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Let a, b ∈ R, a ≤ b. We show that the set A = [a, b] is compact. Let {an} be a sequence in A. Since a ≤ an ≤ b for all n, then the sequence is bounded. By the Bolzano-Weierstrass theorem (Theorem 2.4.1), we can obtain a convergent subsequence {ank}. Say, limk → ∞ank = s. We now must show that s ∈ A.
Any finite topological space, including the empty set, is compact. More generally, any space with a finite topology (only finitely many open sets) is compact; this includes in particular the trivial topology.
Jul 18, 2018 · According to one of the definition I found: A set S ⊂Rn S ⊂ R n is compact if every sequence in S S has a convergent subsequence, whose limit lies in S S.
We say a collection of sets {Dα: α ∈ A} has the finite intersection property if for every finite set B ⊂ A, ⋂ α ∈ BDα ≠ ∅. Show that a set K ⊂ R is compact if and only for any collection {Eα: α ∈ A, Eα = Cα ∩ K where Cα ⊂ R is closed }
16.2 Compact Sets. A set of real numbers S S is said to be covered by a collection O O of open sets, when every element of S S is contained in at least one member of O O. (The members of O O can contain numbers outside of S S as well as those in S S .) S S is said to compact, if, for every covering O O of S S by open sets, S S is covered by ...
In En (∗ and Cn) E n ( ∗ and C n) a set is compact iff it is closed and bounded.
Apr 25, 2024 · A set S of real numbers is called compact if every sequence in S has a subsequence that converges to an element again contained in S.
