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What is the difference between a complete metric space and a closed set?

Can a set be closed but not complete?

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Completeness and closure are not properties of sets; they are properties of metric spaces and of subsets of topological spaces (which include metric spaces), respectively. Context is everything in mathematics. –  Qiaochu Yuan Oct 14 '10 at 11:10

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up vote 24 down vote accepted

A metric space is complete if every Cauchy sequence converges. A subset $F$ of a metric space $X$ is closed if $F$ contains all of its limit points; this can be characterized by saying that if a sequence in $F$ converges to a point $x$ in $X$, then $x$ must be in $F$. It also makes sense to ask whether a subset of $X$ is complete, because every subset of a metric space is a metric space with the restricted metric.

It turns out that a complete subspace must be closed, which essentially results from the fact that convergent sequences are Cauchy sequences. However, closed subspaces need not be complete. For a trivial example, start with any incomplete metric space, like the rational numbers $\mathbb{Q}$ with the usual absolute value distance. Like every metric space, $\mathbb{Q}$ is closed in itself, so there you have a subset that is closed but not complete. If taking the whole space seems like cheating, just take the rationals in $[0,1]$, which will be closed in $\mathbb{Q}$ but not complete.

If $X$ is a complete metric space, then a subset of $X$ is closed if and only if it is complete.

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@Matt: No. If $X$ is a topological space and $A\subseteq X$, then a limit point of $A$ is an element $x$ of $X$ such that for every open subset $U$ of $X$ containing $x$, $(U\setminus\{x\})\cap A\neq \emptyset$, and $A$ is closed if and only if it contains all of its limit points if and only if $X\setminus A$ is open in $X$. This specializes to the case where $X$ is an arbitrary metric space, except that there (or in other first countable spaces) limit points can be characterized in terms of sequences: $x$ is a limit point of $A$ iff there exists a sequence $(x_n)$ in $A$ such that.... –  Jonas Meyer Jan 12 '12 at 20:12
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...$\lim_n x_n=x$ and $x_n\neq x$ for all $n$. Without assuming completeness it can be shown using this characterization of limit points in metric spaces that the statement appearing in Wikipedia is equivalent to the statement that if $A\subseteq X$, then $A$ is closed iff for every sequence $(x_n)$ in $A$ converging to some $x\in X$, it must be the case that $x\in A$. (For one direction, note that if the terms are all different from $x$, then $x$ is a limit point. If not, then $x\in A$ because $(x_n)$ is a sequence in $A$.) Completeness ensures that closed subspaces are the same as... –  Jonas Meyer Jan 12 '12 at 20:16
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@Matt: Completeness is not a relative property, and $\mathbb Q$ is not complete, because it has Cauchy sequences that don't converge, e.g. $x_1=1$, $x_n=\frac{1}{2}\left(x_{n-1}+\frac{2}{x_{n_1}}\right)$ for $n>1$. If you are considering the metric space $\mathbb Q$, there is no such thing as $\sqrt 2$; i.e., there is no missing limit point. However, since $\mathbb Q$ is not complete, it can be imbedded as a proper subspace of its completion, which is generally identified with $\mathbb R$. Then we know that $\mathbb Q$ is not closed in $\mathbb R$, because for example the sequence... –  Jonas Meyer Jan 12 '12 at 20:27
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@mathusiast: You mean a topological vector space? Every topological space is a closed subset of itself, so every incomplete space is an example. E.g., the real vector space of polynomials $\mathbb R[x]$ is a metric space with the distance between $p$ and $q$ being the maximum of the absolute values of the coefficients of $p-q$. This space, as a subspace of itself, provides an example of what you ask. –  Jonas Meyer May 16 '13 at 0:45
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@mathusiast: What is $\|\cdot\|_\infty$ supposed to mean on $C^1(\mathbb R)$? Perhaps you intended to restrict to bounded functions? Or if you don't mind thinking again of polynomials, just take the subspace $x\mathbb R[x]$, or $\mathbb R[x^2]$, etc. In the case of function spaces, you could take say $C[0,1]$ with $d(f,g)=\int_0^1|f(x)-g(x)|dx$, and consider the subspace of functions that vanish at $0$. (Direct products also give easy examples, but I wasn't sure quite what your example means.) –  Jonas Meyer May 16 '13 at 1:27

In some sense, a complete metric space is "universally closed": A metric space $X$ is complete iff its image by any isometry $i : X \to Y$ is closed.

Indeed, if $X$ is complete, $i(X)$ is a complete subspace of $Y$ so $i(X)$ is closed in $Y$; moreover, if $X$ is closed in its completion then $X$ is complete itself.

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