If a cauchy sequence has a convergent subsequence then prove that it is itself convergent
Answers
Answer:
Step-by-step explanation:
can't
Step-by-step explanation:
Certainly not the most elementary proof, but this one feels quite satisfying conceptually: let (X,d) be a metric space and contemplate a Cauchy sequence {xn} with a convergent subsequence, say convergent to L∈X. Now consider the completion
¯
X
of X: by definition every Cauchy sequence in
¯
X
converges, so our sequence {xn} converges in
¯
X
, say to M. But then every subsequence also converges to M and thus M=L. It follows that the original Cauchy sequence is convergent to L!
Added: Really though, the direct proof is also quite simple conceptually: you want to show that the sequence converges to L. For any fixed ϵ we know (i) all pairs of terms with sufficiently large index are within
ϵ
2
of each other, and (ii) there is at least one term of the sequence with sufficiently large index which is within
ϵ
2
of L. Apply the triangle inequality!
.
hope it helps
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