Bounded set

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Let [ilmath](X,d)[/ilmath] be a metric space. Let [ilmath]A\in\mathcal{P}(X)[/ilmath] be an arbitrary subset of [ilmath]X[/ilmath]. Then we say "[ilmath]A[/ilmath] is bounded in [ilmath](X,d)[/ilmath]" if[1]:

  • [ilmath]\exists C<\infty\ \forall a,b\in A[d(a,b)<C][/ilmath] - where [ilmath]C[/ilmath] is real[Note 1]

If a set is not bounded it is "unbounded" (that link redirects to this line)

Equivalent conditions

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 2]:

  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]

See also


  1. [ilmath]C\in\mathbb{R}_{\ge 0} [/ilmath] should do as [ilmath]0[/ilmath] could be a bound, I suppose on a one point set?
  2. 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]"


  1. 1.0 1.1 Functional Analysis - Volume 1: A gentle introduction - Dzung Minh Ha



TODO: Surely this can be generalised to an arbitrary Metric space

Of [ilmath]\mathbb{R}^n[/ilmath]

Given a set [ilmath]A\subseteq\mathbb{R}^n[/ilmath] we say [ilmath]A[/ilmath] is bounded[1] if:

  • [ilmath]\exists K\in\mathbb{R} [/ilmath] such that [ilmath]\forall x\in A[/ilmath] (where [ilmath]x=(x_1,\cdots,x_n)[/ilmath]) we have [ilmath]\vert x_i\vert\le K[/ilmath] for [ilmath]i\in\{1,\cdots,n\} [/ilmath]

Immediate results

(Real line) [ilmath]A\subseteq[-K,K]\subset\mathbb{R} [/ilmath] (where [ilmath]K> 0[/ilmath] and [ilmath][-K,K][/ilmath] denotes a closed interval) if and only if [ilmath]A[/ilmath] is bounded.

This follows right from the definitions, the [ilmath]K[/ilmath] is the bound.

Proof: [ilmath]\implies[/ilmath]

If such a [ilmath]K[/ilmath] exists then it is the bound. We are done.

Proof: [ilmath]\impliedby[/ilmath]

If it is bounded then it is easy to see that given a bound [ilmath]K[/ilmath], [ilmath]A\subseteq[-K,K][/ilmath] (use the implies-subset relation)

Note:[1] stipuates the [ilmath]A\subseteq[-K,K]\implies[/ilmath] bounded direction only.

(Real line) Every closed interval ([ilmath][a,b][/ilmath] for [ilmath]a,b\in\mathbb{R} [/ilmath] and [ilmath]a\le b[/ilmath]) is bounded.[1]

Choose [ilmath]K=\text{Max}(\vert a\vert,\vert b\vert)[/ilmath] - result follows

See also


  1. 1.0 1.1 1.2 Introduction to topology - Bert Mendelson - Third Edition