Question. If I attempt to prove that space $X$ is complete by pursuing the strategy, “Assume $x_n \rightarrow x$; the space $X$ is complete if $x \in X$,” then why is that wrong?
Context. I know the definition of Cauchy sequences and convergent sequences, and that the definition of completeness is that Cauchy sequences in the space converge. And so I know that if one is attempting to prove that a space is complete, then the usual proof should start, “Assume that $x_n$ is a Cauchy sequence; we will show that $x_n$ converges in $X$.”
The misconception I seem to be battling is this: It seems like a Cauchy sequence must converge to something, just that the something might not be in the space. So it seems to me like the question really is, “Is the limit in the space or is it not in the space?” The classic example is the sequence of rationals that converges to $\sqrt{2}$. The sequence is Cauchy within the space of the rationals, and also the sequence does converge, just to a limit that is outside the space in consideration. So recently, I began a proof of completeness with the line, “Assume $f_n \rightarrow f$. We want to show that $f \in X$.” And the feedback was, “Unclear what is being proved. Nothing related to completeness. 0/4 points.” It seems to me that showing that the limit of a convergent sequence resides in the space is equivalent to saying that Cauchy sequences converge. Why is that wrong?
Thank you!
Best Answer
You are correct in the narrow sense that every Cauchy sequence does converge in some space.
To be precise, let $(X;d_1)$ be any metric space with at least two points, let $Y$ be the set of Cauchy sequences in $X$, and define $d_2:Y^2\to\mathbb{R}$; $$d_2(\{x_n\}_n,\{y_n\}_n)=\lim_{n\to\infty}{d_1(x_n,y_n)}$$ Then it is easy (well, a decent homework problem, anyways) to verify that
Thus if we identify $X$ with the embedded subspace of $Z$, then any Cauchy sequence in $X$ converges in $Z$. The end limit might be $X$, or it might not; to show $X$ complete is to show that the end limit is in fact in $X$.
For this reason, $Z$ is called the completion of $X$.
With that said, some space is much more general than you give it credit for.
Given your notation ($f_n\to f$), I think you assumed that the limit of a Cauchy sequence of functions is itself a function. Probably $X$ was the space of continuous functions under the uniform norm, and so you thought that you just had to prove $f$ was itself a continuous function.
But $f$ could be far worse! For example, let $X$ be the space of continuous functions on $[0,1]$ with the following norm: $$d_1(f,g)=\int_0^1{|f(x)-g(x)|\,dx}$$ Then $X$ has a well-known completion: $L^1([0,1])$, the space of Lebesgue-integrable functions on $[0,1]$, up to a.e. equivalence.
If you haven't seen the Lebesgue integral, don't worry; the pathology carries over to Riemann-integrable functions. In particular, one can show that any piecewise-continuous function lies in the equivalent of $(Y;d_2)$ (to borrow notation from above).[*] So, for any $r$, the function $\delta_r$ where $$\delta_r(x)=\begin{cases}r&x=0\\0&x\neq0\end{cases}$$ is in $Y$.
But all those functions collapse to the same point in $(Z;d_2)$. That is, given any convergent sequence $f_n$, we can find some $f_{\infty}$ such that $f_n\to f_{\infty}$…and $f_n\to f_{\infty}+\delta_r$ too!
So we can't define an evaluation map to describe $f_{\infty}(0)$. Indeed, for any fixed point $x$, $f_{\infty}(x)$ is not well-defined.
[*]: As Noiralef pointed out in comments, I'm cheating here. I previously defined $Y$ as an equivalence set of Cauchy sequences; now I'm saying a function $\delta_r$ is in $Y$. What I mean here is, there exists a (uniformly bounded) Cauchy sequence of continuous functions converging pointwise to $\delta_r$.