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Recently I learnt this new characterization of completeness of the system of real numbers:

Absolute convergence of an infinite series implies convergence of the series.

I also found one proof on MSE which uses the above criterion to prove that a Cauchy sequence is convergent. The proof builds a subsequence $x_{n_k} $ of the given Cauchy sequence $x_n$ such that $$|x_{n_{k+1}}-x_{n_k}|<2^{-k}\tag{1}$$ and then considers the series $\sum y_k$ where $y_{k} =x_{n_{k+1}}-x_{n_k}$. The claim is that $\sum y_{k} $ is absolutely convergent because of inequality $(1)$. This is true, but not automatic. It rather involves the Cauchy criterion of convergence which is the thing we are trying to prove here.

I guess there is another way to use inequality $(1)$ to prove absolutely convergence of $\sum y_k$ which does not involve completeness of real numbers (possibly using direct definition of absolute convergence), but I can't think of an obvious way.

Does that mean that the proof linked above is circular? Or perhaps I am mistaken (most likely)? If the proof is indeed circular I would want another proof which is correct.

Update: As explained in the answer from David Ullrich, the statement "absolute convergence of an infinite series implies its convergence" does not appear to be equivalent to the completeness of reals (but perhaps someone can find a novel proof for this equivalence). And to clarify further the proof linked above is a correct proof for a characterization of Banach spaces (ie complete normed vector spaces).

To put the matter in different terms "norm convergence of a series implies convergence" is equivalent to "completeness of a normed vector space" and this equivalence is proved using "completeness of real numbers". A series $\sum x_n$ whose terms $x_n$ belong to a normed vector space $X$ is said to be norm convergent if the series of corresponding norms $\sum ||x_n||$ is convergent.

Further Update: It turns out that the characterization "absolute convergence of a series implies comvergence" is indeed equivalent to the completeness of reals, but the proof is non-trivial and non-obvious.

Paramanand Singh
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1 Answers1

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Yes, it does appear that deducing convergence of $\sum|y_n|$ uses completeness, rendering the argument circular.

Is it actually true that completeness follows from the assertion that every absolutely convergent series is convergent? I don't know, but I can't imagine how you'd prove it.

Two related facts, both of which are actually true:

  1. Say the series $\sum x_j$ is absolutely bounded if there exists $c$ with $\sum_{j=1}^n|x_j|\le c$ for all $n$. If every absolutely bounded series is convergent then completeness follows. The proof is as above, the difference being that "comparison" for absolute boundedness is trivial.

(I tend to suspect that the author of the assertion you ask about was simply overlooking the distinction between absolute convergence and absolute boundedness, since after all they are equivalent in the reals, and this equivalence is something we use so often he may have not realized it's a theorem (depending on completeness).)

  1. Back to business as usual, assuming we know that the reals are complete: If you have a normed vector space $X$ then $X$ is complete if and only if every series $\sum x_n$ such that $\sum||x_n||<\infty$ is convergent. This can be useful.

Note that that proof you say you found on MSE is not and does not claim to be a proof that the reals are complete using this criterion! It's actually a correct proof of (2). The proof of (2) is not circular - there's no problem using the completeness of $\Bbb R$ in proving (2).

David C. Ullrich
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  • Yes even I doubted that statement and was unable to supply a proof. But I needed some sort confirmation from people who are far more knowledgeable than myself in this field. Your answer serves the exact purpose. +1 for now and will accept after waiting for some more answers. – Paramanand Singh Jan 22 '18 at 14:15
  • Your last result is the characterization for Banach space. While searching online about these topics Banach space turned up and I saw your last result mentioned as a famous theorem. So overall the whole exercise did give me an opportunity to learn something new. – Paramanand Singh Jan 22 '18 at 14:20
  • Fully agree with your last note also. Perhaps I was trying to lift that proof and use it to prove something which does not appear to be true. – Paramanand Singh Jan 22 '18 at 14:22
  • I will update my answer with these points to avoid confusion for future readers. – Paramanand Singh Jan 22 '18 at 14:28
  • @ParamanandSingh In your edit you say that I explained that convergence of absolutely convergent series is not equivalent to completeness. I didn't say that! I said that the proof above is wrong, but I also said that I didn't know whether the assertion was true. Please don't say I said things that I never said... – David C. Ullrich Jan 22 '18 at 14:38
  • Sorry I will fix this. Please understand it was not intentional. – Paramanand Singh Jan 22 '18 at 14:40
  • @ParamanandSingh By the way it seems to me that your use of the term "norm convergent" is a very bad idea, because that phrase already means, or in any case can be taken to mean, that the partial sums converge in norm. (As opposed to for example a "weakly convergent" series...) – David C. Ullrich Jan 22 '18 at 14:49
  • Thanks for the feedback. I am new to these topics. So the term "weakly convergent" will be OK? – Paramanand Singh Jan 22 '18 at 14:55
  • @ParamanandSingh "Weakly convergent" seems much worse! When people state (2) they typically don't use a single word for the condition in question, it seems to me they typically state it just as I did. – David C. Ullrich Jan 22 '18 at 15:09