Every sequence in a compact space is a lingering sequence

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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)


Statement

In a metric space (X,d) that is compact every sequence is a lingering sequence, that is to say[1]:

  • (xn)n=1X : xX ϵ>0[|Bϵ(x)(xn)n=1|=0]

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=1X that:

  • xXϵ>0[|Bϵ(x)(xn)n=1|0]
    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:
    • xXϵ>0[|Bϵ(x)(xn)n=1|N] we will assume this is true and reach a contradiction.

Proof:

  • Let (xn)n=1X be given
  1. Notice that for all xX we get an associated ϵx such that the open ball Bϵx(x) contains only finitely many terms of the sequence.
  2. Notice that this family, {Bϵx(x)}xX is actually an open cover of X (a cover of X by sets open in (X,d))
  3. 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
  4. If the n balls in this cover each contain only finitely many terms of the sequence then their sum must be finite! That is:
    • ni=1(|Bϵxi(x)(xk)k=1|)N
      is a finite sum of finite numbers, thus is finite.
  5. But the family of balls cover the space! This suggests there is at most a finite number of terms of the sequence in X
  6. 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 xXϵ>0[|Bϵ(x)(xn)n=1|N]. Thus we must have:

  • xX ϵ>0[|Bϵ(x)(xn)n=1|=0]

Since the sequence was arbitrary, this is true for all/any sequences.

See also

References

  1. Jump up Introduction to Topology - Theodore W. Gamelin & Robert Everist Greene