# Heine–Cantor theorem

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Not to be confused with Cantor's theorem.

In mathematics, the Heine–Cantor theorem, named after Eduard Heine and Georg Cantor, states that if f : MN 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 dM and dN, respectively. Suppose further that ${\displaystyle f:M\to N}$ is continuous, and that M is compact. We want to show that f is uniformly continuous, that is, for every ${\displaystyle \epsilon >0}$ there exists ${\displaystyle \delta >0}$ such that for all points x,y in the domain M, ${\displaystyle d_{M}(x,y)<\delta }$ implies that ${\displaystyle d_{N}(f(x),f(y))<\epsilon }$.

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

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

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

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

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

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

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

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

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

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

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