I've seen a lot of proofs of how to do this and I don't quite like any of them so I tried proving this myself. Let me know if I am making any logical errors here.


Let $\varepsilon > 0$ be given.

Since $\{x_n\}_{n=1}^\infty$ is Cauchy, $\forall \varepsilon >0, \exists N \in \mathbb{N} : |x_n-x_m|<\varepsilon$ when $n,m \geq N$

Since $\{x_n\}$ is Cauchy, $\{x_n\}$ is bounded. We know by Bolzano-Weierstrass this means that $\{x_n\}$ has a convergent subsequence $\{x_{n_k}\}$ that converges to $l$.

Put another way: $\forall \varepsilon > 0, \exists N' \in \mathbb{N}: |x_{n_k}-l|<\varepsilon$ when $k>N'$

Suppose $n >$ max$\{N,N'\}$

Then $|x_n-l|=|x_n-x_{n_k}+x_{n_k}-l| \leq |x_n-x_{n_k}|+|x_{n_k}-l| < \varepsilon + \varepsilon = 2\varepsilon.$ $\square$

(Note: I'm not concerned with the fact that my proof ends with $2\varepsilon$. I just wanna know if my statements are true and if my proof logic is good.)

  • $\begingroup$ Yes, looks great, although I would remove the "for all epsilon > 0" in the sentence following the fixing of a particular epsilon. It is fine to assume familiarity with the definition and just grab the value of N that the definition gives you. There's also a tiny gap between k exceeding N (or N') and $n_k$ exceeding N (which is what you later use) although the fundamentals are sound. $\endgroup$ Feb 1, 2021 at 1:41
  • $\begingroup$ where is the epsilon fixed? I must be misunderstanding $\endgroup$ Feb 1, 2021 at 1:42
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    $\begingroup$ Oh I see. So what you're saying is that any instance of "for all" beyond the first "for all" is redundant since we assumed that to begin with $\endgroup$ Feb 1, 2021 at 1:45
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    $\begingroup$ Exactly. You're fixing a symbol for epsilon and saying everything that you are going to assume about it (that it's positive). From there you can just run with it like it's a particular number, and not something that needs to be quantified with "for alls." $\endgroup$ Feb 1, 2021 at 1:47
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    $\begingroup$ +1 for the OPs responses to comments about sharpening the prose. $\endgroup$ Feb 1, 2021 at 14:32

1 Answer 1


That's fine as long as you have a definition of the real numbers that leads to Bolzano Weierstrass without using Cauchy sequences.


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