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

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

Proof of claims

[<collapsible-expand>]

$1\implies 2)$ $\big(\exists C<\infty\ \forall a,b\in A[d(a,b)<C]\big)\implies\big(\forall x\in X\exists C<\infty\forall a\in A[d(a,x)<C]\big)$, 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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[<collapsible-expand>]

$2\implies 1)$ $\big(\forall x\in X\exists C<\infty\forall a\in A[d(a,x)<C]\big)\implies \big(\exists C<\infty\ \forall a,b\in A[d(a,b)<C]\big)$, 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).
The message provided is:

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