In mathematics, the Heine–Cantor theorem states that a continuous function between two metric spaces is uniformly continuous if its domain is compact. The theorem is named after Eduard Heine and Georg Cantor.

Heine–Cantor theorem—If f : M → N {\displaystyle f\colon M\to N} is a continuous function between two metric spaces M {\displaystyle M} and N {\displaystyle N} , and M {\displaystyle M} is compact, then f {\displaystyle f} is uniformly continuous.

An important special case of the Cantor theorem is that every continuous function from a closed bounded interval to the real numbers is uniformly continuous.

Proof of Heine–Cantor theorem

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

Consider some positive real number ε > 0 {\displaystyle \varepsilon >0} . By continuity, for any point x {\displaystyle x} in the domain M {\displaystyle M} , there exists some positive real number δ x > 0 {\displaystyle \delta _{x}>0} such that d N ( f ( x ) , f ( y ) ) < ε / 2 {\displaystyle d_{N}(f(x),f(y))<\varepsilon /2} when d M ( x , y ) < δ x {\displaystyle d_{M}(x,y)<\delta _{x}} , i.e., a fact that y {\displaystyle y} is within δ x {\displaystyle \delta _{x}} of x {\displaystyle x} implies that f ( y ) {\displaystyle f(y)} is within ε / 2 {\displaystyle \varepsilon /2} of f ( x ) {\displaystyle f(x)} .

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

U x = { y ∣ d M ( x , y ) < 1 2 δ x } . {\displaystyle U_{x}=\left\{y\mid d_{M}(x,y)<{\frac {1}{2}}\delta _{x}\right\}.}

Since each point x {\displaystyle x} is contained in its own U x {\displaystyle U_{x}} , we find that the collection { U x ∣ x ∈ M } {\displaystyle \{U_{x}\mid x\in M\}} is an open cover of M {\displaystyle M} . Since M {\displaystyle M} is compact, this cover has a finite subcover { U x 1 , U x 2 , … , U x n } {\displaystyle \{U_{x_{1}},U_{x_{2}},\ldots ,U_{x_{n}}\}} where x 1 , x 2 , … , x n ∈ M {\displaystyle x_{1},x_{2},\ldots ,x_{n}\in M} . Each of these open sets has an associated radius δ x i / 2 {\displaystyle \delta _{x_{i}}/2} . Let us now define δ = min 1 ≤ i ≤ n δ x i / 2 {\displaystyle \delta =\min _{1\leq i\leq n}\delta _{x_{i}}/2} , i.e. the minimum radius of these open sets. Since we have a finite number of positive radii, this minimum δ {\displaystyle \delta } is well-defined and positive. We now show that this δ {\displaystyle \delta } works for the definition of uniform continuity.

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

d M ( x i , y ) ≤ d M ( x i , x ) + d M ( x , y ) < 1 2 δ x i + δ ≤ δ x i , {\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 {\displaystyle x} and y {\displaystyle y} are both at most δ x i {\displaystyle \delta _{x_{i}}} away from x i {\displaystyle x_{i}} . By definition of δ x i {\displaystyle \delta _{x_{i}}} , this implies that d N ( f ( x i ) , f ( x ) ) {\displaystyle d_{N}(f(x_{i}),f(x))} and d N ( f ( x i ) , f ( y ) ) {\displaystyle d_{N}(f(x_{i}),f(y))} are both less than ε / 2 {\displaystyle \varepsilon /2} . Applying the triangle inequality then yields the desired

d N ( f ( x ) , f ( y ) ) ≤ d N ( f ( x i ) , f ( x ) ) + d N ( f ( x i ) , f ( y ) ) < ε 2 + ε 2 = ε . {\displaystyle d_{N}(f(x),f(y))\leq d_{N}(f(x_{i}),f(x))+d_{N}(f(x_{i}),f(y))<{\frac {\varepsilon }{2}}+{\frac {\varepsilon }{2}}=\varepsilon .}

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