Prove that $d_n$ is a Cauchy sequence in $\mathbb{R}$

Question

Let $(x_n$) and $(y_n)$ be Cauchy sequences in $\mathbb{R}^n$ , i.e. lim$_{n,m}$ |$x_n$ − $x_m$| = $0$ and lim$_{n,m}$ |$y_n$ − $y_m$| = $0$. For each n, let $d_n = |x_n − y_n|$. Prove that $d_n$ is a Cauchy sequence in $\mathbb{R}$.

Solution:

let $|x_n − x_m| < \epsilon/2$ and $|y_n − y_m| < \epsilon/2$

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

let $z = x_n − x_m$ and $w = x_n − x_m$

By Triangle inequality,

$|z + w|\leq|z| + |w|$

$\implies|x_n − x_m + (-y_n + y_m)| \leq |x_n − x_m| + |y_n − y_m| \leq \epsilon/2 + \epsilon/2 = \epsilon$

I know this needs some fixing up, but is this right?

Answer 1

You have the right idea; now you only need to specify how large $m,n$ need to be in order for $|x_n-x_m|<\varepsilon/2$ and $|y_n-y_m|<\varepsilon/2$ to both hold.

Answer 2

let $|x_n − x_m| < \epsilon/2 \,\,\, \forall m,n>N_1$ and $|y_n − y_m| < \epsilon/2\,\,\, \forall m,n>N_2$, this is the the definition of the above limites. Then, take $N_0=\max{N_1,N_2}$, for $m,n>N_0$ we have

$||x_n − y_n| - |x_m − y_m|| \leq |(x_n − y_n)- (x_m − y_m)| = |x_n − x_m + (-y_n + y_m)|\leq |x_n − x_m| + |y_n − y_m| \leq \epsilon/2 + \epsilon/2 = \epsilon$

Hence you have the definition of Cauchy sequence.

Answer 3

More generally, if $d$ is any metric on a set $X$,

$$|d(x,y)-d(z,w)|\leqslant d(x,z)+d(y,w)$$

Setting $x=x_n,y=x_m,z=y_n,w=y_m$, the above inequality proves that if $(x_n),(y_n)$ are Cauchy in $(X,d)$, $a_n=d(x_n,y_n)$ is Cauchy in $\Bbb R$. This is an important fact used in the construction of the completion of any metric space by means of equivalence classes of Cauchy sequences.