# 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)

## Contents

## 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^{[Note 1]}:

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

## 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

**This proof has been marked as an page requiring an easy proof**[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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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**## Notes

- ↑ Just in case the reader isn't sure what this means, if [ilmath]A[/ilmath] and [ilmath]B[/ilmath] are logically equivalent then:
- [ilmath]A\iff B[/ilmath]. In words "[ilmath]A[/ilmath]
*if and only if*[ilmath]B[/ilmath]"

- [ilmath]A\iff B[/ilmath]. In words "[ilmath]A[/ilmath]

## References

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