Heine–Cantor theorem

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 bounded closed interval to the real numbers is uniformly continuous.

Proof

Suppose that $M$ and $N$ are two metric spaces with metrics $d_{M}$ and $d_{N}$ , respectively. Suppose further that $f:M\to N$ is continuous, and that $M$ is compact. We want to show that $f$ is uniformly continuous, that is, for every $\varepsilon >0$ there exists $\delta > 0$ such that for all points $x, y$ in the domain $M$ , $d_{M}(x,y)<\delta$ implies that $d_{N}(f(x),f(y))<\varepsilon$ .

Fix some positive $\varepsilon >0$ . Then by continuity, for any point $x$ in our domain $M$ , there exists a positive real number $\delta _{x}>0$ such that $d_{N}(f(x),f(y))<\varepsilon /2$ when $y$ is within $\delta _{x}$ of $x$ .

Let $U_{x}$ be the open $\delta _{x}/2$ -neighborhood of $x$ , i.e. the set

U_{x}=\left\{y\mid d_{M}(x,y)<{\frac {1}{2}}\delta _{x}\right\}

Since each point $x$ is contained in its own $U_{x}$ , we find that the collection $\{U_{x}\mid x\in M\}$ is an open cover of $M$ . Since $M$ is compact, this cover has a finite subcover. That subcover must be of the form

U_{x_{1}},U_{x_{2}},\ldots ,U_{x_{n}}

for some finite set of points $\{x_{1},x_{2},\ldots ,x_{n}\}\subset M$ . Each of these open sets has an associated radius $\delta _{x_{i}}/2$ . Let us now define $\delta =\min _{1\leq i\leq n}{\frac {1}{2}}\delta _{x_{i}}$ , i.e. the minimum radius of these open sets. Since we have a finite number of positive radii, this number $\delta$ is well-defined and positive. We may now show that this $\delta$ works for the definition of uniform continuity.

Suppose that $d_{M}(x,y)<\delta$ for any two $x, y$ in $M$ . Since the sets $U_{x_{i}}$ form an open (sub)cover of our space $M$ , we know that $x$ must lie within one of them, say $U_{x_{i}}$ . Then we have that $d_{M}(x,x_{i})<{\frac {1}{2}}\delta _{x_{i}}$ . The Triangle Inequality then implies that

d_{M}(x_{i},y)\leq d_{M}(x_{i},x)+d_{M}(x,y)<{\frac {1}{2}}\delta _{x_{i}}+\delta \leq \delta _{x_{i}}

implying that $x$ and $y$ are both at most $\delta _{x_{i}}$ away from $x_{i}$ . By definition of $\delta _{x_{i}}$ , this implies that $d_{N}(f(x_{i}),f(x))$ and $d_{N}(f(x_{i}),f(y))$ are both less than $\varepsilon /2$ . Applying the Triangle Inequality then yields the desired

d_{N}(f(x),f(y))\leq d_{N}(f(x_{i}),f(x))+d_{N}(f(x_{i}),f(y))<{\frac {1}{2}}\varepsilon +{\frac {1}{2}}\varepsilon =\varepsilon

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

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