# Understanding Complete Metric Spaces and Cauchy Sequences

From my own definition, I have concluded that a complete metric space is a set and a metric where the set consists of no holes in it. Book definitions describe that "A complete metric space is a metric space in which every Cauchy sequence is convergent." I understand that a metric is a distance measuring device defined on an arbitrary set, and when speaking of a "metric space" they are talking about a set and a metric defined on that set (X,d). However, I have yet to get an understanding of a "Cauchy sequence" when speaking of a Complete Metric Space. I am seeking an example of a complete metric space, relatively one that I can interpret.

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@Michael how would I go about proving that in an example. If I were to explain to someone else, how could I use what you commented to back if up? –  user58315 Feb 26 '13 at 20:25
I was puzzled by your first sentence so ignored it in my answer. But what do you mean by "From my own definition, I have concluded that a complete metric space is a set and a metric where the set consists of no holes in it." What is your definition of a complete metric space?? –  Pete L. Clark Feb 26 '13 at 20:28
Sorry, I just deleted it because it also came up in the answer. A proof that $\mathbb{R}$ is complete would go something like this: Show that a Cauchy sequence is bounded. By the Bolzano-Weierstrass theorem, a bounded sequence in $\mathbb{R}$ has a convergent subsequence. Finally, show that a Cauchy sequence with a convergent subsequence is convergent and combine these facts. –  Michael Greinecker Feb 26 '13 at 20:29
@MichaelGreinecker can you repost what you deleted? –  user58315 Feb 26 '13 at 22:04
@user58315 Sure: If $X=\mathbb{R}$ and $d$ is given by $d(x,y)=|x-y|$, then $(X,d)$ is a complete metric space. –  Michael Greinecker Feb 26 '13 at 22:11

Example -1: Any set endowed with the discrete metric is complete: every Cauchy sequence is eventually constant, hence convergent.

Example 0: A subset $Y$ of a complete metric space $(X,d)$ is complete with the inherited metric if and only if it is closed.

Example 1: The real numbers $\mathbb{R}$ with $d(x,y) = |x-y|$. (Some people regard using $\mathbb{R}$ as an early example of a metric space to be circular; I am not one of them.)

Example 2: Any compact metric space. (More generally, one has the characterization of compact metric spaces as those which are complete and totally bounded.)

Example 3: a) For any positive integer $n$, if $(X_1,d_1),...,(X_n,d_n)$ are complete metric spaces, and we endow the Cartesian product $X = \prod_{i=1}^n X_i$ with any of several reasonable metrics -- e.g. $d(x,y) = \max_{1 \leq i \leq n} d(x_i,y_i)$ -- then $(X,d)$ is a complete metric space.
b) If $\{(X_n,d_n)\}_{n=1}^{\infty}$ is a sequence of complete metric spaces, and we endow $X = \prod_{i=1}^{\infty} X_i$ with the metric $d(x,y) = \sum_{i=1}^{\infty} \frac{1}{2^i} \frac{ d_i(x_i,y_i)}{1+d_i(x_i,y_i)}$, then $(X,d)$ is a complete metric space.

Example 4: For any metric space $X$, let $C_b(X)$ be the set of bounded, continuous functions $f: X \rightarrow \mathbb{R}$, endowed with the metric $d(f,g) = \sup_{x \in X} |f(x) - g(x)|$. This is a complete metric space and indeed a Banach space.

Example 5: The completion of any metric space. For instance, completing the rational numbers with respect to the $p$-adic metric one gets the field $\mathbb{Q}_p$ of p-adic numbers.

I thought about taking seriously the idea of formalizing "no holes" as a definition of a complete metric space. Here is what I came up with:

Proposition: For a metric space $(X,d)$, the following are equivalent:
(i) For any isometric embedding $\iota: (X,d) \rightarrow (Y,d)$ of $X$ into another metric space $Y$ and any sequence $\{x_n\}$ in $X$, if $\iota(x_n)$ converges in $Y$ then $x_n$ converges in $X$.
(ii) $X$ is complete.

Proof: The basic observations here are that if $\iota: (X,d) \rightarrow (Y,d)$ is an isometric embedding and $\{x_n\}$ is a sequence in $X$, then:
$\bullet$ $\{x_n\}$ is Cauchy iff $\{\iota(x_n)\}$ is Cauchy, hence also
$\bullet$ if $\{ \iota(x_n)\}$ is convergent, then $\{x_n\}$ is Cauchy.
Then (ii) $\implies$ (i) is immediate; to show (i) $\implies$ (ii) look at the completion $\iota: X \rightarrow \tilde{X}$ of $X$.

Thus the "holes" in $X$ are detected by embeddings into larger spaces. I am skeptical though that this definition would be helpful for beginning students: aside from relying on the existence of the completion of a metric space, the idea of considering all possible embeddings of one metric space into another seems relatively abstract and sophisticated.

