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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19$\begingroup$ 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. $\endgroup$– Qiaochu YuanCommented Oct 14, 2010 at 11:10
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$\begingroup$ Related: Why do we want complete spaces? We don't we just use closed spaces? $\endgroup$– MJDCommented Apr 4, 2022 at 22:24
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4 Answers
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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.
answered Oct 14, 2010 at 7:13
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4$\begingroup$ @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.... $\endgroup$ Commented Jan 12, 2012 at 20:12
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1$\begingroup$ ...$\lim_n x_n=x$ 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... $\endgroup$ Commented Jan 12, 2012 at 20:16
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4$\begingroup$ @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... $\endgroup$ Commented Jan 12, 2012 at 20:27
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1$\begingroup$ @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. $\endgroup$ Commented May 16, 2013 at 0:45
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1$\begingroup$ @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.) $\endgroup$ Commented May 16, 2013 at 1:27
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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.
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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.
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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.
