0
$\begingroup$

Show that if $\{x_k\}$ is a Cauchy Sequence then the limit of sup{$\lvert x_i - x_j \lvert : i, j \geq k\}$ when k goes to infinity is equal to 0.

We know that $\{x_k\}$ is a Cauchy Sequence therefore for all l > 0 exists k $\in$ N such that such that if I, j $\geq$ k then $\lvert x_i - x_j \lvert$ < l, now take $l = 1/k_0$ then $\lvert x_i - x_j \lvert$ < $1/k_0$. Therefore, sup{$\lvert x_i - x_j \lvert : i, j \geq k_0\} \leq 1/k_o$. Therefore given l > 0 exists $k_0$ $\in$ N such that $1/k_0$ < l, so $\lvert$ sup{$\lvert x_i - x_j \lvert : i, j \geq k_0\}$ - 0 $\lvert $ $\leq$ $1/k_0$ < l, so we are done.

$\endgroup$

1 Answer 1

2
$\begingroup$

Let $C_k=\sup \{|x_i-x_j|:i,j\geq k\}$. $\{x_k\}$ is bounded since it is Cauchy. So $C_k<\infty$ for all $k$, also note that $\{C_k\}$ is a decreasing sequence of non-negative reals. There exists $C\geq0$ such that $\lim_{k \rightarrow \infty} C_k=C$. If $C>0$, for each $k \in \mathbb{N}$ there exists $i_k>j_k>k$ such that $C_k-\frac{1}{k}<|x_{i_k}-x_{j_k}|~(\leq C_k)$, taking limit as $k \rightarrow \infty$ we get $\lim_{k \rightarrow \infty}C_k=0$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .