Difference between revisions of "Metric space"

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Revision as of 20:33, 23 July 2015

A normed space is a special case of a metric space, to see the relationships between metric spaces and others see: Subtypes of topological spaces

Definition of a metric space

A metric space is a set [math]X[/math] coupled with a "distance function"[1]:

  • [math]d:X\times X\rightarrow\mathbb{R}[/math] or sometimes
  • [math]d:X\times X\rightarrow\mathbb{R}_+[/math][2], Note that here I prefer the notation [math]d:X\times X\rightarrow\mathbb{R}_{\ge 0}[/math]

With the properties that for [math]x,y,z\in X[/math]:

  1. [math]d(x,y)\ge 0[/math] (This is implicit with the [ilmath]d:X\times X\rightarrow\mathbb{R}_{\ge 0}[/ilmath] definition)
  2. [math]d(x,y)=0\iff x=y[/math]
  3. [math]d(x,y)=d(y,x)[/math] - Symmetry
  4. [math]d(x,z)\le d(x,y)+d(y,z)[/math] - the Triangle inequality

We will denote a metric space as [math](X,d)[/math] (as [math](X,d:X\times X\rightarrow\mathbb{R}_{\ge 0})[/math] is too long and Mathematicians are lazy) or simply [math]X[/math] if it is obvious which metric we are talking about on [math]X[/math]

Examples of metrics

Euclidian Metric

The Euclidian metric on [math]\mathbb{R}^n[/math] is defined as follows: For [math]x=(x_1,...,x_n)\in\mathbb{R}^n[/math] and [math]y=(y_1,...,y_n)\in\mathbb{R}^n[/math] we define the Euclidian metric by:

[math]d_{\text{Euclidian}}(x,y)=\sqrt{\sum^n_{i=1}((x_i-y_i)^2)}[/math]

Proof that this is a metric




TODO:



Discrete Metric

Let [ilmath]X[/ilmath] be a set. The discrete[3] metric, or trivial metric[4] 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

Notes

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.

See also

Notes

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

References

  1. Introduction to Topology - Bert Mendelson
  2. Analysis - Part 1: Elements - Krzysztof Maurin
  3. Introduction to Topology - Theodore W. Gamelin & Robert Everist Greene
  4. Functional Analysis - George Bachman and Lawrence Narici