Equivalent conditions to a set being bounded

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Cleanup required. New Metrically bounded set page could link to this in another form. Make sure the two are compatible Alec (talk) 23:12, 18 March 2017 (UTC)

Statement

Let [ilmath](X,d)[/ilmath] be a metric space and let [ilmath]A\in\mathcal{P}(X)[/ilmath] be an arbitrary subset of [ilmath]X[/ilmath]. Then the following are all logical equivalent to each other:

1. [ilmath]\exists C<\infty\ \forall a,b\in A[d(a,b)<C][/ilmath] - [ilmath]A[/ilmath] is bounded (the definition)
2. [ilmath]\forall x\in X\exists C<\infty\forall a\in A[d(a,x)<C][/ilmath]

Proof of claims

[ilmath]1\implies 2)[/ilmath] [ilmath]\big(\exists C<\infty\ \forall a,b\in A[d(a,b)<C]\big)\implies\big(\forall x\in X\exists C<\infty\forall a\in A[d(a,x)<C]\big)[/ilmath], that boundedness implies condition 2

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Easy and routine proof. If stuck see page 13 in Functional Analysis, Dzung Minh Ha

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[ilmath]2\implies 1)[/ilmath] [ilmath]\big(\forall x\in X\exists C<\infty\forall a\in A[d(a,x)<C]\big)\implies \big(\exists C<\infty\ \forall a,b\in A[d(a,b)<C]\big)[/ilmath], that condition 2 implies boundedness

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Grade: C

This page requires one or more proofs to be filled in, it is on a to-do list for being expanded with them.

Please note that this does not mean the content is unreliable. Unless there are any caveats mentioned below the statement comes from a reliable source. As always, Warnings and limitations will be clearly shown and possibly highlighted if very important (see template:Caution et al).
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Easy and routine proof. If stuck see page 13 in Functional Analysis, Dzung Minh Ha

This proof has been marked as an page requiring an easy proof