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Mathematics
Calculus
Proving half of the Heine-Borel theorem
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[QUOTE="Eclair_de_XII, post: 6573895, member: 538457"] I disagree. Let ##U=\{(n-1.5,n+1.5)\}_{n\in\mathbb{Z}}##. Let ##x\in\mathbb{R^+}\cup\{0\}##. If ##x=0##, then take ##n=0##. Otherwise, there is an integer ##N## s.t. ##x\in(N-1,N]\subset(N-1.5,N+1.5)\subset U##. Hence, ##U## must be an open cover for the positive real numbers, and certainly ##X:=[0,1]\cup[2,3]\subset\mathbb{R^+}\cup\{0\}##. There is an open interval in ##U## that owns the infimum of ##X'##, namely ##(-1.5,1.5)##. There is a point in ##X'## that does not belong to ##(-1.5,1.5)##, namely, ##2##, and so there is an open interval in ##U## that owns ##2##, namely, ##(1.5,3.5)##. There are no more points in ##X'## that belong to the complement of ##(-1.5,1.5)\cup(1.5,3.5)##. Hence, this union comprises a finite subcover for ##X'##. Necessarily, if we are constructing an open cover with the minimal number of elements, each open cover must own at least one point of ##X'##. If there are infinitely many open covers, it stands to reason that there are infinitely many limit points. By the Bolzano-Weierstrauss Theorem, there must exist a limit point ##x_0\in X''## with the property that any open set that owns ##x_0## owns infinitely many points of ##X'##. And since ##X'## is closed, it must follow that ##x_0\in X'##, and so, this guarantees the existence of an open set within the cover that owns ##x_0##. This open cover, to be denoted ##U_0##, owns the points that are contained in an infinite subset of the open cover, which means that you could nix that infinite subset from the open cover and replace it with ##U_0## to form a smaller subcover. This contradicts the fact that this open cover had the least number of elements. [/QUOTE]
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Forums
Mathematics
Calculus
Proving half of the Heine-Borel theorem
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