Doubt on proving that $l^1$ is complete.

I have two doubts on the initial statements on this proof of the completeness of $l^1$. I don't understand why we can say that the sequence $x_k^{(n)}$ is a Cauchy sequence in $R$. I understand that $|x_k^{(n)} - x_k^{(m)}| \le \epsilon$, my doubt is that $x^{(n)}$ and $x^{(m)}$ are two different sequences.

When defining a Cauchy sequence in $R$ we say that a sequence is Cauchy if if for every positive real number $\epsilon_1$, there is a positive integer $N$ such that for all natural numbers $m, n > N$ $$|x_m - x_n| < \varepsilon$$

In this case the distance is between elements of the same sequence. In the proof we have that the components of different sequences can be made arbitrary small. I hope I have made clear what is confusing me.

The other doubt is that I do not understand why we suspect $X$ to be the limit of the sequence $X_n$, what makes us suspect that? Maybe solving my previous doubt will help me understand this.

• The first part: It is Cauchy because $|x_k^{(n)}-x_k^{(n)}|\leq\|X_n-X_m\|$ by the definition of $\|X_n-X_m\|$ and it follows since $\{X_i\}$ is Cauchy under the norm metric. – Thomas Andrews Oct 5 '15 at 19:23
• Not sure this can be helpful, but there is a previous question on mine that addresses this proof in some detail (by following Kreyszig's book, which is very nice in terms of notation, that can become a bit cumbersome when addressing this problem): math.stackexchange.com/questions/1096253/… – Kolmin Oct 5 '15 at 19:24
• Do you know the term "metric spaces"? I might be able to clarify if you do. @Monolite – Thomas Andrews Oct 5 '15 at 19:25
• @ThomasAndrews thank you for your answer, a set with a distance function defined on it? I still have the doubt that $x^{(n)}$ and $x^{(m)}$ belong to two different sequences. – Monolite Oct 5 '15 at 19:31
• Yeah, the problem is the notation - dealing with sequence of sequences. – Thomas Andrews Oct 5 '15 at 19:47

I'm going to start from the general, then move to the specific, because part of the confusion is the notation of sequences of sequences.

In general, let $U,V$ be two metric spaces. Let $f:U\to V$ be defined so that $$d_V(f(u_1),f(u_2))\leq d_U(u_1,u_2)$$ for all $u_1,u_2\in U$. It is easy to prove that such a function is continuous.

Under these conditions, if $\{u_i\}$ is a Cauchy sequence in $U$, then we can show that $\{f(u_i)\}$ is a Cauchy sequence in $V$.

Now, in the above case, $U=\ell^1$ and $V=\mathbb R$, and $f=f_k$ is defined on $\ell^1$ as:

$$f_k(\{x_i\}_{i=1}^\infty) = x_k$$

All that the proof above is saying is that $f_k$ has this property, and thus if $X_1,\dots,X_n,\dots$ is Cauchy in $\ell^1$ then $f_k(X_1),\dots,f_k(X_n),\dots$ is Cauchy in the real numbers.

As for the question about why we "suspect." Go back to the general case. If $V$ is complete, and $u_1,\dots,u_n,\dots$ is Cauchy in $U$, then if it converges to $u\in U$, we'd have to have $\lim f(u_n)=f(u)$ since such $f$ is continuous.

In our specific case, if $X_n$ converges to $X\in\ell^1$, we'd have to have $f_k(X_n)\to f_k(X)$ for all $k$ - that is, we'd have to get the same result as component-wise convergence.

That's actually stronger than suspicion, then, because we see that the "component-wise" $X$ is the only possible candidate for the limit. But it still doesn't prove that it is the limit. We have not even yet shown that the component-wise limit is in $\ell^1$.

• (Sorry, originally wrote "vector spaces" when I meant "metric spaces.") – Thomas Andrews Oct 5 '15 at 19:53
• Oh so the sequence that we prove is cauchy is the sequence of elements at the $k$ spot of a given sequence $\in l^1$. Right? Thanks a bunch. And you totally demystified the "suspect" part. – Monolite Oct 5 '15 at 20:05
• Right. If you write each $X_i$ as a row, and write $X_{i+1}$ right under $X_i$, then the $k$ is the column, and we are looking at the sequence going down that column. The above proves that if $\{X_i\}$ is a Cauchy sequence in $\ell^1$, then the columns are Cauchy sequences in $\mathbb R$. – Thomas Andrews Oct 5 '15 at 20:07