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If a set is closed and bounded, then it is compact. If a set S in R n is bounded, then it can be enclosed within an n-box = [,] where a > 0. By the lemma above, it is enough to show that T 0 is compact. Assume, by way of contradiction, that T 0 is not compact.
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 ...
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]
Then Arzelà–Ascoli theorem states that a subset of C(X) is compact if and only if it is closed, uniformly bounded and equicontinuous. [15] This is analogous to the Heine–Borel theorem, which states that subsets of R n are compact if and only if they are closed and bounded. [16]
Consider the real line with its usual Borel topology. Let denote the Dirac measure, a unit mass at the point in .The collection := {|} is not tight, since the compact subsets of are precisely the closed and bounded subsets, and any such set, since it is bounded, has -measure zero for large enough .
As norm-closed, convex sets are weakly closed (Hahn–Banach theorem), norm-closures of convex bounded sets in Hilbert spaces or reflexive Banach spaces are weakly compact. Closed and bounded sets in () are precompact with respect to the weak operator topology (the weak operator topology is weaker than the ultraweak topology which is in turn ...
which is a continuous function from the open interval (−1,1) to itself. Since x = 1 is not part of the interval, there is not a fixed point of f(x) = x. The space (−1,1) is convex and bounded, but not closed. On the other hand, the function f does have a fixed point for the closed interval [−1,1], namely f(1) = 1.
However, bounded and weakly closed sets are weakly compact so as a consequence every convex bounded closed set is weakly compact. As a consequence of the principle of uniform boundedness, every weakly convergent sequence is bounded. The norm is (sequentially) weakly lower-semicontinuous: if converges weakly to x, then