# Norm - Cauchy Sequence

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.

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 $$\lim_{k \rightarrow \infty}C_k=0$$.