Difference between revisions of "Equivalent conditions to a set being bounded"

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Latest revision as of 23:12, 18 March 2017

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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]:

  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][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

Grade: C
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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

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

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Notes

  1. ↑ 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]"

References

  1. ↑ Functional Analysis - Volume 1: A gentle introduction - Dzung Minh Ha
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