Heine–Cantor theorem

Not to be confused with Cantor's theorem.

In mathematics, the Heine–Cantor theorem, named after Eduard Heine and Georg Cantor, states that if M and N are metric spaces and M is compact then every continuous function

f : M → N,

is uniformly continuous.

For instance, if f : [a,b] → R is a continuous function, then it is uniformly continuous.

Proof

Suppose that f is continuous on a compact metric space M but not uniformly continuous, then the negation of

$\forall \varepsilon >0\quad \exists \delta >0$ such that $d(x,y)<\delta \Rightarrow \rho (f(x),f(y))<\varepsilon$ for all x, y in M

is:

$\exists \varepsilon _{0}>0$ such that $\forall \delta >0,\ \exists x,y\in M$ such that $d(x,y)<\delta$ and $\rho (f(x),f(y))\geq \varepsilon _{0}$

where d and $\rho$ are the distance functions on metric spaces M and N, respectively.

Choose two sequences x_n and y_n such that

$d(x_{n},y_{n})<{\frac {1}{n}}$ and $\rho (f(x_{n}),f(y_{n}))\geq \varepsilon _{0}.$

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

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

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.