Кіріспе
Математикада Эдуард Хайн мен Георг Кантордың есімімен аталатын Хайн-Кантор теоремасы, егер f екі метрикалық кеңістік X пен Y арасындағы үздіксіз функция болса және X жиыны компакт болса, онда f функциясы біркелкі үздіксіз болады. Маңызды ерекше жағдай – жабық шектелген интервалдан нақты сандарға дейінгі кез келген үздіксіз функция біркелкі үздіксіз болады.
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.
Дәлел
Егер және екі метрикалық кеңістік болса, олардың метрикалары тиісінше және болады делік. Сондай-ақ, функция үздіксіз және жиынтық деп есептейік. Біз функцияның біркелкі үздіксіз екенін көрсетуіміз керек, яғни, кез келген оң нақты сан үшін, функция доменіндегі барлық нүктелер үшін оң нақты сан бар екенін білдіреді, яғни егер болса, онда .
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.
Кез келген оң нақты сан қарастырайық. Үздіксіздік бойынша, домендегі кез келген нүкте үшін, кейбір оң нақты сан бар, егер болса, онда , яғни, нүктесінен арақашықтығы болса, онда нүктесінен арақашықтығы болады.
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.
нүктесінің ашық маңайы, яғни келесі жиынтық болсын:
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.
Әрбір нүкте өзінің маңында орналасқандықтан, жиынтығы жиынтықтың ашық жабыны болады. Жиынтық жиынтық болғандықтан, бұл жабынның шекті ішкі жабыны бар, онда әрбір ашық жиынға сәйкес радиус бар. Енді осы ашық жиындардың ең кіші радиусын анықтаймыз, яғни . Бізде оң радиустардың саны шекті болғандықтан, бұл минимум жақсы анықталған және оң. Енді осы радиус біркелкі үздіксіздік анықтамасы үшін жұмыс істейтінін көрсетейік. Егер кез келген екі үшін болса, онда бұл сандар кеңістігіміздің ашық (ішкі) жабыны құрайтындықтан, олардың бірінде болуы керек, айталық, онда бізде үшбұрыш теңсіздігі бар, яғни .
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.
Бұл және екеуі де нүктесінен ең көп дегенде арақашықтықта екенін білдіреді. анықтамасы бойынша, бұл және екеуі де арақашықтықта екенін білдіреді. Үшбұрыш теңсіздігін қолдансақ, қажетті нәтижеге жетеміз.
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.
жабық аралық жағдайындағы баламалы дәлелдеу үшін «Стандартты емес есептеу» мақаласын қараңыз.
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.