View source for Discrete metric and topology/Summary

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! Property
! Comment
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! [[Topology induced by a metric|induced topology]]
| [[Discrete topology|discrete topology]] - which is the topology {{M|(X,\mathcal{P}(X))}} (where {{M|\mathcal{P} }} denotes [[Power set|power set]])
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! [[Open ball]]
| {{M|1=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. }}
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! [[Open set|Open sets]]
| Every subset of {{M|X}} is open.<br/>'''Proof outline:''' as for a subset {{M|A\subseteq X}} we can show {{M|1=\forall x\in A\exists r[B_r(x)\subseteq A] }} by choosing {{M|r=\tfrac{1}{2} }} say, that is {{M|A}} contains an ''[[Open ball|open ball]]'' centred at each point in {{M|A}}.
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! [[Connected space|Connected]]
| The [[Topological space|topology]] generated by {{M|(X,d_\text{discrete})}} is '''not''' connected if {{M|X}} has more than one point.<br/>'''Proof outline:'''
:* Let {{M|A}} be any non empty subset of {{M|X}}, then define {{M|1=B:=A^c}} which is also a subset of {{M|X}}, thus {{M|B}} is open. Then {{M|1=A\cap B=\emptyset}} and {{M|1=A\cup B=X}} thus we have found a separation, a partition of non-empty disjoint open sets, that separate the space. Thus it is ''not connected''
:* if {{M|X}} has only one point then we cannot have a partition of non empty disjoint sets. Thus it cannot be not connected, it is connected.
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{{Theorem Of|Topology|Metric Space}}
{{Definition|Topology|Metric Space}}</noinclude>