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:

• [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:

• "If there exists an [ilmath]x\in X[/ilmath] and [ilmath]C\in\mathbb{R}_{>0} [/ilmath] such that for all [ilmath]a\in A[/ilmath] we have [ilmath]d(a,x)<C[/ilmath] then [ilmath]A[/ilmath] is (metrically) bounded" or symbolically:
  • [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]

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.
         • 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
      • 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