Bounded Subspace of Euclidean Space is Totally Bounded

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $\struct {\R^n, \norm {\, \cdot \,} }$ be a Euclidean space, where $\norm {\, \cdot \,}$ denotes the usual metric.

Let $M$ be a metric subspace of $\struct {\R^n, \norm {\, \cdot \,} }$.

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.

\(\displaystyle \norm {y - x}\) \(=\) \(\displaystyle \sqrt {\sum_{i \mathop = 1}^n \paren {y_i - x_i}^2}\)
\(\displaystyle \) \(\le\) \(\displaystyle \sqrt {n \max_i \set {\paren {y_i - x_i}^2} }\)
\(\displaystyle \) \(=\) \(\displaystyle \sqrt n \max_i \size {y_i - x_i}\)

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

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

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

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

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


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

Consider a normal subdivision of $\closedint {c - R} {c + R}$ into $k$ pieces.

The length of each subinterval is then:

$\dfrac {\paren {c + R - \paren {c - R} } } k = 2 R k^{-1}$

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 $2 R k^{-1} \sqrt n$

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

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.

$\blacksquare$


Sources