I have two doubts on the initial statements on this proof of the completeness of $l^1$.

enter image description here

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.

  • $\begingroup$ 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. $\endgroup$ – Thomas Andrews Oct 5 '15 at 19:23
  • 1
    $\begingroup$ 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/… $\endgroup$ – Kolmin Oct 5 '15 at 19:24
  • $\begingroup$ Do you know the term "metric spaces"? I might be able to clarify if you do. @Monolite $\endgroup$ – Thomas Andrews Oct 5 '15 at 19:25
  • $\begingroup$ @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. $\endgroup$ – Monolite Oct 5 '15 at 19:31
  • $\begingroup$ Yeah, the problem is the notation - dealing with sequence of sequences. $\endgroup$ – 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 $\ell^1$.

  • 1
    $\begingroup$ (Sorry, originally wrote "vector spaces" when I meant "metric spaces.") $\endgroup$ – Thomas Andrews Oct 5 '15 at 19:53
  • $\begingroup$ 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. $\endgroup$ – Monolite Oct 5 '15 at 20:05
  • 1
    $\begingroup$ 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 $\mathbb R$. $\endgroup$ – Thomas Andrews Oct 5 '15 at 20:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.