# Incomplete metric space or normed space with only one non-convergent Cauchy sequence

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.

## 2 Answers

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.

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.