# How can one know every Cauchy sequence in a complete metric space converges?

I am new to Cauchy sequences. I stumbled onto them in the process of learning what a Hilbert space is. As I understand it, a Cauchy sequence is a sequence whose elements become arbitrarily close to each other as the sequence progresses.

But consider the following statement:

A metric space $X$ in which every Cauchy sequence converges to an element of $X$ is called complete.

How can we know every Cauchy sequence converges?

• You choose an arbitrary sequence in your space. You show that the arbitrary sequence is a Cauchy sequence. You can conclude that all sequences are Cauchy since the sequence you chose at the beginning was arbitrary. – Sujaan Kunalan Jan 26 '15 at 22:48
• If not every Cauchy sequence converges, then the space is not complete, by definition. For instance, the field of rational numbers is not complete – actually, $\mathbf R$ is the completion of $\mathbf Q$ (the smallest complete space that contains $\mathbf Q$). – Bernard Jan 26 '15 at 22:54
• @Bernard Yeah, that's the definition. But one still has to show a given metric space is complete -- that is, you have to show it satisfies that definition. That's the part I am confused by. – Stan Shunpike Jan 26 '15 at 22:57
• @SujaanKunalan does that mean every sequence in a complete metric space has to be Cauchy? – Stan Shunpike Jan 26 '15 at 22:57
• @StanShunpike: No. It just means that every Cauchy sequence converges. There can still be sequences that are not Cauchy. – Sujaan Kunalan Jan 26 '15 at 22:58

Your comment below the question seems to indicate that you are really concerned about a different question: Given a metric space, how can you prove that it is complete. This is a difficult question in general. One common path to doing this is something like the following:

1. It is relatively easy to prove that if a Cauchy sequence has a convergent subsequence, then the whole sequence converges to the same limit as the subsequence.

2. Now prove that every Cauchy sequence from your space has a convergent subsequence.

For example, to prove that $\mathbb{R}^{n}$ is complete, it is easy to prove that every Cauchy sequence from $\mathbb{R}^{n}$ is bounded, and then it is possible to prove that every bounded sequence has a convergent subsequence.

However, there are some metric spaces where things are different, and proving completeness is not as straightforward. In a finite-dimensional normed linear space, the closed unit ball is compact, but that is not true in infinite dimensional normed linear spaces. This has the consequence that in an infinite dimensional normed linear space, it is always possible to find a bounded sequence which has no convergent subsequence.

• I think you characterized the question I was striving for much better than I did. In fact, I received comments originally saying I made several statements that really were silly for a mathematician to say, but I think this is because I didn't have the words. – Stan Shunpike Jan 27 '15 at 1:53
• I asked this question because I was concerned that there might be a variety of metric spaces and the ability to prove completeness might vary depending on what other additional properties they have. user86419 mentioned compactness. I liked his answer, but I selected this one because it didn't single out one case in particular. – Stan Shunpike Jan 27 '15 at 1:55

$\Bbb R$ is a complicated example. Let's instead consider a simple example. Let $S$ be the space $\{7, 8\}$ with the usual distance. (That is, $d(7,7) = d(8,8) = 0,$ and $d(7,8) = d(8,7) = 1$.) I claim that $S$ is a complete space.

We are not going to show this by considering every cauchy sequence one at a time. As you observed, that is impossible.

Let $C$ be a cauchy sequence and the elements of $C$ are $c_0, c_1, \ldots$. Since $C$ is a cauchy sequence, we have, for any positive $\epsilon$, that elements of the sequence that are far enough out will be closer than $\epsilon$. Formally, for any given positive $\epsilon$ there is some integer $N$ so that for any $i,j>N$, we have $d(c_i, c_j) < \epsilon$.

Take, for example, $\epsilon = \frac12$. Then because $C$ is cauchy, we know there must exist some $N$ such that for any $i,j>N$, we have $d(c_i , c_j) < \frac12$, because that is what it means for $C$ to be cauchy; that is the definition.

But if we have $d(c_i , c_j) < \frac12$ then $d(c_i , c_j) = 0$, because the only possible values of $d(\ldots)$ are $0$ and $1$, and it can't be $1$ because $1$ is not less than $\frac12$. So it must be $0$.

So we conclude that if $C$ is cauchy, then for all sufficiently large $i$, and $j$ we have $d(c_i, c_j) = 0$ and therefore $c_i = c_j$ for all sufficiently large $i$ and $j$. Thus, every cauchy sequence in $S$ is eventually constant. It might have a mixture of $7$s and $8$s at the beginning, but there must come some point (that's $N$) after which all the elements are $7$s or all the elements are $8$s. If this doesn't happen, then $C$ was not a cauchy sequence.

