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Please check my proof :)]

We must show if given $\epsilon >0$ we can find n>N such that $|S_{n}-L_{n}|<\epsilon $

We set up $|S_n-L_n|$ for any uncountable cauch sequence as

$$|S_{1}-L_{1}|<\frac{\epsilon }{2}$$

and for any cauchy sequence of rational number as

$$|S_{2}-L_{2}|< \frac{\epsilon }{2}$$

then

$$|S_{n}-L_{n}| = |S_{1}-L_{1}+S_{2}-L_{2}|\leq |S_{1}-L_{1}|+|S_{2}-L_{2}|\leq \frac{\epsilon }{2}+\frac{\epsilon }{2}=\epsilon $$

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    Uncountable Cauchy sequence? There are no uncountable sequences. And where do you get $\epsilon = |S_1-L_1+S_2-L_2|$ from? – 2017-01-15
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    Um this problem is from the way of analysis by Robert I copied it without edit any word or the question is wrong? – 2017-01-15
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    Let me rephrase, no sequence is uncountable. There are only countably many terms in any sequence. The question is to prove that for each Cauchy sequence $(a_n)$ the set $\{(b_n)\mid (b_n) \ \text{equivalent to}\ (a_n)\}$ is uncountable. You only have _one_ $(L_n)$. – 2017-01-15
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    @Ennar ok thank ^ ^ – 2017-01-15
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    $|S_n-L_n| = |S_1-L_1+S_2-L_2|$ is still wrong. Please rethink what you need to show again. – 2017-01-15
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    @Ennar thank for comment ^ ^. I just tried to mimic the proof from similiar question in the book – 2017-01-15
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    Ask yourself whether you reached conclusion: "There are _uncountably_ many sequences equivalent to my original Cauchy sequence." If you look at your work, there is nothing even remotely closed to that. – 2017-01-15

2 Answers 2

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Suppose there are only a countable number of Cauchy sequences which converge to some real number $r$. Construct a sequence $x_{n} $ as follows. Let $a_n$ be the the $n^{\text{th}} $ term of $n^{\text{th}} $ sequence converging to $r$ and let $x_{n} =a_{n} +1/n$. You can now show that $x_n$ is Cauchy sequence and it converges to $r$.

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    Very nice, I didn't think of using diagonalization. At least not directly. – 2017-01-15
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    @Ennar: Cantor is linked with diagonalization and also with theory of real numbers as Cauchy sequences of rationals. So... – 2017-01-15
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    Well, yes, I just thought of something else and didn't look further. – 2017-01-15
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Hint: Let's say that you have convergent sequence $(a_n)$. Could you find uncountably many sequences $(b_n)$ such that $\lim_na_n=\lim_nb_n$? To further hint toward solution, what do you know about subsequences of convergent sequence and how many of them are there?

Cauchy sequences of rational numbers converge to real numbers, so if you can answer the above question, you are done.