Heine–Cantor theorem

Not to be confused with Cantor's theorem.

In mathematics, the Heine–Cantor theorem, named after Eduard Heine and Georg Cantor, its statement is as follows:

Theorem. If M and N are metric spaces and M is compact then every continuous function f : M → N, is uniformly continuous.

Corollary. If f : [a,b] → R is a continuous function, then it is uniformly continuous.

Proof

First note that uniformly continuous for a function f is stated as follows:

$\forall \varepsilon >0\ \exists \delta >0\ \forall x,y\in M:\left(d_{M}(x,y)<\delta \Rightarrow d_{N}(f(x),f(y))<\varepsilon \right),$

where d_M, d_N are the distance functions on metric spaces M and N, respectively. Now suppose f is continuous on the compact metric space M but not uniformly continuous, then the negation of uniform continuity is

$\exists \varepsilon _{0}>0\ \forall \delta >0\ \exists x,y\in M:\left(d_{M}(x,y)<\delta \wedge d_{N}(f(x),f(y))\geq \varepsilon _{0}\right).$

Fixing ε₀, for every positive number we have a pair of points in M with the above properties. So we can create two sequences {x_n}, {y_n} such that

$d_{M}(x_{n},y_{n})<{\frac {1}{n}}\wedge d_{N}(f(x_{n}),f(y_{n}))\geq \varepsilon _{0}.$

As M is compact there exist two converging subsequences ($x_{n_{k}}$ to x₀ and $y_{n_{k}}$ to y₀), so

$d_{M}(x_{n_{k}},y_{n_{k}})<{\frac {1}{n_{k}}}\Rightarrow d_{N}(f(x_{n_{k}}),f(y_{n_{k}}))\geq \varepsilon _{0}$

but as f is continuous and $x_{n_{k}}$ and $y_{n_{k}}$ converge to the same point, this statement is impossible.

For an alternative proof in the case of M = [a, b] a closed interval, see the article on non-standard calculus.

External links

• "Heine–Cantor theorem". PlanetMath.
• "Proof of Heine–Cantor theorem". PlanetMath.