Every sequence in a compact space is a lingering sequence
From Maths
- Note: this page is taking about a metric space not a topological space (at least at this time)
TODO: Does this apply at all to topological spaces? (Maybe, but with adaptation, as open balls do not exist in general top. spaces)
Contents
[hide]Statement
In a metric space (X,d) that is compact every sequence is a lingering sequence, that is to say[1]:
- ∀(xn)∞n=1⊆X : ∃x∈X ∀ϵ>0[|Bϵ(x)∩(xn)∞n=1|=ℵ0]
Note that such an x exists, suggesting the space is complete. See Equivalent statements to compactness of a metric space for information
Proof
Suppose that it is untrue (this is a proof by contradiction), then we want to show that, given a (xn)∞n=1⊆X that:
- ∀x∈X∃ϵ>0[|Bϵ(x)∩(xn)∞n=1|≠ℵ0]
- Obviously as the intersection of sets is a subset of each set we see that Bϵ(x)∩(xn)∞n=1⊆(xn)∞n=1 and a sequence contains a countably infinite amount of terms. As a subset has cardinality less than or equal to a set we see that $\ne\aleph_0$ means showing that the intersection is finite, or:
- ∀x∈X∃ϵ>0[|Bϵ(x)∩(xn)∞n=1|∈N] we will assume this is true and reach a contradiction.
- Obviously as the intersection of sets is a subset of each set we see that Bϵ(x)∩(xn)∞n=1⊆(xn)∞n=1 and a sequence contains a countably infinite amount of terms. As a subset has cardinality less than or equal to a set we see that $\ne\aleph_0$ means showing that the intersection is finite, or:
Proof:
- Let (xn)∞n=1⊆X be given
- Notice that for all x∈X we get an associated ϵx such that the open ball Bϵx(x) contains only finitely many terms of the sequence.
- Notice that this family, {Bϵx(x)}x∈X is actually an open cover of X (a cover of X by sets open in (X,d))
- However X is compact by assumption, this means every open cover has a finite subcover. Thus:
- ∃ {x1,…,xn}⊆X such that {Bϵxi(xi)}ni=1 is an open cover of X
- If the n balls in this cover each contain only finitely many terms of the sequence then their sum must be finite! That is:
- n∑i=1(|Bϵxi(x)∩(xk)∞k=1|)∈Nis a finite sum of finite numbers, thus is finite.
- n∑i=1(|Bϵxi(x)∩(xk)∞k=1|)∈N
- But the family of balls cover the space! This suggests there is at most a finite number of terms of the sequence in X
- This is obviously a contradiction, as there are countably many terms of the sequence in the space!
Thus we have shown if X is compact then we cannot have ∀x∈X∃ϵ>0[|Bϵ(x)∩(xn)∞n=1|∈N]. Thus we must have:
- ∃x∈X∀ ϵ>0[|Bϵ(x)∩(xn)∞n=1|=ℵ0]
Since the sequence was arbitrary, this is true for all/any sequences.
See also
References
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