Heine–Cantor theorem

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.

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 ho (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 $$d(x,y) < delta and $$ho (f(x) , f(y) ) ge varepsilon_0.where "d" and $$ho are the distance functions on metric spaces "M" and "N", respectively.

Choose two sequences "x"_"n" and "y"_"n" such that:$$d(x_n, y_n) < frac {1}{n} and $$ho ( f (x_n), f (y_n)) ge varepsilon_0.

As the metric space is compact there exist two converging subsequences ($$x_{n_k} to "x"₀ and $$y_{n_k} to "y"₀), so:$$d(x_{n_k}, y_{n_k}) < frac{1}{n_k} Rightarrow ho ( f (x_{n_k}), f (y_{n_k})) ge varepsilon_0but as "f" is continuous and $$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.

ee also

* Georg Cantor

External links

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