An archived instance of discourse for discussion in undergraduate Real Analysis I.

Review Day 1 #4 Prob 2.6.5

lhouse1

a.) Assuming that $x_n$ and $y_n$ are cauchy, prove that $(x_n + y _n)$ are cauchy.
Proof:
Let $\epsilon > 0$.
Choose an $N$ such that when $n,m \geq N$, $|x_n - x_m| < \frac{\epsilon}{2}$ and $|y_n - y_m| < \frac{\epsilon}{2}. $
Therefore,
$|(x_n + y_n) - (x_m +y_m)| $
$\leq |x_n - x_m| + |y_n -y_m)|$
$< \frac{\epsilon}{2} + \frac{\epsilon}{2}$
$ = \epsilon$







This is what I was thinking for this, any objections?

b.) Assuming that $x_n$ and $y_n$ are cauchy, prove that $x_ny_n$ is cauchy.

Has anyone had any lucky with part b?

lhouse1

b.)
$|(x_n y_n) - (x_my_m)| $
$\leq |x_n y_n - x_ny_m + x_ny_m -x_my_m|$
$\leq |x_n y_n - x_ny_m| + |x_ny_m -x_my_m|$
$= |x_n| |y_n-y_m| + |y_m| | x_n - x_m| $



From here can we conclude that $|x_n| |y_n-y_m| < \frac{\epsilon}{2}$ and $|y_m| | x_n - x_m| < \frac{\epsilon}{2} $
Therefore we get $ |x_n| |y_n-y_m| + |y_m| | x_n - x_m| = \epsilon$

not sure?????sugestions??

violincounter

My only small objection to your proof of a is that it seems like you are referring semantically to two different $N$'s.

I would say something like, choose $N$ such that when $n, m \geq N$, $|x_n - x_m| < \dfrac{\epsilon}{2}$ and $|y_n - y_m| < \dfrac{\epsilon}{2}$.

mark

@lhouse1 I gave this a like, but I also like @violincounter's suggestion. I think it's really key that you choose $N$ in terms of $(x_n)$ and $(y_n)$ - simultaneously. Also, I'm not sure about your first choice of $N$ - you've got $Choose an $N$ such that $n,m\geq N$". That's just kinda hanging there. I think it needs more.

lhouse1

Thanks @mark and @violincounter I added the suggestions!

violincounter

Perhaps you can shed some light on this @mark . I'm noticing in the textbook that the author will often choose $N_1$ and $N_2$ for two different inequalities involving different sequences and then in the final stages of a proof choose $N = $max$(\{N_1, N_2\})$ when he puts the inequalities together.

Are there situations where you must use this approach instead of just choosing one $N$ that satisfies multiple inequalities? Or is this just an aesthetic difference?