Normal order

Template:Distinguish In mathematics, the Heine–Cantor theorem, named after Eduard Heine and Georg Cantor, states that if f : M → N is a continuous function between two metric spaces, and M is compact, then f is uniformly continuous. An important special case is that every continuous function from a closed interval to the real numbers is uniformly continuous.

Proof

Uniform continuity 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 assume for a contradiction that f is continuous on the compact metric space M but not uniformly continuous; in this case, the negation of uniform continuity for f is that

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

Fixing ε₀, for every positive number δ we have a pair of points x and y in M with the above properties. Setting δ = 1/n for n = 1, 2, 3, ... gives 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, the Bolzano–Weierstrass theorem shows the existence of two converging subsequences ($x_{n_{k}}$ to x₀ and $y_{n_{k}}$ to y₀) of these two sequences. It follows that

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

But as f is continuous and $x_{n_{k}}$ and $y_{n_{k}}$ converge to the same point, this statement is impossible. The contradiction proves that our assumption that f is not uniformly continuous cannot be true, so f must be uniformly continuous as the theorem states.

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

External links

• Template:Planetmath reference
• Template:Planetmath reference

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