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  2. Heine–Borel theorem - Wikipedia

    en.wikipedia.org/wiki/Heine–Borel_theorem

    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.

  3. Compact space - Wikipedia

    en.wikipedia.org/wiki/Compact_space

    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]

  4. Bolzano–Weierstrass theorem - Wikipedia

    en.wikipedia.org/wiki/Bolzano–Weierstrass_theorem

    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 ...

  5. Weak topology - Wikipedia

    en.wikipedia.org/wiki/Weak_topology

    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 ...

  6. Equicontinuity - Wikipedia

    en.wikipedia.org/wiki/Equicontinuity

    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]

  7. Tightness of measures - Wikipedia

    en.wikipedia.org/wiki/Tightness_of_measures

    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 .

  8. Totally bounded space - Wikipedia

    en.wikipedia.org/wiki/Totally_bounded_space

    [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.

  9. Weak convergence (Hilbert space) - Wikipedia

    en.wikipedia.org/wiki/Weak_convergence_(Hilbert...

    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