5
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The caveat is that I cannot use the Cauchy Criterion in the proof, i.e. I cannot use the fact that A sequence converges if and only if it is a Cauchy sequence in the proof.

So, what I can work with: A Cauchy Sequence is one such that $\forall \epsilon > 0, \exists N \in \mathbb{N} $ so that whenever $m,n \geq N$, it follows that $|a_n - a_m|< \epsilon$

This is where I'm at in the proof: Given: $x_n$ is Cauchy, so there is an $m_1,n_1 \geq N_1$ such that $|x_n - x_m|< \dfrac{\epsilon} {2}$

and

$y_n$ is Cauchy, so there is an $m_2,n_2 \geq N_2$ such that $|y_n - y_m|< \dfrac{\epsilon} {2}$

Using the triangle inequality,

$|(x_n + y_n) - (x_m + y_m)|=$

$|x_n - x_m) + (y_n - y_m)| \leq$

$|x_n - x_m| + |y_n - y_m| < \dfrac {\epsilon} {2} + \dfrac {\epsilon} {2} = \epsilon$

For $\max[N_1, N_2]$

I think I have it, but I'm wondering if there is a better way of doing it. Thanks!

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  • 4
    $\begingroup$ That is the way to do it. $\endgroup$ – Friedrich Philipp Mar 11 '16 at 4:57
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    $\begingroup$ That looks fine to me. What kind of improvement are you hoping for? $\endgroup$ – Thomas Mar 11 '16 at 4:57
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    $\begingroup$ That's pretty much it: just chase definitions. $\qquad$ $\endgroup$ – Michael Hardy Mar 11 '16 at 4:57
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    $\begingroup$ It seemed to be the only way @Thomas, since I couldn't use convergence to prove it. I was just making sure. Thanks. $\endgroup$ – Jabernet Mar 11 '16 at 5:09
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    $\begingroup$ By the way, the Cauchy Criterion does not work under some circumstances, but this argument continues to hold. $\endgroup$ – Henricus V. Mar 11 '16 at 5:23

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