Heine–Cantor theorem

In mathematics, the Heine–Cantor theorem, named after Eduard Heine and Georg Cantor, states that if $f:M\to N$ is a continuous function between two metric spaces $M$ and $N$, and $M$ is compact, then $f$ is uniformly continuous. An important special case is that every continuous function from a closed bounded 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 a function $f:M\to N$ is continuous and $M$ is compact. We want to show that $f$ is uniformly continuous, that is, for every positive real number $\epsilon >0$ there exists a positive real number $\delta >0$ such that for all points $x,y$ in the function domain $M$, $d_M(x,y)<\delta$ implies that $d_N(f(x),f(y))<\epsilon$.

Consider some positive real number $\epsilon >0$. By continuity, for any point $x$ in the domain $M$, there exists some positive real number ${\delta }_x>0$ such that $d_N(f(x),f(y))<\epsilon /2$ when $d_M(x,y)<{\delta }_x$, i.e., a fact that $y$ is within ${\delta }_x$ of $x$ implies that $f(y)$ is within $\epsilon /2$ of $f(x)$.

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

$U_x=\{y\mid d_M(x,y)<\frac{1}{2}{\delta }_x\}.$

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 $\{U_{x_1},U_{x_2},\dots ,U_{x_n}\}$ where $x_1,x_2,\dots ,x_n\in M$. Each of these open sets has an associated radius ${\delta }_{x_i}/2$. Let us now define $\delta =\underset{1\le i\le n}{\min }{\delta }_{x_i}/2$, i.e. the minimum radius of these open sets. Since we have a finite number of positive radii, this minimum $\delta$ is well-defined and positive. We 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

$d_M(x_i,y)\le d_M(x_i,x)+d_M(x,y)<\frac{1}{2}{\delta }_{x_i}+\delta \le {\delta }_{x_i},$

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 $\epsilon /2$. Applying the triangle inequality then yields the desired

$d_N(f(x),f(y))\le d_N(f(x_i),f(x))+d_N(f(x_i),f(y))<\frac{\epsilon }{2}+\frac{\epsilon }{2}=\epsilon .$

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

See also

• Cauchy-continuous function

External links

• Heine–Cantor theorem at PlanetMath.org.
• Proof of Heine–Cantor theorem at PlanetMath.org.