Discrete metric and topology

Metric space definition

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

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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

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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

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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

• Every map from a space with the discrete topology is continuous
• Every continuous map from a non-empty connected space to a discrete space is constant