Completeness and Category Are Related, but Not Identical
The Baire Category Theorem shows that every complete metric space is a Baire space: countably many open dense sets have dense intersection. Completeness is therefore a sufficient condition for an important category property. It is not the same property, however. An open subset of a complete metric space may be incomplete and still be a Baire space. To use category results accurately, we need to understand how completeness behaves when the space is replaced by a subspace.
Throughout, a subset \(Y\subseteq X\) carries the metric inherited from \(X\), unless another metric is specified. Completeness is always a property of a metric, not just of the set of points: the same set with a different metric may have different completeness behavior. Category is also relative to the ambient space, because closure and interior are taken in that space.
Closed Subspaces and Complete Subspaces
One basic way to obtain new complete spaces is to take a closed subset of a complete one. The converse is also useful: a subspace that is complete in the inherited metric must be closed in its ambient metric space. Together, these facts give an exact characterization when the ambient space is already complete.
Proof. First suppose \(X\) is complete and \(Y\) is closed. Let \((y_n)\) be a Cauchy sequence in \(Y\). It is also Cauchy in \(X\), since the distances are given by the same metric. Completeness of \(X\) gives a point \(x\in X\) such that \(y_n\to x\) in \(X\). Every \(y_n\) lies in \(Y\), so \(x\) is in the closure of \(Y\). Since \(Y\) is closed, \(x\in Y\). Thus the sequence converges in \(Y\), proving that \(Y\) is complete.
For the converse, suppose \(Y\) is complete and let \(x\in\overline{Y}\), where the closure is taken in \(X\). For each positive integer \(n\), choose \(y_n\in Y\) such that \(d(x,y_n)<1/n\). Such a point exists because every open ball around \(x\) meets \(Y\). For any \(m,n\), the triangle inequality gives \(d(y_m,y_n)\leq d(y_m,x)+d(x,y_n)<1/m+1/n\). Therefore \((y_n)\) is Cauchy in \(Y\). By completeness of \(Y\), it converges to some \(y\in Y\). It also converges to \(x\) in \(X\), because \(d(x,y_n)<1/n\). Limits in a metric space are unique, so \(x=y\in Y\). We have shown \(\overline{Y}\subseteq Y\), and hence \(Y\) is closed. \(\square\)
The second direction does not require \(X\) to be complete. The condition that \(Y\) be complete refers to its inherited metric; the proof shows that a limit in the ambient space of a sequence from \(Y\) must already belong to \(Y\).
Worked Example: The Parabola Is Complete in the Plane
Let \(P=\{(t,t^2):t\in\mathbb{R}\}\subseteq\mathbb{R}^2\), with the Euclidean metric inherited from the plane. The plane is complete, so it suffices to verify that \(P\) is closed.
Suppose \((t_n,t_n^2)\in P\) converges in \(\mathbb{R}^2\) to \((a,b)\). Coordinate convergence gives \(t_n\to a\) and \(t_n^2\to b\). Since the squaring function is continuous, \(t_n^2\to a^2\). Uniqueness of limits in \(\mathbb{R}\) then gives \(b=a^2\), so \((a,b)=(a,a^2)\in P\). Thus \(P\) is closed. By the theorem, \(P\) is complete with its inherited metric.
This argument also identifies exactly which metric is being used. It proves completeness for the Euclidean distance restricted to \(P\), not for an arbitrary distance one might define on the same set.
A Nested-Set Criterion for Completeness
The nested-ball construction in the Baire Category Theorem uses completeness to produce a point in all the closed balls. A related statement gives a useful characterization of completeness using arbitrary nested closed sets. The shrinking-diameter condition ensures that the sets cannot retain multiple distinct common points.
