Does every Cauchy sequence converge to *something*, just possibly in a different space?

Solution 1:

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

1. $Y$ is not a metric space under $d_2$; different points of $Y$ might be distance-$0$ from each other.
2. For each $y\in Y$, there exists an equivalence class $c(y)=\{z:d_2(y,z)=0\}$. Let $Z$ be the set of all equivalence classes, i.e. $Z=\{c(y):y\in Y\}$. Then $d_2$ extends to $Z^2\to\mathbb{R}$ in the natural way.
3. $(Z;d_2)$ is a metric space.
4. $(X;d_1)$ embeds homeomorphically into $(Z;d_2)$ via $x\mapsto c(x,x,x,\dots)$.
5. $(Z;d_2)$ is complete.

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 $\delta_r$.