Heine–Cantor theorem

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

f : M → N,

where N is a metric space, is uniformly continuous.

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

This is not Cantor's theorem.

[edit] 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) ) \ge \varepsilon_0.

where d and ρ are the distance functions on metric spaces M and N, respectively.

Choose two sequences xn and yn such that

d(x_n, y_n) < \frac {1}{n} and \rho ( f (x_n), f (y_n)) \ge \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})) \ge \varepsilon_0

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

[edit] See also

[edit] External links

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