Since $C$ is eventually constant, there is some value, either $7$ or $8$, call it $L$, such that all sufficiently late elements of $c$ are equal to $L$. That is, there is some $N$ such that for any $i>N$, $c_i = L$, where $L$ is either 7, or 8, depending on which $C$ we actually chose.

Is is then trivial to show that $C$ converges to the value $L$. To show this we need to show that for any positive $\epsilon$, the terms of the sequence are eventually within $\epsilon$ of $L$. In this case we can do even better; we know that the terms of $C$ are eventually equal to $L$. $C$ is therefore a convergent sequence.

Therefore, all cauchy sequences in $S$ converge, and $S$ is therefore a complete metric space.

The same argument goes through just as well if we take the more complicated space $\Bbb Z=\{\ldots, -1, 0, 1, 2, 3\ldots\}$ with the usual distance function $d(i,j) = |i-j|$; the proof is exactly the same.

Mathematics is not (usually) about considering a lot of examples. It is about what properties logically imply which other properties. Just as we can prove that every square number is positive without considering each square individually, we can also prove that every cauchy sequence of elements of $\Bbb Z$ must converge, even without considering each cauchy sequence individually.

It is a definition.

To show that a metric space is complete requires you to prove that every sequence satisfying $$\forall \epsilon>0 \exists N\in \mathbb{N}: d(x_i,x_j)\leq \epsilon,\ i,j\geq N$$

also satisfies

$$\exists \bar{x}\in X: (\forall \epsilon>0\exists N\in \mathbb{N}: d(x_i,\bar{x})<\epsilon, i\geq N)$$

To see how this is done in practice, you can examine these proofs showing that the space $\mathbb{R}$ equipped with the metric $d(x,y)=|x-y|$ is complete.

Another interesting proof that $\mathbb{Q}$ is not complete can be found here.

As I see it, the question is, "How can the criterion in the definition of completeness ever be verified (in principle, to say nothing of in practice), since the definition makes an assertion about arbitrary Cauchy sequences, and most spaces admit A Lot of Cauchy sequences?"

If one examines the proofs for, say, compact metric spaces or Euclidean spaces, there's a common theme: Show that every Cauchy sequence in $(X, d)$ has a convergent subsequence. It's a general fact in a metric space $(X, d)$ that a Cauchy sequence with a convergent subsequence is itself convergent in $(X, d)$.

Since a compact metric space is sequentially compact (every sequence has a convergent subsequence), a compact metric space is complete.

Since every real sequence has a monotone subsequence,[*] and every Cauchy sequence of reals is bounded (in the usual metric), and every bounded, monotone sequence of reals converges to a real limit, we conclude that every Cauchy sequence of reals has a convergent subsequence, and therefore converges.

And so forth. The point is, one doesn't explicitly inspect every Cauchy sequence; one proves that an arbitrary Cauchy sequence converges (e.g., by constructing a convergent subsequence), using nothing but the Cauchy property and properties of the "ambient" space $(X, d)$.

[*] Spivak's Calculus has a beautiful, elementary proof based on the concept of a peak point of a sequence $(x_{k})$, an index $n$ such that $x_{n+k} < x_{n}$ for all $k > 0$. If a sequence has finitely many peak points, one inductively constructs a non-decreasing subsequence; if instead there are infinitely many peak points, one constructs a strictly decreasing subsequence.

A metric space $X$ in which every Cauchy sequence that has a limit converges to an element of $X$ is called complete.
And you sense of what "Cauchy" means is a bit vague. For example, in the space $\Bbb{R}$ (the real line), the sequence $H_n \equiv \sum_{i=1}^n \frac{1}{i}$ has arbitrarily small distances between successive members, but is not is Cauchy because there for any given $m$ there is an $H_m$ with $m>n$ that is arbitrarily far from $H_m$. So this sequence it is not considered in deciding whether $\Bbb{R}$ is a complete space.
One way to prove that all Cauchy sequences that have a limit converge to an element of $X$ is to show that $X$ is compact, for that is a property of a compact space. This is easy to see: If $X$ is compact, then it contains all its limit points, and the limit of a Cauchy sequence is a limit point.
It turns out that compactness is a bit of a stronger condition than is needed for completeness. For example, $\Bbb{R}$ is not compact. But since it does contain open sets including each of the points it contains, it is easy to see that $\Bbb{R}$ contains all its limit points, and therefore the limit of a Cauchy sequence of elements of $\Bbb{R}$ is in $\Bbb{R}$ so $\Bbb{R}$ must be complete.
• The statement is correct as it stands: A complete metric space is one in which every Cauchy sequence converges. The point of the definition is that Cauchy sequences act like convergent sequences; complete spaces are ones in which they actually do converge. Unless you're implicitly considering some embedding of $X$ into its completion, your bolded edit makes every space complete. (How would a sequence have a limit but not be convergent, again unless you're implicitly considering $X$ as a subspace of some larger space?) – anomaly Jan 26 '15 at 23:58