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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?

Verónica Rmz.
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ABC
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4 Answers4

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A metric space is complete if every Cauchy sequence converges (to a point already in the space). 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.

Jonas Meyer
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  • @JonasMeyer So when Wikipedia writes "a set is closed if and only if it contains all of its limit points", then this is under the assumption that the space is a complete metric space? I'm asking because then the Wikipedia entry would be wrong. – Rudy the Reindeer Jan 12 '12 at 19:58
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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
  • ...subspaces that are complete in the restricted metric. – Jonas Meyer Jan 12 '12 at 20:17
  • @JonasMeyer Thank you! My confusion arose from the fact that I considered $\sqrt{2}$ a limit point of $\mathbb{Q}$ in $\mathbb{Q}$ I think. And as anon stated concisely. "A limit point of X in Y has to actually exist in Y.". – Rudy the Reindeer Jan 12 '12 at 20:22
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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
  • ...$(x_n)$ defined above converges to $\sqrt 2\not\in \mathbb Q$. But, every metric space is a closed subset of itself. And yes, the quote of anon you gave is the point. – Jonas Meyer Jan 12 '12 at 20:28
  • Prof. Meyer would you care to explain what you mean by "A metric space is complete if every Cauchy sequence converges." I thought Cauchy sequences always converged from their definition... – phoenix Mar 14 '13 at 06:28
  • You should read: http://en.wikipedia.org/wiki/Cauchy_sequence – Seirios Mar 14 '13 at 06:44
  • Consider a Cauchy sequence of rational numbers converging to $\pi$. Now consider this sequence in $\Bbb Q$. – Asaf Karagila Mar 14 '13 at 06:44
  • @user65165: No, that is not the definition. My second to last comment gave an example of a Cauchy sequence in $\mathbb Q$ that does not converge in the metric space $\mathbb Q$. Alternatively, in the metric space $\mathbb R\setminus{0}$ with the usual absolute value distance, the sequence $(1,1/2,1/3,1/4,\ldots)$ is a Cauchy sequence that does not converge. – Jonas Meyer Mar 14 '13 at 14:03
  • @JonasMeyer: is there an example of an incomplete but closed subspace of an incomplete vector space over $\mathbb R$? – Leo May 15 '13 at 22:08
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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 00:45
  • @JonasMeyer: thank you for the response and sorry for being imprecise. I meant a proper subspace. So an example of an incomplete but closed proper subspace of an incomplete vector space over $\mathbb R$? Could this work: direct product of $\mathbb R^2$ and $(C^1(\mathbb R), ||·||_\infty)$. Then the resulting vector space is incomplete. And a proper subspace (say, direct product of the $y$-axis and the whole $C^1$) is closed yet incomplete. Right? – Leo May 16 '13 at 01:21
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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 01:27
  • @JonasMeyer: you are perfectly right, I intended to say $(C^1([a,b],\mathbb R), ||·||_\infty)$. Would my example work then? I wanted to construct a simple one for my studies of Analysis. Since I didn't have Functional Analysis yet, your examples are too advanced for me. – Leo May 16 '13 at 01:41
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    @mathusiast: I don't see how polynomials are more advanced than $C^1$ functions in this context. Yes, your example would work, perhaps more simply as ${0}\times C^1[a,b]\subset \mathbb R\times C^1[a,b]$. Or you could take ${f\in C^1[a,b]: f(a)=0}\subset C^1[a,b]$ (with sup norm). – Jonas Meyer May 16 '13 at 01:44
  • Please I don't see the reasoon why $\mathbb{Q}$ has to be closed in itself, since it cannot contain its limit points as $\sqrt{2}$ is definitely one of them. – Machinato Jan 06 '19 at 23:12
  • @Machinato: You have given a reason why $\mathbb Q$ is not closed in $\mathbb R$. Every space is closed in itself as a direct consequence of the definition of closed. For $A\subset \mathbb Q$ to not be closed in $\mathbb Q$ there would have to be an $x\in \mathbb Q$ that is a limit point of $A$ but with $x\not\in A$. This can't happen with $A = \mathbb Q$. – Jonas Meyer Jan 08 '19 at 00:57
  • @JonasMeyer This topic is for me very confusing. Suppose I have a subset of $\mathbb{Q}$ which contains all finite truncated decimal expansions of the square root of 2, then what is the limit point for this subset in a sense of complete metric space Q? Maybe I am missing something.... – Machinato Jan 09 '19 at 06:06
  • @Machinato: $\mathbb Q$ is not a complete metric space. The sequence of truncated decimal expansions of $\sqrt 2$ is a Cauchy sequence in $\mathbb Q$ that does not converge in $\mathbb Q$, giving an example of why $\mathbb Q$ is not complete. Within $\mathbb Q$, that sequence simply has no limit. Within $\mathbb R$, that sequence has a limit not in $\mathbb Q$, showing that $\mathbb Q$ is not closed in $\mathbb R$. – Jonas Meyer Jan 10 '19 at 21:39
  • Respectfully, I think this answer could be greatly clarified with two edits. (1) The first sentence would be much clearer if you add "A metric space is complete if every Cauchy sequence converges to a point in the space." After all, every Cauchy sequence in any metric space $M$ converges to some point in the completion of $M$. This proviso may seem to you to be implicit in the definition of the word "converges", but I think it's the source of many people's confusion so it's worth stating explicitly. – tparker Jul 18 '20 at 13:53
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    (2) It's worth clarifying that the concepts of "complete" and "closed" are fundamentally different for the following reason: completeness is a standalone property of a metric space itself, with no reference needed to any other space. However, closure (of any topological set, not just a metric space) is a relative property. Strictly speaking, it's meaningless to say that "a set is closed"; you can only say that "a subset of a topological space $T$ is closed in $T$". Closure is not a property of sets; it's a property of subsets of topological spaces with respect to the overall space. – tparker Jul 18 '20 at 13:59
  • The connection between the notions of "complete" and "closed" is the following: every metric space (like any topological space) is closed within itself, but not every metric space is closed within its completion. On the other hand, a complete metric space is closed within any metric superspace that contains it (and the converse is true as well). – tparker Jul 18 '20 at 14:05
  • @tparker: Thank you for the thoughtful comments. I can see your point with (1)--it isn't logically needed, but the extra emphasis may help some readers. I'll consider adding that. Regarding (2), your comments are interesting and may help some readers, but I don't currently agree that adding them to this answer would improve it. – Jonas Meyer Aug 04 '20 at 22:00
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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.

