In mathematics, the Heine–Cantor theorem, named after Eduard Heine and Georg Cantor, states that if
is a continuous function between two metric spaces
and
, and
is compact, then
is uniformly continuous. An important special case is that every continuous function from a closed bounded interval to the real numbers is uniformly continuous.
Suppose that
and
are two metric spaces with metrics
and
, respectively. Suppose further that a function
is continuous and
is compact. We want to show that
is uniformly continuous, that is, for every positive real number
there exists a positive real number
such that for all points
in the function domain
,
implies that
.
Consider some positive real number
. By continuity, for any point
in the domain
, there exists some positive real number
such that
when
, i.e., a fact that
is within
of
implies that
is within
of
.
Let
be the open
-neighborhood of
, i.e. the set
![{\displaystyle U_{x}=\left\{y\mid d_{M}(x,y)<{\frac {1}{2}}\delta _{x}\right\}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b5091d99df5a01d42559cf6fd06a652432dce611)
Since each point
is contained in its own
, we find that the collection
is an open cover of
. Since
is compact, this cover has a finite subcover
where
. Each of these open sets has an associated radius
. Let us now define
, i.e. the minimum radius of these open sets. Since we have a finite number of positive radii, this minimum
is well-defined and positive. We now show that this
works for the definition of uniform continuity.
Suppose that
for any two
in
. Since the sets
form an open (sub)cover of our space
, we know that
must lie within one of them, say
. Then we have that
. 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}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6d7746b824c33c4dfe4a00520241b600e52dcf71)
implying that
and
are both at most
away from
. By definition of
, this implies that
and
are both less than
. 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 {\varepsilon }{2}}+{\frac {\varepsilon }{2}}=\varepsilon .}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0adeaad44b9173c49f16d57668b4eba0782d748a)
For an alternative proof in the case of
, a closed interval, see the article Non-standard calculus.
See also[edit]
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