I'm trying to prove that a real Cauchy sequence is convergent, but I need some help for a step.

We have the following statements:

  • $\{ s_i\}$ is a real Cauchy sequence, i.e. $\forall\epsilon>0,\exists N\geq 1,\forall n\geq N,\forall k\geq 1, \mid s_i-s_{i+k}\mid<\epsilon$

  • Each $s_i$ is an equivalence class of rational Cauchy sequence,

  • For each $i$, we define $\epsilon_i=\frac{1}{2i}$,

  • For each $i$, we choose $N_i$ using the definition of a Cauchy sequence $$\exists N_i\geq1,\forall n\geq N_i,\forall k\geq 1, \mid s_{i,n}-s_{i,n+k}\mid\leq\frac{1}{2i}:=\epsilon_i$$

  • We define $v_i:=s_{i,N_i}$

  • We already shown that

    • $\mid v_i -s_i\mid<\frac{1}{i}$

    • $\{v_i\}$ is a rational Cauchy sequence

  • We define $s$ as an equivalence class $s:=\overline{\{v_n\}}$

Now I need to show that $\mid v_i -s\mid<3\epsilon$


We can show that $\mid v_i-v_{i+k}\mid<2\epsilon$ with

$\mid v_i-v_{i+k}\mid=\mid v_i-s_i+s_i-s_{i+k}+s_{i+k}-v_{i+k}\mid$ $\leq\mid v_i-s_i\mid+\mid s_i-s_{i+k}\mid+\mid s_{i+k}-v_{i+k}\mid<\frac{1}{i}+\epsilon+\frac{1}{i+k}<2\epsilon$


1 Answer 1


It's important to keep the distinction between equivalence class and reprentative clear to show this.

The real number $s$ is an equivalence class of rational Cauchy sequences that are equivalent to $\{v_n\}$. Let $\{q_n\}$ be any representative. Then $\{q_n\} \sim \{v_n\}$ and given $\epsilon > 0$ we have $|q_n - v_n| < \epsilon$ for $n \geq N'$.

The rational number $v_i$, considered as a real number, is the equivalence class of the sequence $(v_i,v_i,v_i, \ldots).$ Hence, $|v_i - s|$ is a real number with a representative Cauchy sequence $\{|v_i - q_n|\}.$

Then for $n \geq N'$

$$|v_i - q_n| \leqslant |v_i - v_n| + |v_n - q_n| < \epsilon + |v_i - v_n|.$$

Also $\{v_i\}$ is itself a rational Cauchy sequence with $|v_i - v_{i+k}| < 2\epsilon$, as shown in the EDIT, for $i$ sufficiently large. Then $|v_i -v_n| < 2 \epsilon$ if $n > \max(N',i)$.

Therefore, every representative of $|v_i - s|$ satsifies $|v_i - q_n| < 3\epsilon$ for sufficiently large $i$.

  • $\begingroup$ Thank you for the answer! Just a detail maybe it's better to say $\mid v_i-v_n\mid<2\epsilon$ rather than $\mid v_i-v_n\mid\leq 2\epsilon$ and also $N'$ rather that $N$ beacause it's not the same $N$ as in the begin of my question. $\endgroup$ Dec 16, 2014 at 21:59
  • $\begingroup$ You're welcome. Thanks for the suggestions - I added them above. $\endgroup$
    – RRL
    Dec 16, 2014 at 22:07

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