Seirios
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Completeness asks for a space: does every Cauchy sequence converge to a limit in that space? This is the case for R.

A subset of a space is closed if it contains its limit points. It should be intuitive that if you are a subset of R, then any sequence in your subset that converges must converge in R. Now the question is: will that point still be in my set? If so, it is closed. That's why, by definition, a closed interval in a complete space must be complete.

Steps towards showing that any closed interval in R is complete.

mrmagicfluffyman
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An interesting thing about you question need to be noticed. A complete metric space $Y$ is a metric space $(Y,d_Y)$ such that every Cauchy sequence determined by the metric $d_Y$ is convergent for some point of $Y$. A closed subset $Y$ of a metric space $X$ is a set such that every convergent sequence of points in $Y$ converges to a point of $Y$ - with the metric $(X,d).$

This raise a funny observation. Let us prove that every complete subspace $Y$ is closed in $X$; that is, if $Y$ is endowed with the same metric as X and is complete relative to its metric, then $Y$ is closed in $X.$

Proof: Let $d_Y: Y \times Y \rightarrow \mathbb{R}$ be the induced metric of $d: X \times X \rightarrow \mathbb{R}.$ Then if $y \in X$ is limit of a convergent sequence of points in $Y$ with the metric $d$, say $\left(y^{k}\right)_{n\in\mathbb{N}}$, then for each $\epsilon>0$ there exist a $N_{\epsilon}$ such that $d(y_m,y_n)<\epsilon$ for all $n,m>N_{\epsilon}.$ Now, note that we can evaluate the function $d_Y$ at $(y_m,y_n),$ since each coordinate belongs to $Y$. Then for each $\epsilon>0$ there exist a $N_{\epsilon}$ such that $d_{Y}(y_m,y_n)<\epsilon$ for all $n,m>N_{\epsilon}.$ Hence $\left(y^{k}\right)_{n\in\mathbb{N}}$ is Cauchy and there exist a point $\bar{y} \in Y$ such that $d_Y(y^{k}, \bar{y}) \rightarrow 0$. Since $d$ is equal to $d_Y$ for every point of $Y$, then $d(\bar{y},y) \leq d(\bar{y},y^{k})+d(y^{k},y)\rightarrow 0$. Finally, we can conclude that $\bar{y}=y$ with $\bar{y} \in Y$ and $y$ belongs to $Y.$

This is a detailed proof. I found it quite funny to prove it without using a circular argument.

R. W. Prado
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