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翻訳と辞書　辞書検索 [ 開発暫定版 ] スポンサード リンク Cantor's intersection theorem ： ウィキペディア英語版 Cantor's intersection theorem In real analysis, a branch of mathematics, Cantor's intersection theorem, named after Georg Cantor, is a theorem related to compact sets of a compact space $S$. It states that a decreasing nested sequence of non-empty compact subsets of $S$ has nonempty intersection. In other words, supposing is a sequence of non-empty, closed and totally bounded sets satisfying :$C\_0\; \backslash supseteq\; C\_1\; \backslash supseteq\; \backslash cdots\; C\_k\; \backslash supseteq\; C\_\; \backslash cdots,\; \backslash ,$ it follows that :$\backslash left(\backslash bigcap\_\; C\_k\backslash right)\; \backslash neq\; \backslash emptyset.\; \backslash ,$ The result is typically used as a lemma in proving the Heine–Borel theorem, which states that sets of real numbers are compact if and only if they are closed and bounded. Conversely, if the Heine–Borel theorem is known, then it can be restated as: a decreasing nested sequence of non-empty, compact subsets of a compact space has nonempty intersection. As an example, if ''C''''k'' = (), the intersection over is . On the other hand, both the sequence of open bounded sets ''C''''k'' = (0, 1/''k'') and the sequence of unbounded closed sets ''C''''k'' = [''k'', ∞) have empty intersection. All these sequences are properly nested. The theorem generalizes to R''n'', the set of ''n''-element vectors of real numbers, but does not generalize to arbitrary metric spaces. For example, in the space of rational numbers, the sets : $C\_k\; =\; [\backslash sqrt,\; \backslash sqrt+1/k]\; =\; (\backslash sqrt,\; \backslash sqrt+1/k)$ are closed and bounded, but their intersection is empty. A simple corollary of the theorem is that the Cantor set is nonempty, since it is defined as the intersection of a decreasing nested sequence of sets, each of which is defined as the union of a finite number of closed intervals; hence each of these sets is non-empty, closed, and bounded. In fact, the Cantor set contains uncountably many points. == Proof == Assume, by way of contradiction, that $\backslash bigcap\; C\_n=\backslash emptyset$. For each n, let $U\_n=X\backslash setminus\; C\_n$. Since $\backslash bigcup\; U\_n=X\backslash setminus\backslash bigcap\; C\_n$ and $\backslash bigcap\; C\_n=\backslash emptyset$, thus $\backslash bigcup\; U\_n=X$. Since $X$ is compact and $(U\_n)$ is an open cover of it, we can extract a finite cover. Let $U\_k$ be the largest set of this cover; then $X=U\_k$. But then $C\_k=X\backslash setminus\; U\_k=\backslash emptyset$, a contradiction. 抄文引用元・出典: フリー百科事典『 ウィキペディア（Wikipedia）』 ■ウィキペディアで「Cantor's intersection theorem」の詳細全文を読む スポンサード リンク 翻訳と辞書 : 翻訳のためのインターネットリソース Copyright(C) kotoba.ne.jp 1997-2016. All Rights Reserved.