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Problem:
If we denote real numbers as Cauchy sequences and:
$$[\{a_i\}]+[\{b_i\}] = [\{a_i+b_i\}] ; i∈ N$$
Show that "$+$" is well defined under real numbers.

My try:
Assume that:
$$ \begin{cases} [\{a_i\}]=[\{a_i'\}]\\ \\ [\{b_i\}]=[\{b_i'\}] \end{cases} (I)$$
We must show that: $$[\{a_i+b_i\}] = [\{a_i'+b_i'\}] $$
From LHS:
$$[\{a_i+b_i\}]=[\{a_i\}+\{b_i\}]$$ $$=[\{a_i\}]+[\{b_i\}]$$ From (I)$$[\{a_i\}]+[\{b_i\}]=[\{a_i'\}]+[\{b_i'\}]$$ $$=[\{a_i'\}+\{b_i'\}]$$ $$=[\{a_i'+b_i'\}]$$

Is that correct?

I also checked this link but I wanted to try another approach.

Asaf Karagila
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Arian Ghasemi
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  • What does the notation mean here, are these equivalence classes of Cauchy sequences? In that case, it looks like your argument is circular. You cannot use $[{a_i} + {b_i}] = [{a_i}] + [{b_i}]$ before having proved that that definition even makes sense. – sbares Mar 01 '21 at 17:24
  • @sbares Yes. Cauchy sequences.I edited that. – Arian Ghasemi Mar 01 '21 at 17:26
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    At one point you wrote the step $[{a_i}]+[{b_i}]=[{a_i'}]+[{b_i'}]$, but that's precisely what you're being asked to prove. At some point the definition of the equivalence classes needs to be used. – Greg Martin Mar 01 '21 at 17:33
  • @sbares So can you help me how to make my approach correct? – Arian Ghasemi Mar 01 '21 at 17:57
  • @GregMartin So can you help me how to make my approach correct? – Arian Ghasemi Mar 01 '21 at 17:58
  • You should prove that if ${a_i} \sim {a_i'}$ and ${b_i} \sim {b_i'}$ then ${a_i + b_i} \sim {a_i' + b_i'}$ (here $\sim$ is the equivalence relation that defines the equivalence classes). That will ensure that $[{a_i + b_i}]$ does not depend on the representatives of $[{a_i}]$ and $[{b_i}]$. To do this you need to use the definition of the equivalence relation, and probably do an $\varepsilon$-$N$ proof. The exact details depend on how the equivalence relation is defined. – sbares Mar 01 '21 at 18:13
  • No. You are using in your argument what is required to prove. I agree with @sbares ... use $\epsilon$-type proof for this. – GEdgar Mar 01 '21 at 18:18

1 Answers1

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As written in the comments, your proof is not fine as it's using the statement to prove.

Let us denote the sequences by $a=\{a_i\}$ and similarly for $a',\ b,\ b'$.

We have $[x]=[y]$ iff the sequence of the differences of the elements of $x$ and $y$ converges to $0$, i.e. $y_n-x_n\to 0$.

Now, if $[a]=[a']$ and $[b]=[b']$, that means $a_n'-a_n\to 0$ and $b_n'-b_n\to 0$, so using the additive property of limits, we get $c_n:=(a_n'-a_n)\,+\,(b_n'-b_n)\ \to 0$.

But observe that $c_n=(a_n'+b_n')\,-\,(a_n+b_n)=(a'+b')_n\,-\,(a+b)_n$, and this shows $[a+b]=[a'+b']$.

Berci
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