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

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]

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TODO: Do this, it's easy enough see Metric space page for outline

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