Heine-Cantor Theorem

Theorem

Let $M_1$ and $M_2$ be metric spaces.

Let $f: M_1 \to M_2$ be a continuous mapping.

If $M_1$ is compact, then $f$ is uniformly continuous on $M_1$.

Proof 1

Let $M_1 = \left({A_1, d_1}\right)$ and $M_2 = \left({A_2, d_2}\right)$ be metric spaces such that $M_1$ is compact.

Let $f: M_1 \to M_2$ be continuous.

Let $\epsilon > 0$.

Then by definition:

$\forall x \in M_1: \exists \delta \left({x}\right) > 0: \forall y \in M_1: d_1 \left({x, y}\right) < 2 \delta \left({x}\right) \implies d_2 \left({f \left({x}\right), f \left({y}\right)}\right) < \dfrac \epsilon 2$

The set $\left\{{N_{\delta \left({x}\right)} \left({x}\right): x \in M_1}\right\}$ is an open cover for $M_1$.

As $M_1$ is compact, there is a finite subcover $\left\{{N_{\delta \left({x_1}\right)} \left({x_1}\right), N_{\delta \left({x_2}\right)} \left({x_2}\right), \ldots, N_{\delta \left({x_r}\right)} \left({x_r}\right)}\right\}$.

Now let $\delta = \min \left\{{\delta \left({x_1}\right), \delta \left({x_2}\right), \ldots, \delta \left({x_r}\right)}\right\}$.

Consider any $x, y \in M_1$ which satisfy $d_1 \left({x, y}\right) < \delta$.

There is some $i \in \left\{{1, 2, \ldots, r}\right\}$ such that $d_1 \left({x, x_i}\right) < \delta \left({x_i}\right)$.

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle d_1 \left({y, x_i}\right)\)	\(\le\)	\(\displaystyle d_1 \left({y, x}\right) + d_1 \left({x, x_i}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Triangle Inequality
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(<\)	\(\displaystyle \delta + \delta \left({x_1}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\le\)	\(\displaystyle 2 \delta \left({x_1}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

Thus:

• From $d_1 \left({x, x_i}\right) < \delta \left({x_i}\right)$ we have:

$d_2 \left({f \left({x}\right), f \left({x_1}\right)}\right) \le \dfrac \epsilon 2$

• From $d_1 \left({y, x_i}\right) \le 2 \delta\left({x_i}\right)$ we have:

$d_2 \left({f \left({y}\right), f \left({x_1}\right)}\right) \le \dfrac \epsilon 2$

So:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle d_2 \left({f \left({x}\right), f \left({y}\right)}\right)\)	\(\le\)	\(\displaystyle d_2 \left({f \left({x}\right), f \left({x_1}\right)}\right) + d_2 \left({f \left({x_1}\right), f \left({y}\right)}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Triangle Inequality
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\le\)	\(\displaystyle \frac \epsilon 2 + \frac \epsilon 2\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \epsilon\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

$\blacksquare$

Proof 2

By Sequence of Implications of Metric Space Compactness Properties, we can assume that $M_1$ is sequentially compact.

Suppose $f: M_1 \to M_2$ is continuous but not uniformly continuous.

Then for some $\epsilon > 0$ and each number $\dfrac 1 n$, we can select points $x_n, y_n \in M_1$ such that $ d_1 \left({x_n, y_n}\right) < \dfrac 1 n$ but $d_2 \left({f \left({x_n}\right), f \left({y_n}\right)}\right) \geq \epsilon$.

Since $M_1$ is sequentially compact, some subsequence $x_{\phi \left({n}\right)}$ of $x_n$ converges to a point $x$.

Since $d_1 \left({y_{\phi \left({n}\right)}, x}\right) \leq d_1 \left({y_{\phi \left({n}\right)}, x_{\phi \left({n}\right)} }\right) + d_1 \left({x_{\phi \left({n}\right)}, x}\right) \to 0$, we see that $y_{\phi \left({n}\right)}$ converges to $x$ as well.

Since continuous functions send convergent sequences to convergent sequences (see: Limit of Image of Sequence), both $f \left({x_{\phi \left({n}\right)} }\right)$ and $f \left({y_{\phi \left({n}\right)} }\right)$ both converge to $f \left({x}\right)$, hence they must grow arbitrarily close to each other.

But this contradicts the fact that we chose $f \left({x_n}\right)$ and $f \left({y_n}\right)$ to always be at least $\epsilon$ apart.

$\blacksquare$

Warning

If a mapping is uniformly continuous it is not necessarily compact.

For example, the identity mapping is (trivially) uniformly continuous on any metric space, whether compact or not.

Source of Name

This entry was named for Heinrich Eduard Heine and Georg Cantor.