# Bounded Subspace of Euclidean Space is Totally Bounded

## Theorem

Let $\left({\R^n,\Vert \cdot \Vert}\right)$ be a Euclidean space, where $\Vert \cdot \Vert$ is the usual metric.

Let $M$ be a metric subspace of the Euclidean space.

Let $M$ be bounded.

Then $M$ is totally bounded.

## Proof

As $M$ is bounded, it has a finite diameter $R$.

Consider arbitary $x,y \in \R^n$, expressed in their usual coordinates.

 $\displaystyle \left \Vert { y - x }\right \Vert$ $=$ $\displaystyle \sqrt{\sum_{i \mathop = 1}^n \left({y_i - x_i}\right)^2}$ $\displaystyle$ $\le$ $\displaystyle \sqrt{n \max_i \left\{ { \left({y_i - x_i}\right)^2 }\right\} }$ $\displaystyle$ $=$ $\displaystyle \sqrt n \max_i \left \vert y_i - x_i \right \vert$

Thus, up to a positive constant multiple, the inequality:

$\displaystyle \sup_{x,y \mathop \in M} \max_i \left \vert y_i - x_i \right \vert \le R$

implies that $M$ is bounded with radius $R$.

But this inequality is a characterization of an $n$-cube.

Because $R \le 2R\sqrt{n}$ for all $R > 0$, $M$ can be placed inside an $n$-cube of the form $\left[ {c-R\,.\,.\,c+R} \right]^n$, by diameter of $n$-cube.

For any $\epsilon > 0$, pick an integer $k > \sqrt{n}R \epsilon^{-1}$

Consider a normal subdivision of $\left[ {c-R\,.\,.\,c+R} \right]$ into $k$ pieces.

The length of each subinterval is then:

$\dfrac{\left({c+R - (c-R)}\right)} {k} = 2Rk^{-1}$

For any subinterval chosen this way, the product of $n$ copies of such an interval will itself be a cube, because the subintervals were chosen as to all be the same length.

By diameter of $n$-cube, each sub-cube has diameter $2Rk^{-1}\sqrt{n}$

By the choice of $k$, we have $2Rk^{-1}\sqrt{n} < 2\epsilon$

That means that each sub-cube can fit inside an open ball $n$-ball of radius $\epsilon$.

This in turn implies that $M$ can be covered by a finite number of balls with radius $\epsilon$.

Thus $M$ is totally bounded by definition, as $\epsilon$ was arbitrary.

$\blacksquare$