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!

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    $\begingroup$ I know comments aren't the place for idle remarks, but I just wanted to commend you on the depth of thought portrayed in this question! To me at least, the idea of considering the "completion" of a space - the space of equivalence classes of Cauchy sequences - was a terribly difficult jump in understanding. Keep up the good work! $\endgroup$ Commented Oct 11, 2021 at 17:29
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    $\begingroup$ The important detail that you're missing is that "converges" is defined relative to the space that the sequence and limit will live in. For example, the sequence $x_1=1;x_{n+1}=\frac{x_n+2/x_n}{2}$ does not converge if you consider it as living in $\mathbb{Q}$. $\endgroup$
    – Ian
    Commented Oct 11, 2021 at 19:09
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    $\begingroup$ That said, if you have a problem like "show that $C([0,1])$ with the uniform metric is complete", recasting things to the way you wrote is a sensible thing to do. But there is a step to achieve that. Specifically, you say "for each $x \in [0,1]$, $f_n(x)$ is Cauchy so by completeness of $\mathbb{R}$ it has a limit which we can call $f(x)$" and now you've "assembled" the $f$ that the sequence would have to converge to, and so it suffices to show that this $f$ is in fact in the space you're looking at. $\endgroup$
    – Ian
    Commented Oct 11, 2021 at 19:11
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    $\begingroup$ Aside: the set $\mathbb{R}$ of real numbers is sometimes said to be defined as the completion of $\mathbb{Q}$. That is, we assume any Cauchy sequence of rationals converges to some value, and define what it means for a pair of Cauchy sequences to converge to the same value, and then real analysis is about determining the properties of that set of values. $\endgroup$
    – aschepler
    Commented Oct 12, 2021 at 2:58
  • $\begingroup$ @aschepler $\mathbf R$ is the completion of $\mathbf Q$ with respect to a certain metric (or absolute value). The field $\mathbf Q$ has many other completions (the various $p$-adic fields for prime $p$, for instance) and a choice of completion depends on the metric under consideration. $\endgroup$
    – KCd
    Commented Oct 12, 2021 at 5:38

1 Answer 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 $\delta_r$.

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    $\begingroup$ Finally, finally, a really nice question went into the hot question queue and got what it deserved : heat. +1. $\endgroup$ Commented Oct 11, 2021 at 16:24
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    $\begingroup$ "the space of Lebesgue-integrable functions on [0,1], up to a.e. equivalence." Does "a.e." stand for "almost everywhere"? I wouldn't expect the OP to be familiar with that abbreviation. $\endgroup$ Commented Oct 12, 2021 at 2:46
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    $\begingroup$ +1 but (1) in the first equation, the $|\cdots|$ should be a $d_1$, (2) in the last part, some more explanation might be needed about how a piecewise continuous function can be element of $Y$ (which was previously defined to be the set of Cauchy sequences in $X$) $\endgroup$
    – Noiralef
    Commented Oct 12, 2021 at 14:29
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    $\begingroup$ Another enveloping space which is sometimes used in constructing a completion is $C(X)$, the space of continuous functions $X \to \mathbb{R}$. This is always complete regardless of whether $X$ is complete or not; you can embed $X$ in it via $x \mapsto d(x, {-})$; and thus you can construct the completion of $X$ as the closure of the image of $X$. (Of course, that doesn't help if you want to construct $\mathbb{R}$ as a completion of $\mathbb{Q}$ in the first place; and also I don't think it will work for completions of general uniform spaces.) $\endgroup$ Commented Oct 12, 2021 at 16:36
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    $\begingroup$ For (1) note that X should contain at least two elements. $\endgroup$
    – user10575
    Commented Oct 15, 2021 at 12:19

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