Введение
Ограниченная последовательность в конечномерном евклидовом пространстве имеет сходящуюся подпоследовательность. В математике, в частности в вещественном анализе, теорема Болцано — Вейерштрасса, названная в честь Бернарда Болцано и Карла Вейерштрасса, является фундаментальным результатом об сходимости в конечномерном евклидовом пространстве. Теорема утверждает, что любая бесконечная ограниченная последовательность в имеет сходящуюся подпоследовательность. Эквивалентная формулировка состоит в том, что подмножество является последовательно компактным тогда и только тогда, когда оно замкнуто и ограничено. Теорему иногда называют теоремой о последовательной компактности.
In mathematics, specifically in real analysis, the Bolzano–Weierstrass theorem, named after Bernard Bolzano and Karl Weierstrass, is a fundamental result about convergence in a finite dimensional Euclidean space The theorem states that each infinite bounded sequence in has a convergent subsequence. An equivalent formulation is that a subset of is sequentially compact if and only if it is closed and bounded. The theorem is sometimes called the sequential compactness theorem.
История и значение
Теорема Болцано-Вейерштрасса названа в честь математиков Бернарда Болцано и Карла Вейерштрасса. Фактически, впервые она была доказана Болцано в 1817 году как лемма при доказательстве теоремы о промежуточных значениях. Примерно через пятьдесят лет результат был признан значимым сам по себе и повторно доказан Вейерштрассом. С тех пор она стала основополагающей теоремой математического анализа.
Доказательство
Сначала мы докажем теорему для $\mathbb{R}$ (множества всех действительных чисел), в этом случае упорядочение на $\mathbb{R}$ может быть использовано. Действительно, у нас есть следующий результат:
Lemma: Every infinite sequence in has an infinite monotone subsequence (a subsequence that is either non decreasing or non increasing). Proof: Let us call a positive integer valued index of a sequence a "peak" of the sequence when for every Suppose first that the sequence has infinitely many peaks, which means there is a subsequence with the following indices and the following terms So, the infinite sequence in has a monotone (non increasing) subsequence, which is But suppose now that there are only finitely many peaks, let be the final peak if one exists (let otherwise) and let the first index of a new subsequence be set to Then is not a peak, since comes after the final peak, which implies the existence of with and Again, comes after the final peak, hence there is an where with Repeating this process leads to an infinite non decreasing subsequence , thereby proving that every infinite sequence in has a monotone subsequence. Now suppose one has a bounded sequence in ; by the lemma proven above there exists a monotone subsequence, likewise also bounded. It follows from the monotone convergence theorem that this subsequence converges. Finally, the general case , can be reduced to the case of as follows: given a bounded sequence in , the sequence of first coordinates is a bounded real sequence, hence it has a convergent subsequence. One can then extract a sub subsequence on which the second coordinates converge, and so on, until in the end we have passed from the original sequence to a subsequence times—which is still a subsequence of the original sequence—on which each coordinate sequence converges, hence the subsequence itself is convergent.
Лемма: Каждая бесконечная последовательность в $\mathbb{R}$ имеет бесконечную монотонную подпоследовательность (подпоследовательность, которая либо не убывает, либо не возрастает). Доказательство: Будем называть положительный целочисленный индекс последовательности "пиком", если $a_n \ge a_{n+1}$ для каждого $n$. Предположим сначала, что последовательность имеет бесконечно много пиков, что означает, что существует подпоследовательность с индексами $n_1 < n_2 < \dots$ и членами $a_{n_1} \ge a_{n_2} \ge \dots$. Итак, бесконечная последовательность в $\mathbb{R}$ имеет монотонную (не возрастающую) подпоследовательность.
Lemma: Every infinite sequence in has an infinite monotone subsequence (a subsequence that is either non decreasing or non increasing). Proof: Let us call a positive integer valued index of a sequence a "peak" of the sequence when for every Suppose first that the sequence has infinitely many peaks, which means there is a subsequence with the following indices and the following terms So, the infinite sequence in has a monotone (non increasing) subsequence, which is But suppose now that there are only finitely many peaks, let be the final peak if one exists (let otherwise) and let the first index of a new subsequence be set to Then is not a peak, since comes after the final peak, which implies the existence of with and Again, comes after the final peak, hence there is an where with Repeating this process leads to an infinite non decreasing subsequence , thereby proving that every infinite sequence in has a monotone subsequence. Now suppose one has a bounded sequence in ; by the lemma proven above there exists a monotone subsequence, likewise also bounded. It follows from the monotone convergence theorem that this subsequence converges. Finally, the general case , can be reduced to the case of as follows: given a bounded sequence in , the sequence of first coordinates is a bounded real sequence, hence it has a convergent subsequence. One can then extract a sub subsequence on which the second coordinates converge, and so on, until in the end we have passed from the original sequence to a subsequence times—which is still a subsequence of the original sequence—on which each coordinate sequence converges, hence the subsequence itself is convergent.
Но предположим теперь, что существует лишь конечное число пиков. Пусть $n_k$ – последний пик, если он существует (иначе пусть $n_k$ не определено), и пусть первый индекс новой подпоследовательности $m_1$ будет равен $n_k + 1$. Тогда $m_1$ не является пиком, так как $m_1$ следует за последним пиком, что подразумевает существование $m_2 > m_1$ такого, что $a_{m_1} < a_{m_2}$. Опять же, $m_2$ следует за последним пиком, следовательно, существует $m_3 > m_2$ такое, что $a_{m_2} < a_{m_3}$. Повторение этого процесса приводит к бесконечной не убывающей подпоследовательности, тем самым доказывая, что каждая бесконечная последовательность в $\mathbb{R}$ имеет монотонную подпоследовательность.
