Введение
В математике теорема Хайне — Кантора, названная в честь Эдуарда Хайне и Георга Кантора, утверждает, что если f — непрерывная функция между двумя метрическими пространствами X и Y, и X компактно, то f равномерно непрерывна. Важным частным случаем является то, что любая непрерывная функция из замкнутого ограниченного интервала в множество действительных чисел равномерно непрерывна.
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.
Доказательство
Предположим, что и являются двумя метрическими пространствами с метриками и , соответственно. Предположим далее, что функция непрерывна и компактна. Мы хотим показать, что равномерно непрерывна, то есть для каждого положительного действительного числа существует положительное действительное число такое, что для всех точек из области определения функции , влечет .
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
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
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
For an alternative proof in the case of , a closed interval, see the article Non standard calculus.
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
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
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
For an alternative proof in the case of , a closed interval, see the article Non standard calculus.
Рассмотрим некоторое положительное действительное число . По непрерывности, для любой точки в области определения , существует некоторое положительное действительное число такое, что при , то есть, тот факт, что находится на расстоянии не более от влечет, что находится на расстоянии не более от .
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
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
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
For an alternative proof in the case of , a closed interval, see the article Non standard calculus.
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
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
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
For an alternative proof in the case of , a closed interval, see the article Non standard calculus.
Пусть – открытое -окрестность , то есть множество
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
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
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
For an alternative proof in the case of , a closed interval, see the article Non standard calculus.
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
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
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
For an alternative proof in the case of , a closed interval, see the article Non standard calculus.
Поскольку каждая точка содержится в своей собственной , мы находим, что множество является открытым покрытием . Поскольку компактна, это покрытие имеет конечное подпокрытие , где . Каждое из этих открытых множеств имеет связанный радиус. Давайте теперь определим , то есть минимальный радиус этих открытых множеств. Поскольку у нас конечное число положительных радиусов, этот минимум хорошо определен и положителен. Теперь мы покажем, что это подходит для определения равномерной непрерывности. Предположим, что для любых двух . Поскольку множества образуют открытое (под)покрытие нашего пространства , мы знаем, что должна лежать в одном из них, скажем, . Тогда у нас есть, что . Треугольное неравенство тогда влечет, что
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
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
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
For an alternative proof in the case of , a closed interval, see the article Non standard calculus.
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
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
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
For an alternative proof in the case of , a closed interval, see the article Non standard calculus.
что влечет, что и оба находятся на расстоянии не более от . По определению , это влечет, что и оба меньше, чем . Применение треугольного неравенства тогда дает желаемое.
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
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
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
For an alternative proof in the case of , a closed interval, see the article Non standard calculus.
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
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
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
For an alternative proof in the case of , a closed interval, see the article Non standard calculus.
Для альтернативного доказательства в случае , замкнутого интервала, см. статью Нестандартный анализ.
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
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
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
For an alternative proof in the case of , a closed interval, see the article Non standard calculus.
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
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
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
For an alternative proof in the case of , a closed interval, see the article Non standard calculus.