A set is bounded if and only if for all points in the space there is a positive real such that the distance from that point to any point in the set is less than the positive real

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This is a part of the new metrically bounded set page group. It is one of the claims that forms a component of the currently not-updated Equivalent conditions to a set being bounded which should be moved to Equivalent conditions to a set being metrically bounded Alec (talk) 00:41, 19 March 2017 (UTC)
• also it actually has the proofs Alec (talk) 00:41, 19 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 we claim[1]:

• [ilmath]\big(\underbrace{\exists C\in\mathbb{R}_{>0}\forall a,b\in A[d(a,b)<C]}_{\text{Bounded definition} }\big)\iff\big(\underbrace{\forall x\in X\exists C\in\mathbb{R}_{>0}\forall a\in A[d(a,x)<C]}_{\text{Claim} }\big)[/ilmath]
• In words: [ilmath]A[/ilmath] is a metrically bounded set if and only if for all points in [ilmath]X[/ilmath] there exists a positive real [ilmath]C[/ilmath] such that the distance (as measured by the metric) between that first point and any point in the set [ilmath]A[/ilmath] is strictly less than [ilmath]C[/ilmath]

Related statement

There is an almost-corollary to this, which is very almost exactly the [ilmath]\impliedby[/ilmath] part of our statement:

This is useful as it shows we get to pick a point such that the distance between that point and any [ilmath]a\in A[/ilmath] is less than some bound; this means it is bounded, which means for all points there is a bound on the distance from that point to any point in the subset. i.e.:

• [ilmath]\big(\exists x\in X\exists C\in\mathbb{R}_{>0}\forall a\in A[d(a,x)<C]\big)\implies\big(\exists C\in\mathbb{R}_{>0}\forall a,b\in A[d(a,b)<C]\big)[/ilmath] [ilmath]\underbrace{\implies}_{\iff}\big(\forall x\in X\exists C\in\mathbb{R}_{>0}\forall a\in A[d(a,x)<C]\big) [/ilmath]

Proof

• Boundedness [ilmath]\implies[/ilmath] claim - [ilmath]\big(\exists C\in\mathbb{R}_{>0}\forall a,b\in A[d(a,b)<C]\big)\implies\big(\forall x\in X\exists C\in\mathbb{R}_{>0}\forall a\in A[d(a,x)<C]\big)[/ilmath]
• Let [ilmath]x\in X[/ilmath] be given
1. If [ilmath]A\eq\emptyset[/ilmath]:
• Define [ilmath]C[/ilmath] to be the [ilmath]C\in\mathbb{R}_{>0} [/ilmath] that exists by boundedness
• Recall from rewriting for-all and exists within set theory that [ilmath]\forall a\in A[d(a,x)<C][/ilmath] is logically equivalent to [ilmath]\forall a[a\in A\implies d(a,x)<C][/ilmath]
• Let [ilmath]a[/ilmath] be given
• We must have [ilmath]a\notin A[/ilmath] as [ilmath]A[/ilmath] is the empty set
• By the nature of logical implication we do not care whether [ilmath]d(a,x)[/ilmath] is true or false (or even defined) in this case, we say the implication holds
• So this case holds
2. If [ilmath]A\neq\emptyset[/ilmath]
• Note that for some fixed [ilmath]b\in A[/ilmath] that [ilmath]d(x,a)\le d(x,b)+d(b,a)[/ilmath] by the triangle inequality property of a metric
• Note also that [ilmath]d(b,a)<C'[/ilmath] (where [ilmath]C'\in\mathbb{R}_{>0} [/ilmath] exists by the "bounded" hypothesis such that [ilmath]\forall a,b\in A[d(a,b)<C'][/ilmath]
• Thus [ilmath]d(x,a)< d(x,b) + C[/ilmath]
• Let (or choose) [ilmath]b\in A[/ilmath] be given (and arbitrary) - there is such a [ilmath]b[/ilmath] as we are in the [ilmath]A\neq\emptyset[/ilmath] case[Note 1].
• Define [ilmath]C:\eq C'+d(x,b)[/ilmath] where [ilmath]C'\in\mathbb{R}_{>0} [/ilmath] comes from boundedness as mentioned above (which is: [ilmath]\exists C'\in\mathbb{R}_{>0}\forall a,b\in A[d(a,b)<C'][/ilmath])
• Let [ilmath]a\in A[/ilmath] be arbitrary
• Now [ilmath]d(x,a)\le d(x,b)+d(b,a)[/ilmath] by the triangle inequality property of metrics
[ilmath]\eq d(x,b) + d(a,b)[/ilmath]
[ilmath]< d(x,b) + C'[/ilmath]
[ilmath]\eq: C[/ilmath]
• Thus [ilmath]d(x,a)<C[/ilmath] - as required
• So the statement holds in the case [ilmath]A\neq\emptyset[/ilmath] also
• Since [ilmath]x\in X[/ilmath] was arbitrary we're done
• Claim [ilmath]\implies[/ilmath] boundedness - [ilmath]\big(\exists C\in\mathbb{R}_{>0}\forall a,b\in A[d(a,b)<C]\big)\impliedby\big(\forall x\in X\exists C'\in\mathbb{R}_{>0}\forall a\in A[d(a,x)<C']\big)[/ilmath] (notice the use of [ilmath]C'[/ilmath])
• This will come down to showing [ilmath]d(a,b)[/ilmath] is less than something, notice that [ilmath]d(a,b)\le d(a,x)+d(x,b)<2C'[/ilmath], should [ilmath]C'[/ilmath] be given. This is the core of our proof.
• Let [ilmath]x\in X[/ilmath] again[Note 2]
• Define [ilmath]C:\eq 2C'[/ilmath] where [ilmath]C'\in\mathbb{R}_{>0} [/ilmath] exists per our claim (as above)
• Let [ilmath]a,b\in A[/ilmath] be given
• [ilmath]d(a,b)\le d(a,x)+d(x,b)[/ilmath] by the triangle inequality property of a metric
[ilmath]< C' + C'[/ilmath]
[ilmath]\eq 2C'[/ilmath]
[ilmath]\eq: C[/ilmath]
• Thus [ilmath]d(a,b)<C[/ilmath] as required.

This completes the proof

Notes

1. See that this is both a "forall" and an "exists" qualifier, as we can choose [ilmath]b\in A[/ilmath] or let it be any member of [ilmath]A[/ilmath]. We know [ilmath]A\neq\emptyset[/ilmath] so [ilmath]X\neq\emptyset[/ilmath] here so either works
2. As the other note touches on, we can also choose (existential rather than for-all's "let") [ilmath]x\in X[/ilmath] provided [ilmath]X\neq\emptyset[/ilmath], otherwise there is no [ilmath]x[/ilmath] to choose and thus the statement would be false.