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Interesting numbering. –  Michael Greinecker Feb 26 '13 at 20:24
@Michael: it's hard to defend, but it seemed reasonable at the time. –  Pete L. Clark Feb 26 '13 at 20:25
Is it because the first example is underground, and never heard of it? –  Pedro Tamaroff Feb 26 '13 at 20:41
I really like your characterization of completeness. –  Michael Greinecker Feb 26 '13 at 22:31
@Michael: Thanks, I liked it too. Here's a question: using this alternate definition can one show that every metric space isometrically embeds in a complete metric space without rehashing the standard theory of Cauchy sequences and equivalence classes thereof? –  Pete L. Clark Feb 27 '13 at 6:42

Saying that it has no holes in it may not be a very accurate description. As an example, for any open set in $\mathbb{R}$ there exists a metric (possibly different from the usual metric) that generates the usual topology, and is complete. Even for a set as $\mathbb{R}\setminus\{0\}$, which seemingly has a whole within it. However, note (!) that this set is not a complete metric space with the usual metric. Let me explain why so in a moment.

So, how could one define this notion of completeness more precisely?

• If $(X,d)$ is a metric space, we say that a sequence $(x_{n})_{n=1}^{\infty}\subseteq X$ is a Cauchy sequence, if for every $\varepsilon>0$ there exists $n_{\varepsilon}\in\mathbb{N}$ so that $d(x_{n},x_{m})<\varepsilon$ for all $n,m\geq n_{\varepsilon}$. So in other words, if for any fixed positive constant there exists an index from which onwards the members of the sequence are closer than this fixed constant from each other.

• We say that a metric space $(X,d)$ is complete if every Cauchy sequence $(x_{n})_{n=1}^{\infty}\subseteq X$ converges to some $x\in X$.

In the case of the usual metric $d(x,y)=|x-y|$ defined on $\mathbb{R}\setminus\{0\}$, one may note that $(\frac{1}{n})_{n=1}^{\infty}$ is a Cauchy sequence that does not converge to any point in $\mathbb{R}\setminus\{0\}$. Hence this metric space is not complete.

A common example of a complete metric space is $\mathbb{R}^{n}$ for any $n\in\mathbb{N}$ with the usual metric, or any closed subset of theirs.

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The essential point in the definition of Cauchy sequences is the following: You can test whether a sequence $(x_n)_{n\geq0}$ is Cauchy just by looking at the $x_n$ themselves; you don't have to know their limit, which maybe doesn't even exist.

Now a Cauchy sequence is a sequence that hopefully converges, because ultimately the terms of the sequence are very near to each other. A metric space where all Cauchy sequences indeed do converge is called complete.

The metric space ${\mathbb R}$ is complete. This is is a deep theorem about the fine structure of ${\mathbb R}$. Other theorems about limits, like $$a_n\to a,\quad b_n\to b\quad \Rightarrow\quad a_n+b_n\to a+b$$ are simple consequences of the definitions and of the continuity of the arithmetic operations.

So the question arises whether incomplete metric spaces can arise in a natural way. Consider the open interval $I:=\ ]0,1[\ \subset{\mathbb R}$. It inherits the metric from ${\mathbb R}$ and is a metric space in its own right. The sequence $$x_n:={1\over n}\in I\qquad(n\geq1)$$ converges to $0\in{\mathbb R}$; therefore it is a Cauchy sequence. On the other hand it cannot converge to some other point $\xi\in {\mathbb R}$, so in any case it is divergent in $I$. This shows that $I$ is not complete, even though nobody has placed any holes there.

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Before moving on, you have probably seen the concept of a Cauchy sequence in the Euclidean spaces $\mathbb{R}$:

A sequence $\langle x_n \rangle_{n=1}^\infty$ in $\mathbb{R}$ is a Cauchy sequence if for any $\epsilon > 0$ there is a $N \in \mathbb{N}$ such that $\| x_m - x_n \| < \epsilon$ for all $m,n \geq N$. (Where $\| \cdot \|$ is the Euclidean norm on the underlying space; for $n = 1$ this is just the absolute value.)

Moving from the Euclidean spaces to general metric spaces just involves replacing the "$| \cdot - \cdot |$" in the definition above with "$d ( \cdot , \cdot )$", where $d$ is the metric under consideration.

So what is a Cauchy sequence? At an intuitive level it means that the points of the sequence eventually "cluster together" (i.e., become arbitrarily close to each other). To check that such a sequence converges we need to somehow find some point about which all of the sequence elements eventually cluster. Perhaps in this sense we can say that there are no "holes": any sequence that eventually clusters is in fact eventually clustering about a point of the space.

But before you become too attached to this idea, consider the following degenerate example (also essentially given in other answers).

Consider the set $\mathbb{Z}$ of integers with the metric inhereted a subset of $\mathbb{R}$: $d ( m , n ) = | m - n |$. It is not hard to show that it is complete: the only Cauchy sequences are the eventually constant sequences, which trivially converge. (If $\langle x_n \rangle_{n=1}^\infty$ is a Cauchy sequence in $\mathbb{Z}$, consider the least $N$ such that $d ( x_m , x_n ) < 1$ for all $m,n \geq N$).

However I doubt people look at the above space and say that there are no holes; I think most would actually say that it is mostly holes!

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