I know that a metric space $X$ is called complete if and only if every Cauchy sequence in $X$ converges to a point in $X$.

But we can also say something about Cauchy sequences in incomplete spaces. For example, we can say that a Cauchy sequence cannot diverge (tend to $\infty$). Also, it cannot oscillate (unless the oscillation gets smaller and smaller). But the limit, let's call it $y$, might not be in $X$. Yet there is always a point such that the sequence "tends" to it (I write tends in quotes since it is not really defined if the "limit" is not in the space.)

What is the mathematically correct way to express these facts about Cauchy sequences in metric spaces? Can we say if $x_n$ is a Cauchy sequence in $X$ then there (always) exists a point $y$ such that for $n$ large enough all $x_n$ are epsilon close to $y$ when $y \notin X$?

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    $\begingroup$ If you are going to use quotes in your second paragraph, then they actually belong around "point" rather than around "tends". The problem is that there may be no such point. It usually makes no sense to talk about "$y\notin X$ in the absence of a superset of $X$, which is not given. $\endgroup$ – Arturo Magidin Jul 8 '12 at 6:42
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    $\begingroup$ Perhaps what you want to look at is the process of "completion" of a metric space. In that completion, every Cauchy sequence converges; in particular, every Cauchy sequeence in $X$ converges in the completion, which gives you the $y$. $\endgroup$ – Arturo Magidin Jul 8 '12 at 6:43
  • $\begingroup$ @ArturoMagidin Thank you. Your second comment answers my question. $\endgroup$ – Rudy the Reindeer Jul 8 '12 at 7:03
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    $\begingroup$ With the metric $d(x,y) = |\arctan(x)-\arctan(y)|$ on $\mathbb{R}$, the sequence $x_n = n$ is cauchy, but clearly tends to $\infty$. $\endgroup$ – copper.hat Jul 8 '12 at 7:31
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    $\begingroup$ I was just illustrating that there is no obvious point to which the sequence converges. With an example such as $(0,1]$ (and usual metric), the sequence $x_n=\frac{1}{n}$ has an obvious limit in the ambient space. Point being that we need to construct something new, in general. In this case, the usual construction is the equivalence class of sequences. $\endgroup$ – copper.hat Jul 8 '12 at 7:50

Generally, one does not have a metric space embedded in some larger space where the "limits" of sequences that do not converge in $X$ may converge. So it really makes no sense to talk about "points to which the series converges but are not in $X$". As I noted in comments, if you really want to put (scare) quotes, they belong around the word "point", in so far as there is no such "point" to which these Cauchy sequences converge.

The solution to this is essentially the same as the one used to construct the reals from the rationals by considering all "possible limits" of Cauchy sequences: one constructs the completion of the space $X$ and embeds $X$ into that space. This completion, $Y$, is a complete metric space that comes equipped with an embedding $X\hookrightarrow Y$ such that (i) the embedding is uniformly continuous; (ii) [the image of] $X$ is dense in $Y$; and (iii) given any uniformly continuous function $f\colon X\to N$ into a complete metric space $N$, there is a unique uniformly continuous extension of $f$ to $\mathfrak{f}\colon Y\to N$. Viewing $X$ as a subspace of that $Y$, then you can talk about the limits of these Cauchy sequences in $X$ much like we can talk about the real limits of Cauchy sequences of rationals.

The universal property given above implies that if any such $Y$ exists, then it is unique up to a uniformly continuous homeomorphism, by the usual abstract nonsense arguments. So it suffices to construct any one such space. The standard construction mimics, as I mentioned above, the construction of $\mathbb{R}$ from $\mathbb{Q}$, specifically the construction of $\mathbb{R}$ as equivalence classes of rational Cauchy sequences. Namely, we let $C$ be the set of all Cauchy sequences of elements of $X$, and we define an equivalence relation on $C$ by letting $(x_n)\sim (y_n)$ if and only if $\lim\limits_{n\to\infty}d(x_n,y_n) = 0$. Then we let $Y$ be the quotient $C/\sim$, and define the metric by $$D\left(\overline{(x_n)},\overline{(y_n)}\right) = \lim_{n\to\infty}d(x_n,y_n).$$ One embeds $X$ into $Y$ by mapping $x$ to the class of the constant sequence $(x)$, and proves it has the appropriate properties.

Once you have this completion $Y$, now you can talk about those $X$-Cauchy sequences converging to points in $Y$ that are not in $X$; they converge in $Y$ because $Y$ is complete.

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    $\begingroup$ I think the first paragraph is too pessimistic. Given a metric space $M$, we may say that a sequence $a_n$ in $M$ potentially converges if there exists a metric space $N$ and an isometry $f : M \to N$ such that $f(a_n)$ converges in $N$; the limit of $f(a_n)$ may then be regarded as a "potential limit" of the sequence $a_n$. Then the statement you want to prove is that a sequence potentially converges if and only if it is Cauchy (one direction is clear and the other follows by the construction of the completion). $\endgroup$ – Qiaochu Yuan Jul 9 '12 at 0:26
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    $\begingroup$ @QiaochuYuan: I guess the point is that we really need to set a context before we can start talking about "points to which the sequence converges but are not in $X$". What you suggest is certainly one way to establish such a context; what I objected to is the tacit assumption that there are some "points" that just happen to be not in $X$. Essentially, thinking that the situation is like that of $\mathbb{Q}$ as a subset of $\mathbb{R}$, but before we even know that $\mathbb{R}$ exists (let alone that it is complete and we can embed $\mathbb{Q}$ into it). $\endgroup$ – Arturo Magidin Jul 9 '12 at 0:36

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