_\text{discrete})[/ilmath] consist of every subset of [ilmath]X[/ilmath] (the power set of [ilmath]X[/ilmath]) - this is how the topology induced by the metric may be denoted [ilmath](X,\mathcal{P}(X))[/ilmath]
Every subset of [ilmath]X[/ilmath] is an open set
- Let [ilmath]A[/ilmath] be a subset of [ilmath]X[/ilmath], we will show that [ilmath]\forall x\in A\exists r>0[B_r(x)\subseteq A][/ilmath]
- Let [ilmath]x\in A[/ilmath] be given
- Choose [ilmath]r=\tfrac{1}{2}[/ilmath]
- Now [ilmath]B_r(x)=\{x\}\subseteq A[/ilmath]
- This must be true as we know already that [ilmath]x\in A[/ilmath] (to show this formally use the implies-subset relation)
- Now [ilmath]B_r(x)=\{x\}\subseteq A[/ilmath]
- We have shown that given an [ilmath]x\in A[/ilmath] we can find an open ball about [ilmath]x[/ilmath] entirely contained within [ilmath]A[/ilmath]
- Choose [ilmath]r=\tfrac{1}{2}[/ilmath]
- We have shown that for any [ilmath]x\in A[/ilmath] we can find an open ball about [ilmath]x[/ilmath] entirely contained within [ilmath]A[/ilmath]
- Let [ilmath]x\in A[/ilmath] be given
This completes the proof
Discrete topology
The discrete topology on [ilmath]X[/ilmath] is the topology that considers every subset to be open. We may write [ilmath]X[/ilmath] imbued with the discrete topology as:
- [ilmath](X,\mathcal{P}(X))[/ilmath] where [ilmath]\mathcal{P} [/ilmath] denotes power set
TODO: find reference - even though it is obvious as I show above that every subset is open
Notes
- ↑ Note the strictly greater than 0 requirement for [ilmath]v[/ilmath]
References
- ↑ Introduction to Topology - Theodore W. Gamelin & Robert Everist Greene
- ↑ Functional Analysis - George Bachman and Lawrence Narici
