3
$\begingroup$

Let $\{X_n\}$ and $\{Y_n\}$ be Cauchy sequences and let $a$ and $b$ be nonzero real numbers. Prove that $\{aX_n$ + b$Y_n\}$ is a Cauchy sequence.

So far I have Since $\{X_n\}$ is Cauchy then for every $\epsilon>0$ there exists an $N_1 \in \mathbb{N}$ s.t. when $m,n\geq N_1$ then $|X_n-X_M|< \frac{\epsilon}{ab}$(?).

Since $\{Y_n\}$ is Cauchy then for every $\epsilon>0$ there exists an $N_2 \in \mathbb{N}$ s.t. when $m,n\geq N_2$ then $|Y_n-Y_M|< \frac{\epsilon}{ab}$(?). Choose $N=\max\{N_1,N_2\}$, so for every $\epsilon>0$, when $n,m\geq N$ then $|a(X_n+Y_n) - b(X_m+Y_m)|=|aX_n+aY_n-bX_n-bY_m|$ but something is missing?

$\endgroup$
  • $\begingroup$ Since you are new to math stack exchange, you would probably benefit from taking a look at this handy guide to formatting mathematics: math.stackexchange.com/help/notation $\endgroup$ – TravisJ Apr 30 '15 at 1:57
  • $\begingroup$ It's a silly technical point, but, you want to be careful about dividing by $a$ and $b$, as, they could be 0. $\endgroup$ – James Apr 30 '15 at 2:02
  • $\begingroup$ okay thank you all, your answers really helped :) $\endgroup$ – Sandra Eades Apr 30 '15 at 2:23
  • $\begingroup$ @James, either by edit (or by oversight--I've done it frequently) there is the hypothesis that $a$ and $b$ are non-zero. But you are right on that you must be careful to not divide by $0$... in this case, there is no worry. $\endgroup$ – TravisJ Apr 30 '15 at 3:09
  • $\begingroup$ @TravisJ it was most likely oversight! Well spotted. $\endgroup$ – James Apr 30 '15 at 13:22
2
$\begingroup$

Assuming your sequences are in $\mathbb{R}$ then you know that they converge (because they are Cauchy). Say $x_{n}\to x$ and $y_{n}\to y$. Then $ax_{n}+by_{n}\to ax+by$. Convergent sequences are Cauchy.

If you really want to bring out the $\epsilon$'s, then you could say that since $x_n\to x$ there is an $N_x$ so that if $n\geq N_x$ then $|x_n - x|<\frac{\epsilon}{2|a|}$ Similarly, there is an $N_y$ so that for $n\geq N_y$ you have that $|y_n-y|<\frac{\epsilon}{2|b|}$. Then, if $n\geq \max\{N_x, N_y\}$ it follows that

\begin{align*} |ax_{n}+by_{yn}-(ax+by)| &\leq |ax_{n}-ax| + |by_{n}-by| \\ &= |a||x_n - x|+|b||y_n-y| \\ &<|a|\frac{\epsilon}{2|a|}+|b|\frac{\epsilon}{2|b|} \\ &=\epsilon \end{align*}

$\endgroup$
0
$\begingroup$

do it in two phases.

$X_n$ Cauchy implies $aX_n$ is.

$X_n, Y_n$ Cauchy implies that $X_n+Y_n$ is .

For the first, given $\epsilon >0,$ there exist $N$ st $m,n>N$ implies $$ |X_n-X_m| < \epsilon/a. $$ So $$ |aX_n-aX_m| < \epsilon, $$ and we are done.

For the second, find $M,N$ so that $$ |X_n-X_m| < \epsilon/2. $$ with $n,m > M,$ $$ |Y_n-Y_n| < \epsilon/2 $$ with $n,m> N.$

Then the condition holds via the triangle inequality for $n,m > \max(N,M).$

$\endgroup$
0
$\begingroup$

Let $M=\max\{|a|,|b|\}$. Since $(x_n)$ and $(y_n)$ are Cauchy sequences, for every $\varepsilon>0$ there exists $n_1,n_2\in\mathbb{N}$ such that if $n,m>n_1,n_2$, $$|x_n-x_m|<\dfrac{\varepsilon}{2M}\ \ \ \text{and}\ \ \ |y_n-y_m|<\dfrac{\varepsilon}{2M}$$ Let $N=\max\{n_1,n_2\}$, then if $n,m>N$, \begin{align*} |(ax_n+by_n)-(ax_m+by_m)|&=|a(x_n-x_m)+b(y_n-y_m)|\\ &\leqslant |a||x_n-x_m|+|b||y_n-y_m|\\ &\leqslant M(|x_n-x_m|+|y_n-y_m|) \\ &< M\left(\dfrac{\varepsilon}{2M}+\dfrac{\varepsilon}{2M}\right)=\varepsilon \end{align*} So $(ax_n+by_n)$ is Cauchy.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.