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Let $(x_n)_{n∈ℕ}$ and $(y_n)_{n∈ℕ}$ be Cauchy sequences of real numbers.

Show, without using the Cauchy Criterion, that if $z_n=x_n+y_n$, then $(z_n)_{n∈ℕ}$ is a Cauchy sequence of real numbers.

Here's my attempt at a proof:

Let $(x_n)$ and $(y_n)$ be Cauchy sequences. Let $(z_n)$ be a sequence and let $z_n=x_n+y_n$.

Since $(x_n)$ and ($y_n)$ are Cauchy, $∃N∈ℕ$ such that $|x_n-x_m|<ε/2$ and $|y_n-y_m|<ε/2$ for $n,m≥N$.

Let $n,m≥N$ and let $z_n,z_m∈(z_n)$. Then, $$|z_n-z_m|=|x_n-x_m|+|y_n-y_m| <ε/2+ε/2=ε.$$ Therefore, $|z_n-z_m|<ε$ for all $n,m≥N$ and hence, $(z_n)$ is a Cauchy sequence of real numbers.

Is this correct? Any input is appreciated.

Thanks.

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  • $\begingroup$ It looks OK the proof! $\endgroup$ – ILoveMath Oct 17 '12 at 2:04
  • $\begingroup$ It's correct except for a minor detail. The triangle inequality results in a $\leq$ not a $=$. $\endgroup$ – copper.hat Oct 17 '12 at 2:16
  • $\begingroup$ Thanks for pointing that out. $\endgroup$ – Alti Oct 17 '12 at 2:35
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The only thing you have to correct is $$|z_n-z_m|=|(x_n-x_m)+(y_n-y_m)|\le|x_n-x_m|+|y_n-y_m| <ε/2+ε/2=ε.$$

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This proof looks correct to me.

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