Difference between revisions of "Discrete metric and topology"

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(Created page with "==Metric space definition== {{:Discrete metric and topology/Metric space definition}} {{Definition|Topology|Metric Space}}")
 
m (Added related theorem)
 
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__TOC__
 
==Metric space definition==
 
==Metric space definition==
 
{{:Discrete metric and topology/Metric space definition}}
 
{{:Discrete metric and topology/Metric space definition}}
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==Metric summary==
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{{:Discrete metric and topology/Summary}}
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==Metric objects==
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===Open balls===
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The [[Open ball|open balls]] of {{M|X}} with the discrete topology are entirely {{M|X}} or a single point, that is:
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{{Begin Theorem}}
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* {{MM|1=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.}}
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{{Begin Proof}}
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: By definition {{MM|1=B_r(x):=\{p\in X\vert\ d(x,p)<r\} }} note that for:
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:* {{M|1=r\le 1}} we have {{M|1=B_r(x)=\{x\} }} as
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:**  {{M|1=d(x,p)< r \le 1}} so {{M|1=d(x,p)<1}} only when {{M|1=x=p}}, as if {{M|x\ne p}} then {{M|1=d(x,p)=1\not<1}} (proof by [[Contrapositive|contrapositive]])
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:* {{M|1=r> 1}} we have {{M|1=B_r(X)=X}} as
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:** {{M|d(x,y)\le 1}} always, so we have {{M|\forall x,y\in X[d(x,y)\le 1< r]}} so {{M|\forall x,y\in X[d(x,y)< r]}} thus the ball contains every point in {{M|X}}
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This completes the proof
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{{End Proof}}{{End Theorem}}
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===Open sets===
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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))}}
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{{Begin Theorem}}
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Every subset of {{M|X}} is an open set
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{{Begin Proof}}
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: 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]}}
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:* Let {{M|x\in A}} be given
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:** Choose {{M|1=r=\tfrac{1}{2} }}
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:**: Now {{M|1=B_r(x)=\{x\}\subseteq A}}
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:**:* 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]])
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:** We have shown that given an {{M|x\in A}} we can find an open ball about {{M|x}} entirely contained within {{M|A}}
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:* We have shown that for any {{M|x\in A}} we can find an open ball about {{M|x}} entirely contained within {{M|A}}
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This completes the proof
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{{End Proof}}{{End Theorem}}
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==Discrete topology==
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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:
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* {{M|(X,\mathcal{P}(X))}} where {{M|\mathcal{P} }} denotes [[Power set|power set]]
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{{Todo|find reference - even though it is obvious as I show above that every subset is open}}
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==Related theorems==
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* [[Every map from a space with the discrete topology is continuous]]
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* [[Every continuous map from a non-empty connected space to a discrete space is constant]]
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==Notes==
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<references group="Note"/>
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==References==
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<references/>
  
 
{{Definition|Topology|Metric Space}}
 
{{Definition|Topology|Metric Space}}

Latest revision as of 18:10, 25 April 2017

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.

Note: however in proofs we shall always use the case [ilmath]v=1[/ilmath] for simplicity

Metric summary

Property Comment
induced topology discrete topology - which is the topology [ilmath](X,\mathcal{P}(X))[/ilmath] (where [ilmath]\mathcal{P} [/ilmath] denotes power set)
Open ball [ilmath]B_r(x):=\{p\in X\vert\ d(p,x)< r\}=\left\{\begin{array}{lr}\{x\} & \text{if }r\le 1 \\ X & \text{otherwise}\end{array}\right.[/ilmath]
Open sets Every subset of [ilmath]X[/ilmath] is open.
Proof outline: as for a subset [ilmath]A\subseteq X[/ilmath] we can show [ilmath]\forall x\in A\exists r[B_r(x)\subseteq A][/ilmath] by choosing say, that is [ilmath]A[/ilmath] contains an open ball centred at each point in [ilmath]A[/ilmath].
Connected The topology generated by [ilmath](X,d_\text{discrete})[/ilmath] is not connected if [ilmath]X[/ilmath] has more than one point.
Proof outline:
  • Let [ilmath]A[/ilmath] be any non empty subset of [ilmath]X[/ilmath], then define [ilmath]B:=A^c[/ilmath] which is also a subset of [ilmath]X[/ilmath], thus [ilmath]B[/ilmath] is open. Then [ilmath]A\cap B=\emptyset[/ilmath] and [ilmath]A\cup B=X[/ilmath] thus we have found a separation, a partition of non-empty disjoint open sets, that separate the space. Thus it is not connected
  • if [ilmath]X[/ilmath] has only one point then we cannot have a partition of non empty disjoint sets. Thus it cannot be not connected, it is connected.

Metric objects

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]

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)
    • 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]
  • 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]

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


Related theorems

Notes

  1. ↑ Note the strictly greater than 0 requirement for [ilmath]v[/ilmath]

References

  1. ↑ Introduction to Topology - Theodore W. Gamelin & Robert Everist Greene
  2. ↑ Functional Analysis - George Bachman and Lawrence Narici
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