why is triangle inequality still needed to prove completeness in $l^p$ spaces

Question

I read from the usual proofs that show that $l$ spaces, i.e. $l^{\infty}$, $l^1$, $l^2$ are complete - use these steps:

1) define a Cauchy sequence $\textbf{x}_n$

2) show that the pointwise limit of $\textbf{x}_n$, i.e. $\textbf{x}$, is in $l$

3) use the triangle inequality and the distance function to show that $\textbf{x}_n \rightarrow \textbf{x}$

Part (2) is usually done by using the boundedness of a Cauchy sequence, i.e. since all Cauchy sequences are bounded, it follows that the pointwise limit $\textbf{x}$ is finite, so that $\textbf{x} \in l$...

But am not sure why (3) is still needed... isn't it true that all Cauchy sequences are bound to converge to its limit? since we already showed that the limit $\textbf{x}$ is in $l$ (using step 2), why do we still need the triangle inequality to show that $\textbf{x}_n$ converges to $\textbf{x}$ (since we already assumed that it is Cauchy at the beginning)...

Answer 1

Well the result is true so its a little hard to see what could go wrong. In broad strokes, the proof says, "look for convergence in a weaker topology" (convergence in $\ell^1$ implies componentwise convergence but not vice versa), and then "upgrade this to convergence in $\ell^1$".

In a possible failed attempt of the proof, it could be that you chose such a weak topology that limits are not unique. For example, say that $x_n$ "Obviously" converges to $x$ if the first component of $x_n$ converges to the first component of $x$, $x_n(1) \to x(1)$. Then, trying to go through the proof scheme:

1. Take $x_n$ Cauchy in $\ell^1$.
2. Note that $x_n$ "Obviously" converges to $x:=(\lim_{n\to\infty} x_n(1) , 0,0,0,\dots)$.
3. Conclude that $x_n$ converges to $x$ in $\ell^1$...??

So part of the reason it works is because the proof uses pointwise instead of some other badly chosen convergence. In more general spaces, its plausible that the correct weaker notion of convergence is not easy to find, and before you have the proof, its not obvious what to use. Hence you should check (however trivial) that pointwise convergence of the Cauchy sequence does in fact imply $\ell^1$ convergence.