Proof. Suppose first that \(X\) is complete, and let \((F_n)\) satisfy the stated assumptions. Choose \(x_n\in F_n\) for each \(n\). If \(m\geq n\), nesting gives \(x_m\in F_m\subseteq F_n\), and also \(x_n\in F_n\). Hence \(d(x_m,x_n)\leq\operatorname{diam}(F_n)\). Since the diameters tend to zero, \((x_n)\) is Cauchy. Completeness gives a limit \(x\in X\). Fix \(n\). For every \(m\geq n\), \(x_m\in F_n\), and \(F_n\) is closed. The tail of the sequence therefore has its limit in \(F_n\). This holds for every \(n\), so \(x\in\bigcap_{n=1}^{\infty}F_n\).
For uniqueness, if \(x,y\in\bigcap_n F_n\), then \(d(x,y)\leq\operatorname{diam}(F_n)\) for every \(n\). Taking the limit as \(n\to\infty\) gives \(d(x,y)=0\), so \(x=y\).
Conversely, suppose every nested sequence of nonempty closed sets with diameters tending to zero has nonempty intersection. Let \((x_n)\) be any Cauchy sequence in \(X\), and define \(F_n=\overline{\{x_k:k\geq n\}}\), with closure taken in \(X\). Each \(F_n\) is nonempty and closed, and \(F_{n+1}\subseteq F_n\). We claim that \(\operatorname{diam}(F_n)\to0\). Given \(\varepsilon>0\), the Cauchy property provides \(N\) such that \(d(x_j,x_k)<\varepsilon/2\) whenever \(j,k\geq N\). Thus the tail \(\{x_k:k\geq N\}\) has diameter at most \(\varepsilon/2\), and its closure has diameter at most \(\varepsilon/2\). In particular, \(\operatorname{diam}(F_N)<\varepsilon\). The diameters therefore tend to zero.
By the assumed nested-set property, choose \(x\in\bigcap_n F_n\). Since \(x_n\in F_n\) and \(x\in F_n\), we have \(d(x_n,x)\leq\operatorname{diam}(F_n)\). The right-hand side tends to zero, so \(x_n\to x\). Every Cauchy sequence in \(X\) converges in \(X\); hence \(X\) is complete. \(\square\)
Worked Example: Shrinking Closed Sets in the Rationals
The rational numbers with the usual metric are not complete. The nested-set criterion makes the failure visible using closed sets in the rational space, rather than closed sets in the real line.
For each \(n\), let \(a_n=\lfloor 10^n\sqrt{2}\rfloor/10^n\) and \(b_n=a_n+2\cdot10^{-n}\). These are rational numbers, and \(a_n\leq\sqrt{2}<a_n+10^{-n}<b_n\). The truncations satisfy \(a_{n+1}\geq a_n\) and \(a_{n+1}-a_n\leq9\cdot10^{-(n+1)}\). Consequently, \(b_{n+1}=a_{n+1}+2\cdot10^{-(n+1)} \leq a_n+11\cdot10^{-(n+1)} <a_n+20\cdot10^{-(n+1)}=b_n\). Thus the real intervals \([a_n,b_n]\) are nested.
Set \(F_n=\mathbb{Q}\cap[a_n,b_n]\). Each \(F_n\) is nonempty because \(a_n\in F_n\). It is closed in \(\mathbb{Q}\), since it is the intersection of \(\mathbb{Q}\) with a closed real interval. The sets are nested, and their diameters satisfy \(\operatorname{diam}(F_n)\leq b_n-a_n=2\cdot10^{-n}\to0\).
There is no rational number in every \(F_n\). Indeed, if \(q\) belonged to all of them, then \(|q-\sqrt{2}|\leq2\cdot10^{-n}\) for every \(n\). Letting \(n\to\infty\) would give \(q=\sqrt{2}\), which is impossible for rational \(q\). This sequence of nonempty nested closed sets has shrinking diameters but empty intersection, as the criterion predicts for an incomplete space.
