Analysis-2-Complete Metric Spaces, guaranteed and verified 100% Pass

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Complete Metric Spaces


Def. A metric space 𝑀 is said to be complete if every Cauchy sequence in 𝑀
converges to a point in 𝑀.


Ex. ℝ𝑛 is complete with the standard metric


Ex. (0,1) is not complete with the standard metric.
ℚ is not complete with the standard metric.


Theorem: Let 𝑀, 𝑑 be a complete metric space and 𝐴 ⊆ 𝑀 a subset of 𝑀. Then
𝐴, 𝑑 is a complete metric space if and only if 𝐴 is closed in 𝑀.


Proof. Assume that 𝐴, 𝑑 is complete and let 𝑥 ∈ 𝑀 be any limit point of 𝐴.
Let {𝑥𝑛 } be a sequence in 𝐴 that converges to 𝑥 ∈ 𝑀.

𝑀
Since {𝑥𝑛 } converges it is a Cauchy sequence in 𝐴.
{𝑥𝑛 }
𝐴 is complete so 𝑥 ∈ 𝐴. 𝐴
Hence 𝐴 is closed.


Now let’s assume that 𝐴 is closed in 𝑀 and show
that 𝐴 is complete.
Let {𝑥𝑛 } be a Cauchy sequence in 𝐴.
Then {𝑥𝑛 } is also a Cauchy sequence in 𝑀, and hence 𝑥𝑛 → 𝑥 ∈ 𝑀, since 𝑀 is
complete.
But 𝐴 is closed so 𝑥 ∈ 𝐴, and 𝐴 is complete.

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Ex. [0,1], [0, ∞), ℤ, and [1,5] ∪ {16} are all complete metric spaces with the
standard metric on ℝ.


Notice that if 𝑀 is complete and totally bounded then every totally bounded
sequence in 𝑀 has a convergent subsequence.
In particular, any closed, bounded subset of ℝ (i.e. compact subsets of ℝ) is both
complete and totally bounded. Thus, for example, every sequence in [𝑎, 𝑏] has a
convergent subsequence.


How you measure distances can determine whether a metric space is complete.
For example, you can have two metric spaces with the same underlying points but
one is complete and the other isn’t.


Ex. Let 𝑀1 = [1, ∞) with 𝑑1 (𝑥, 𝑦) = |𝑥 − 𝑦|
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𝑀2 = [1, ∞) with 𝑑2 (𝑥, 𝑦) = | 𝑥 − 𝑦 | .

The sequence {1,2,3,4, … } is a Cauchy sequence in 𝑀2 (but not in 𝑀1 ), but doesn’t
converge in 𝑀2.
Thus 𝑀1 is a complete metric space (a closed subset of ℝ with the standard
metric), but 𝑀2 is not complete because it has a Cauchy sequence that does not
converge in 𝑀2.