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Lemma: A closed subset of a compact set is compact. Let K be a closed subset of a compact set T in R n and let C K be an open cover of K. Then U = R n \ K is an open set and = {} is an open cover of T. Since T is compact, then C T has a finite subcover ′, that also covers the smaller set K.
The interval C = (2, 4) is not compact because it is not closed (but bounded). The interval B = [0, 1] is compact because it is both closed and bounded. In mathematics, specifically general topology, compactness is a property that seeks to generalize the notion of a closed and bounded subset of Euclidean space. [1]
The Heine–Borel theorem asserts that a subset of the real line is compact if and only if it is both closed and bounded. Correspondingly, a metric space has the Heine–Borel property if every closed and bounded set is also compact. The concept of a continuous function can likewise be generalized.
This form of the theorem makes especially clear the analogy to the Heine–Borel theorem, which asserts that a subset of is compact if and only if it is closed and bounded. In fact, general topology tells us that a metrizable space is compact if and only if it is sequentially compact, so that the Bolzano–Weierstrass and Heine–Borel theorems ...
It is possible to turn the real line into a compact space by adding a single "point at infinity" which we will denote by ∞. The resulting compactification is homeomorphic to a circle in the plane (which, as a closed and bounded subset of the Euclidean plane, is compact). Every sequence that ran off to infinity in the real line will then ...
Let X be a compact Hausdorff space, and equip C(X) with the uniform norm, thus making C(X) a Banach space, hence a metric space. Then Arzelà–Ascoli theorem states that a subset of C(X) is compact if and only if it is closed, uniformly bounded and equicontinuous.
[0, 1] 2 is a totally bounded space because for every ε > 0, the unit square can be covered by finitely many open discs of radius ε. A metric space (,) is totally bounded if and only if for every real number >, there exists a finite collection of open balls of radius whose centers lie in M and whose union contains M.
In particular, a subset of the continuous dual is weak* compact if and only if it is weak* closed and norm-bounded. [1] This implies, in particular, that when X is an infinite-dimensional normed space then the closed unit ball at the origin in the dual space of X does not contain any weak* neighborhood of 0 (since any such neighborhood is norm ...