totally bounded subset of a metric space is bounded

Theorem 1.

Every totally bounded subset of a metric space is bounded.

Proof.

Let $K$ be a totally bounded subset of a metric space. Suppose $x,y\in K$. We will show that there exists $M>0$ such that for any x,y we have . From the definition of totally bounded, we can find an $\epsilon >0$ and a finite subset $\{x_1,x_2\mathrm{\dots },x_n\}$ of $K$ such that $K\subseteq {\bigcup }_{k=1}^{n}B(x_k,\epsilon )$, so $x\in B(x_i,\epsilon )$,$y\in B(x_l,\epsilon )$, $i,l\in \{1,2,\mathrm{\dots }n\}$. So we have that

$d(x,y)$ $\le$ $d(x,x_i)+d(x_i,x_l)+d(x_l,y)$

$\epsilon +\underset{1\le s,t\le n}{\max }d(x_s,x_t)+\epsilon =M$

∎

Title	totally bounded subset of a metric space is bounded
Canonical name	TotallyBoundedSubsetOfAMetricSpaceIsBounded
Date of creation	2013-03-22 15:25:27
Last modified on	2013-03-22 15:25:27
Owner	georgiosl (7242)
Last modified by	georgiosl (7242)
Numerical id	12
Author	georgiosl (7242)
Entry type	Theorem
Classification	msc 54E35