2
$\begingroup$

Does there exist an incomplete metric space with exactly one non-convergent Cauchy sequence?

What about normed spaces? If $(x_n)_{n=1}^\infty$ is a non-convergent Cauchy sequence in a normed space, then $(\alpha x_n)_{n=1}^\infty$ is also Cauchy and non-convergent, for every non-zero scalar $\alpha$. Is it possible for a non-Banach space to have only one non-convergent Cauchy sequence, up to multiplication by scalar?

I haven't got enough reputation to comment so:

@Marios Gestas I know that you can always find many convergent Cauchy sequences. I asked if it was possible to find a space with only one non-convergent Cauchy sequence.

@Joey Zou I completely missed the fact that you can take subsequences. It seems so obvious now.

@EugenR This immediately gave an answer to my follow up question: it seems that there also doesn't exist a normed space where all non-convergent Cauchy sequences are subsequences or permutations of each other. I upvoted your answer.

$\endgroup$
2
$\begingroup$

A Cauchy sequence with only finitely many distinct terms must be eventually constant, and hence converges; thus a Cauchy sequence which does not converge must contain infinitely many distinct terms. Any subsequence of a non-convergent Cauchy sequence is also a non-convergent Cauchy sequence.

Thus, if there exists a non-convergent Cauchy sequence, then it must contain infinitely many distinct terms and hence has infinitely many subsequences, and each such subsequence must also be a non-convergent Cauchy sequence.

$\endgroup$
1
$\begingroup$

Sum of convergent and nonconvergent sequence is nonconvergent. Thus I would assume such a space does not exist.

Example: $(x_n)_n$ is nonconvergent, then for any sequence $(a_n)_n$ in $\mathbb{R}$ such that $a_n \to 0$ for $n\to\infty$, new sequence $$ \bigl((1+a_n)x_n\bigr)_n $$ is nonconvergent.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.