Does sequentially compact implies compact?
Does sequentially compact implies compact?
Theorem: A subset of a metric space is compact if and only if it is sequentially compact.
Are compactness and sequential compactness equivalent?
Compactness is equivalent to sequential compactness. Any Cauchy se- quence in a compact metric space has a convergent subsequence, and thus it converges.
Is sequentially compact?
In mathematics, a topological space X is sequentially compact if every sequence of points in X has a convergent subsequence converging to a point in X.
Is the empty set sequentially compact?
By convention, the empty set ∅ is considered sequentially compact.
Are sequentially compact metric space is?
A metric space is called sequentially compact if every sequence in has a convergent subsequence. A subset Y ⊂ X is called sequentially compact if the metric subspace is sequentially compact.
How do you prove something is sequentially compact?
A compact metric space is sequentially compact. Let A be an infinite set in a compact metric space X. To prove that A has a limit point we must find a point p for whicch every open neighbourhood of p contains infinitely many points of A. X and A is infinite.
What is sequentially compact metric space?
Definition 1. A metric space is sequentially compact if every sequence has a convergent subsequence. Definition 2. A metric space is complete if every Cauchy sequence con- verges.
How do you prove a set is sequentially compact?
To prove it is closed suppose {an} is in S and converges to a. Since S is sequentially compact some subsequence of {an} converges to something in S but that subsequence must converge to a so a ∈ S. To prove it is bounded suppose not, so for all n there is some an > n.
How do you show sequential compactness?
Is a convergent sequence compact?
A useful property of compact sets in a metric space is that every sequence has a convergent subsequence. Such sets are sometimes called sequentially compact. Let us prove that in the context of metric spaces, a set is compact if and only if it is sequentially compact.
Does every sequence have a convergent subsequence?
Proof: Every sequence in a closed and bounded subset is bounded, so it has a convergent subsequence, which converges to a point in the set, because the set is closed. Conversely, every bounded sequence is in a closed and bounded set, so it has a convergent subsequence.
Can a bounded sequence have a divergent subsequence?
If all subsequences of (an) were bounded then (an) would itself be bounded, in which case a divergent subsequence of (an) must contain two subsequences convergent to different limits.