Lemma: Every infinite sequence in has an infinite monotone subsequence (a subsequence that is either non decreasing or non increasing). Proof: Let us call a positive integer valued index of a sequence a "peak" of the sequence when for every Suppose first that the sequence has infinitely many peaks, which means there is a subsequence with the following indices and the following terms So, the infinite sequence in has a monotone (non increasing) subsequence, which is But suppose now that there are only finitely many peaks, let be the final peak if one exists (let otherwise) and let the first index of a new subsequence be set to Then is not a peak, since comes after the final peak, which implies the existence of with and Again, comes after the final peak, hence there is an where with Repeating this process leads to an infinite non decreasing subsequence , thereby proving that every infinite sequence in has a monotone subsequence. Now suppose one has a bounded sequence in ; by the lemma proven above there exists a monotone subsequence, likewise also bounded. It follows from the monotone convergence theorem that this subsequence converges. Finally, the general case , can be reduced to the case of as follows: given a bounded sequence in , the sequence of first coordinates is a bounded real sequence, hence it has a convergent subsequence. One can then extract a sub subsequence on which the second coordinates converge, and so on, until in the end we have passed from the original sequence to a subsequence times—which is still a subsequence of the original sequence—on which each coordinate sequence converges, hence the subsequence itself is convergent.
Теперь предположим, что у нас есть ограниченная последовательность в $\mathbb{R}$. По доказанной выше лемме, существует монотонная подпоследовательность, также ограниченная. Из теоремы о монотонной сходимости следует, что эта подпоследовательность сходится.
Lemma: Every infinite sequence in has an infinite monotone subsequence (a subsequence that is either non decreasing or non increasing). Proof: Let us call a positive integer valued index of a sequence a "peak" of the sequence when for every Suppose first that the sequence has infinitely many peaks, which means there is a subsequence with the following indices and the following terms So, the infinite sequence in has a monotone (non increasing) subsequence, which is But suppose now that there are only finitely many peaks, let be the final peak if one exists (let otherwise) and let the first index of a new subsequence be set to Then is not a peak, since comes after the final peak, which implies the existence of with and Again, comes after the final peak, hence there is an where with Repeating this process leads to an infinite non decreasing subsequence , thereby proving that every infinite sequence in has a monotone subsequence. Now suppose one has a bounded sequence in ; by the lemma proven above there exists a monotone subsequence, likewise also bounded. It follows from the monotone convergence theorem that this subsequence converges. Finally, the general case , can be reduced to the case of as follows: given a bounded sequence in , the sequence of first coordinates is a bounded real sequence, hence it has a convergent subsequence. One can then extract a sub subsequence on which the second coordinates converge, and so on, until in the end we have passed from the original sequence to a subsequence times—which is still a subsequence of the original sequence—on which each coordinate sequence converges, hence the subsequence itself is convergent.
Наконец, общий случай $\mathbb{R}^n$ можно свести к случаю $\mathbb{R}$ следующим образом: для ограниченной последовательности в $\mathbb{R}^n$ последовательность первых координат является ограниченной последовательностью в $\mathbb{R}$, следовательно, она имеет сходящуюся подпоследовательность. Затем можно извлечь подпоследовательность, на которой сходятся вторые координаты, и так далее, пока в конце мы не перейдем от исходной последовательности к подпоследовательности, извлеченной $n$ раз – которая все еще является подпоследовательностью исходной последовательности – на которой каждая последовательность координат сходится, следовательно, сама подпоследовательность является сходящейся.
Lemma: Every infinite sequence in has an infinite monotone subsequence (a subsequence that is either non decreasing or non increasing). Proof: Let us call a positive integer valued index of a sequence a "peak" of the sequence when for every Suppose first that the sequence has infinitely many peaks, which means there is a subsequence with the following indices and the following terms So, the infinite sequence in has a monotone (non increasing) subsequence, which is But suppose now that there are only finitely many peaks, let be the final peak if one exists (let otherwise) and let the first index of a new subsequence be set to Then is not a peak, since comes after the final peak, which implies the existence of with and Again, comes after the final peak, hence there is an where with Repeating this process leads to an infinite non decreasing subsequence , thereby proving that every infinite sequence in has a monotone subsequence. Now suppose one has a bounded sequence in ; by the lemma proven above there exists a monotone subsequence, likewise also bounded. It follows from the monotone convergence theorem that this subsequence converges. Finally, the general case , can be reduced to the case of as follows: given a bounded sequence in , the sequence of first coordinates is a bounded real sequence, hence it has a convergent subsequence. One can then extract a sub subsequence on which the second coordinates converge, and so on, until in the end we have passed from the original sequence to a subsequence times—which is still a subsequence of the original sequence—on which each coordinate sequence converges, hence the subsequence itself is convergent.
Применение в экономике
В экономике существует несколько важных понятий равновесия, доказательства существования которых часто требуют различных вариантов теоремы Болцано — Вейерштрасса. Одним из примеров является существование распределения, оптимального по Парето. Распределение представляет собой матрицу наборов потребления для агентов в экономике, и такое распределение считается оптимальным по Парето, если не существует изменений, которые ухудшили бы положение какого-либо агента, при этом улучшив положение хотя бы одного агента (в этом случае строки матрицы распределения должны быть упорядочиваемыми в соответствии с отношением предпочтения). Теорема Болцано — Вейерштрасса позволяет доказать, что если множество распределений компактно и не пусто, то система имеет распределение, оптимальное по Парето.