View source for Discrete metric and topology

==Metric space definition==
{{:Discrete metric and topology/Metric space definition}}
==Open balls==
The [[Open ball|open balls]] of {{M|X}} with the discrete topology are entirely {{M|X}} or a single point, that is:
{{Begin Theorem}}
* {{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.}}
{{Begin Proof}}
: By definition {{MM|1=B_r(x):=\{p\in X\vert\ d(x,p)<r\} }} note that for:
:* {{M|1=r\le 1}} we have {{M|1=B_r(x)=\{x\} }} as
:**  {{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]])
:* {{M|1=r> 1}} we have {{M|1=B_r(X)=X}} as
:** {{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}}
This completes the proof
{{End Proof}}{{End Theorem}}
==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))}}
{{Begin Theorem}}
Every subset of {{M|X}} is an open set
{{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]}}
{{Todo|Do this, it's easy enough see [[Metric space]] page for outline}}
{{End Proof}}{{End Theorem}}

==Notes==
<references group="Note"/>
==References==
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{{Definition|Topology|Metric Space}}