View source for Given two open balls sharing the same centre but with differing radius then the one defined to have a strictly smaller radius is contained in the other

{{Stub page|grade=A*|msg=I cannot be bothered, the outline is here. 
* Precursor to: [[For any two open balls sharing the same centre point then one or both of them contains the other]]
* Demote to {{C|B}} once links have been added}}
==Statement==
Let {{M|(X,d)}} be a metric space. Let {{M|x\in X}} be given and let {{M|r_1,r_2\in\mathbb{R}_{>0} }} be given such that {{M|r_1<r_2}}. Then:
* {{M|B_{r_1}(x)\subseteq B_{r_2}(x)}} (where {{M|B_\epsilon(x)}} denotes the [[open ball]] of radius {{M|\epsilon>0}} centred at {{M|x\in X}})
==Caveat==
It would perhaps be better phrased as follows:
* {{M|\forall x\in X\forall r_1,r_2\in\mathbb{R}_{>0}[(r_1<r_2)\implies B_{r_1}(x)\subseteq B_{r_2}(x)]}}
{{XXX|Adapt the proof to the caveat}} - this basically means saying "suppose {{M|r_1\ge r_2}}, then by the nature of [[logical implication]] the statement holds regardless of the truth or falsity of the right hand side" then proceeding to the other case. As usual.
==Proof==
* Let {{M|x\in X}} be given.
** Let {{M|r_1,r_2\in\mathbb{R}_{>0} }} be given such that {{M|r_1<r_2}}
*** We need to show: {{M|B_{r_1}(x)\subseteq B_{r_2}(x)}}
**** By the [[implies-subset relation]] this is equivalent to: {{M|\forall p\in B_{r_1}(x)[p\in B_{r_2}(x)]}}
*** Proof of: {{M|\forall p\in B_{r_1}(x)[p\in B_{r_2}(x)]}}
**** Let {{M|p\in B_{r_1}(x)}} be given
***** Then by definition of {{M|p}} we see {{M|d(p,x)<r_1}}
***** But {{M|r_1<r_2}}
***** By [[transitivity]] of a [[strict partial order]] we see:
****** {{M|d(p,x)<r_1<r_2}} so just: {{M|d(p,x)<r_2}}
***** But {{M|[d(p,x)<r_2]\iff[p\in B_{r_2}(x)]}} 
***** So we see that {{M|p\in B_{r_2}(x)}} - as required.
**** Since {{M|p\in B_{r_1}(x)}} was arbitrary we have shown it for all
*** We have now shown {{M|B_{r_1}(x)\subseteq B_{r_2}(x)}}
** Since {{M|r_1}} and {{M|r_2}} were arbitrary, we have shown it for all radiuses
* Since {{M|x\in X}} was arbitrary we have shown it for all open balls of strictly differing radiuses. 
==Notes==
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==References==
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{{Theorem Of|Metric Space|Topology}}