# Proving that $l_\infty$ is complete

I'm learning about Hilbert spaces and operators theory, from some book. I came across the following question -

What I don't understand in the proof -

Why can we understand that each sequence is Cauchy?

Moreover, why the following inequallity is true?

• Every sequence is not Cauchy. Completeness is a condition on Cauchy sequences, that is why the proof only needs to work with those that are so. – user647486 Apr 12 at 17:24
• @user647486 I meant, every sequence in $x^{(k)}$ is Cauchy. Why? – ChikChak Apr 12 at 17:28
• In the chain of inequalities, the first one is triangle inequality $|x_j|=|x_j-x_j^{(k)}+x_j^{(k)}|\leq |x_j-x_j^{(k)}| + |x_j^{(k)}|$. The second comes from choosing $\epsilon <1$. The last one is from the definition of $\|x^{(k)}\|_{\infty}$, which is the suppremum of $|x_j^{(k)}|$ for all $j$. – user647486 Apr 12 at 17:28

If $$\sup_n|x_n^{(k)}-x_n^{(m)}|\leq\epsilon$$ then it follows that for each $$n\in\mathbb{N}$$ we have $$|x_n^{(k)}-x_n^{(m)}|\leq\epsilon$$, because this expression is not bigger than the supremum on $$n$$. It follows that $$(x_n^{(k)})_{k=1}^\infty$$ is Cauchy for each $$n$$. (Cauchy with respect to the usual metric in $$\mathbb{C}$$).
As for the second question: the first inequality is the triangle inequality, the second follows from the fact that $$x_j^{(k)}\to x_j$$, the third follows from the definition of supremum norm.
• Well why can we assume that $\sup_n|x_n^{(k)}-x_n^{(m)}|\leq\epsilon$? We are taking some arbitrary Cauchy sequence in $l_\infty$ i.e., each element of $x_k$ is a sequence in $l_\infty$. So how can we know that $\sup_n|x_n^{(k)}-x_n^{(m)}|\leq\epsilon$? – ChikChak Apr 12 at 17:45
• We take a Cauchy sequence $(x^{(k)})_{k=1}^\infty$ of elements in $l_\infty$. This is the only thing we assume. Now, an element in $l_\infty$ is a sequence of complex numbers. So for each $k\in\mathbb{N}$ we have that $x^{(k)}$ is a complex sequence $(x_n^{(k)})_{n=1}^\infty$. Now what does it mean that $(x^{(k)})_{k=1}^\infty$ is Cauchy? It means that for each $\epsilon>0$ there is $N\in\mathbb{N}$ such that for all $k,m>N$ we have $||x^{(k)}-x^{(m)}||_\infty<\epsilon$. And now just use the definition of the infinity norm. For a sequence $x=(x_n)$ the definition is $||x||_\infty=\sup_n|x_n|$. – Mark Apr 12 at 17:53