Open Subspaces Can Be Baire Without Being Complete
Closed subspaces of complete metric spaces remain complete, but open subspaces need not. Nevertheless, category behaves well for open subspaces: every open subset of a Baire space is itself a Baire space, with its relative topology. This result explains why completeness is sufficient for the Baire property without being necessary for an individual subspace.
Proof. Let \(V_1,V_2,\ldots\) be open dense subsets of \(U\), where openness and density are relative to \(U\). To show their intersection is dense in \(U\), take any nonempty open subset \(W\) of \(U\). Since \(U\) is open in \(X\), \(W\) is open in \(X\) as well.
For each \(n\), relative openness gives an open set \(O_n\subseteq X\) such that \(V_n=U\cap O_n\). Define \(G_n=O_n\cup(X\setminus\overline{U})\). This is open in \(X\). It is also dense in \(X\): density of \(V_n\) in \(U\) implies \(U\subseteq\overline{O_n}\), so \(\overline{G_n}\) contains \(\overline{U}\); it also contains \(X\setminus\overline{U}\). These two sets together cover \(X\).
Because \(X\) is a Baire space, \(\bigcap_nG_n\) is dense in \(X\). It therefore meets the nonempty open set \(W\). Any point in \(W\cap\bigcap_nG_n\) lies in \(U\), and \(G_n\cap U=O_n\cap U=V_n\) for each \(n\). Thus \(W\) meets \(\bigcap_nV_n\). Since \(W\) was arbitrary, that intersection is dense in \(U\), proving that \(U\) is a Baire space. \(\square\)
Worked Example: The Open Interval Is Baire but Not Complete
Consider \(U=(0,1)\) with the usual metric. The sequence \(x_n=1/(n+1)\) is Cauchy: for \(m,n\geq N\), \(|x_m-x_n|\leq1/(N+1)\), which can be made arbitrarily small by increasing \(N\). In \(\mathbb{R}\), it converges to \(0\), but \(0\notin U\). It cannot converge to a different point of \(U\), since limits in \(\mathbb{R}\) are unique. Thus \(U\) is not complete.
On the other hand, \(\mathbb{R}\) is complete and hence a Baire space by the Baire Category Theorem. The interval \(U\) is open in \(\mathbb{R}\), so the open-subspace theorem shows that \(U\) is Baire. Therefore the Baire property does not, by itself, imply completeness.
The ambient space matters in this conclusion. The assertion is that \((0,1)\) is Baire in its relative topology. It does not claim that the interval is complete in the inherited metric.
Using Completeness and Category Carefully
These results give several reliable routes through problems involving category. If a space is complete, the Baire Category Theorem applies. If a subspace is closed in a complete space, it is complete and the theorem applies there too. If the subspace is open, it is Baire even if it is not complete. The nested closed-set criterion offers another way to test completeness directly, especially when a construction naturally produces nested sets of decreasing diameter.
A common pitfall is to treat “Baire” and “complete” as interchangeable. Completeness depends on whether Cauchy sequences have limits in the space; the Baire property concerns intersections of open dense sets. The complete metric on \((0,1)\) does not exist as its inherited metric, but its relative topology still has the Baire property. Likewise, the rationals illustrate how an incomplete space can fail the shrinking nested-set condition.
Always identify the metric and the ambient space before applying these ideas. A set may be closed in one space but not another, and category properties are defined using relative closure and interior. Once those choices are explicit, completeness and category fit together without being confused.
Check Your Understanding
Use the results and examples in this tutorial to answer the following questions.
- Why does a complete subspace of a metric space have to be closed, even if the ambient space is not complete?
- In the nested closed-set criterion, how does completeness ensure that the limit belongs to every \(F_n\)?
- How do the sets \(F_n=\mathbb{Q}\cap[a_n,b_n]\) show that the rationals are incomplete?
- Why is the open interval \((0,1)\) Baire even though the sequence \(1/(n+1)\) has no limit in that interval?
- What distinction should you keep in mind when applying the Baire Category Theorem to a subspace?