Cauchy sequences is my proof correct?

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SUMMARY

The proof presented demonstrates that the sum of two Cauchy sequences, (xn) and (yn), results in another Cauchy sequence (zn) defined as zn = xn + yn. The proof correctly establishes that for any ε > 0, there exists an N such that |zn - zm| < ε for all n, m ≥ N. The discussion emphasizes the need to utilize the maximum of the indices N1 and N2 corresponding to the Cauchy properties of (xn) and (yn) to ensure the inequality holds, rather than relying solely on equality.

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Homework Statement



Let (xn)n[itex]\in[/itex]ℕ and (yn)n[itex]\in[/itex]ℕ be Cauchy sequences of real numbers.

Show, without using the Cauchy Criterion, that if zn=xn+yn, then (zn)n[itex]\in[/itex]ℕ is a Cauchy sequence of real numbers.

Homework Equations





The Attempt at a Solution


Here's my attempt at a proof:

Let (xn) and (yn) be Cauchy sequences. Let (zn) be a sequence and let zn=xn+yn.

Since (xn) and (yn) are Cauchy, [itex]\exists[/itex]N[itex]\in[/itex]ℕ such that,
|xn-xm|<ε/2, and
|yn-ym|<ε/2 for n,m≥N.

Let n,m≥N and let zn,zm[itex]\in[/itex](zn).
Then,
|zn-zm|=|xn-xm|+|yn-ym|
<ε/2+ε/2=ε.
Therefore,
|zn-zm|<ε for all n,m≥N and hence, (zn) is a Cauchy sequence of real numbers.

Is this correct?
Any input is appreciated.

Thanks.
 
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Your first step should be an inequality and not an equality, other than that seems ok, you just need to take the maximum of N1 and N2 where these indices are the ones for which

[tex]\forall n,m \geq N_1 \ |x_n-x_m|\leq \epsilon/2[/tex]
[tex]\forall n,m \geq N_2 \ |y_n-y_m| \leq \epsilon /2[/tex]
 

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