# Proving every Cauchy sequence in $\mathbb{R}$ is convergent

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.

Proof

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.)

• 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. Feb 1, 2021 at 1:41
• where is the epsilon fixed? I must be misunderstanding Feb 1, 2021 at 1:42
• 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 Feb 1, 2021 at 1:45
• 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." Feb 1, 2021 at 1:47
• +1 for the OPs responses to comments about sharpening the prose. Feb 1, 2021 at 14:32