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

Question. If I attempt to prove that space X is complete by pursuing the strategy, “Assume xnx; the space X is complete if xX,” 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 xn is a Cauchy sequence; we will show that xn 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 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 fnf. We want to show that fX.” 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!


You are correct in the narrow sense that every Cauchy sequence does converge in some space.

To be precise, let (X;d1) be any metric space with at least two points, let Y be the set of Cauchy sequences in X, and define d2:Y2R; d2({xn}n,{yn}n)=lim 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.

Source : Link , Question Author : 1Teaches2Learn , Answer Author : Jacob Manaker

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