Bounded set

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Definition

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

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

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

Equivalent conditions

Let $(X,d)$ be a metric space and let $A\in\mathcal{P}(X)$ be an arbitrary subset of $X$. Then the following are all logical equivalent to each other:

1. $\exists C<\infty\ \forall a,b\in A[d(a,b)<C]$ - $A$ is bounded (the definition)
2. $\forall x\in X\exists C<\infty\forall a\in A[d(a,x)<C]$

See also

• Diameter (set)

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Notes

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

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Of $\mathbb{R}^n$

Given a set $A\subseteq\mathbb{R}^n$ we say $A$ is bounded if:

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

Immediate results

[Expand]

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

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This follows right from the definitions, the $K$ is the bound.

Proof: $\implies$

If such a $K$ exists then it is the bound. We are done.

Proof: $\impliedby$

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

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

[Expand]

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

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Choose $K=\text{Max}(\vert a\vert,\vert b\vert)$ - result follows

See also

• Topological property theorems