Why a Closed Cover Must Have a Substantial Member
A countable family of sets can cover a space even when every member is small in some sense. The Baire theorem says that this cannot happen if “small” means closed with empty interior and the space is nonempty and complete. The conclusion is not merely that one member is nonempty: it contains an open ball. This gives a useful test for proposed covers and explains why the completeness and nonemptiness assumptions matter.
Throughout, \((X,d)\) is a metric space with its metric topology. For \(x\in X\) and \(r>0\), write \(B(x,r)=\{y\in X:d(x,y)<r\}\) for the open ball and \(\overline{B}(x,r)=\{y\in X:d(x,y)\le r\}\) for the closed ball. A metric space is complete if every Cauchy sequence in it converges to a point of the space. A subset has nonempty interior if it contains a nonempty open set; in a metric space, this means that it contains some open ball.
The theorem concerns the interior in \(X\), not the interior in some larger space in which \(X\) might sit. A set \(F_n\) may be a proper subset of \(X\) and still have nonempty interior: the conclusion says it contains a whole open ball of \(X\). We prove the theorem by showing that if every \(F_n\) has empty interior, one can construct a point outside every \(F_n\), contradicting the cover.
Proof by Nested Closed Balls
We use two elementary metric facts in the construction. If \(G\) is open and \(x\in G\), some \(\delta>0\) satisfies \(B(x,\delta)\subseteq G\); therefore \(\overline{B}(x,r)\subseteq G\) whenever \(0<r<\delta\). Also, closed balls are closed: if \(d(y,x)>r\), then for every \(z\in B(y,(d(y,x)-r)/2)\), the triangle inequality gives \(d(z,x)\ge d(y,x)-d(z,y)>r\), so this open ball misses \(\overline{B}(x,r)\).
Proof. We prove the theorem by contradiction. Suppose every \(F_n\) has empty interior. Since \(X\) is nonempty and \(F_1\) contains no nonempty open set, \(X\setminus F_1\) is a nonempty open set. Choose \(x_1\in X\setminus F_1\). By openness, there is \(\delta_1>0\) such that \(B(x_1,\delta_1)\subseteq X\setminus F_1\). Choose \(r_1\) with \(0<r_1<\delta_1\) and \(r_1\le 2^{-1}\). Then the closed ball \(C_1=\overline{B}(x_1,r_1)\) is contained in \(X\setminus F_1\).
Suppose \(C_n=\overline{B}(x_n,r_n)\) has been chosen. The open ball \(B(x_n,r_n)\) is nonempty. It cannot be contained in \(F_{n+1}\), because \(F_{n+1}\) has empty interior. Thus \(G=B(x_n,r_n)\setminus F_{n+1}\) is nonempty. It is open because it is the intersection of the open ball with the open set \(X\setminus F_{n+1}\). Choose \(x_{n+1}\in G\), and choose \(\delta_{n+1}>0\) such that \(B(x_{n+1},\delta_{n+1})\subseteq G\). Choose \(r_{n+1}\) with \(0<r_{n+1}<\delta_{n+1}\) and \(r_{n+1}\le 2^{-(n+1)}\). Then \(\displaystyle C_{n+1}=\overline{B}(x_{n+1},r_{n+1})\subseteq B(x_n,r_n)\setminus F_{n+1}\subseteq C_n\setminus F_{n+1}.\) In particular, the closed balls are nested and \(C_{n+1}\) misses \(F_{n+1}\).
We now use completeness. If \(m\ge n\), nesting gives \(x_m\in C_n\), so \(d(x_m,x_n)\le r_n\). If \(m,k\ge n\), the triangle inequality gives \(d(x_m,x_k)\le d(x_m,x_n)+d(x_n,x_k)\le 2r_n\le 2^{1-n}\). Since \(2^{1-n}\) tends to zero, \((x_n)\) is Cauchy. Completeness supplies a limit \(x\in X\). For each fixed \(n\), all \(x_m\) with \(m\ge n\) lie in the closed set \(C_n\); hence their limit \(x\) lies in \(C_n\). But \(C_n\cap F_n=\varnothing\), so \(x\notin F_n\). This holds for every \(n\), contradicting \(X=\bigcup_{n\in\mathbb{N}}F_n\). Therefore at least one \(F_n\) has nonempty interior. By the definition of the metric topology, that interior contains an open ball. \(\square\)
The proof also shows why closedness is important: it ensures that the limit of the constructed sequence remains in each closed ball. Completeness is what guarantees that the Cauchy sequence has a limit in \(X\). Without either ingredient, the construction need not produce a point of \(X\) outside every set.
Reading the Conclusion in Concrete Covers
Worked Example: A Cover of the Real Line by Bounded Closed Sets
Give a closed countable cover of \(\mathbb{R}\) and identify a ball contained in one member. The real line has its standard topology and is complete with its usual metric. Set \(F_n=[-n,n]\) for \(n\in\mathbb{N}\). Each \(F_n\) is closed in \(\mathbb{R}\). For any \(x\in\mathbb{R}\), choose an integer \(n\ge1\) with \(n\ge |x|\); then \(-n\le x\le n\), so \(x\in F_n\). Thus \(\mathbb{R}=\bigcup_{n\in\mathbb{N}}F_n\).
For every \(n\ge1\), the open interval \((-n,n)\) is contained in \(F_n\). In particular, \(B(0,1/2)=(-1/2,1/2)\subseteq F_1=[-1,1]\). This verifies the conclusion directly for this cover. The general theorem is stronger: given any countable closed cover of \(\mathbb{R}\), not just this one, some member must contain an open interval.
Check: \(F_1\) is closed because its complement \((-\infty,-1)\cup(1,\infty)\) is open, and the displayed interval is contained in \(F_1\) because every \(x\) with \(|x|<1/2\) also satisfies \(|x|\le1\).
Worked Example: A Relative Closed Cover of the Unit Interval
Let \(I=[0,1]\) with the metric inherited from \(\mathbb{R}\). For \(n\in\mathbb{N}\) with \(n\ge2\), define \(F_n=[0,1-1/n]\cup\{1\}\), regarded as a subset of \(I\). Each \(F_n\) is closed in \(I\): both \([0,1-1/n]\) and \(\{1\}\) are closed in \(I\), and a finite union of closed sets is closed.
These sets cover \(I\). The point \(1\) belongs to every \(F_n\). If \(0\le x<1\), choose an integer \(n\ge2\) such that \(n\ge1/(1-x)\). Since \(1-x>0\), this gives \(1/n\le1-x\), and hence \(x\le1-1/n\). Therefore \(x\in F_n\). For each \(n\ge2\), the set \([0,1-1/n)\) is open in \(I\), since it equals \(I\cap(-1,1-1/n)\); it is contained in \(F_n\) and is nonempty. Thus every \(F_n\) has nonempty interior relative to \(I\).
Check: the strict inequality in \([0,1-1/n)\) means the endpoint \(1-1/n\) is not in this open subset, although it is in \(F_n\). The endpoint \(1\) is also in \(F_n\), but the example does not require it to be an interior point.
Why the Hypotheses Cannot Be Dropped
Worked Example: A Closed Cover of an Incomplete Metric Space
Take \(X=\mathbb{Q}\) with the metric inherited from \(\mathbb{R}\), and enumerate its elements as \(q_1,q_2,\ldots\). Set \(F_n=\{q_n\}\). Each \(F_n\) is closed in \(X\): its complement is open in the subspace topology, because for each \(q\in\mathbb{Q}\setminus\{q_n\}\), a sufficiently small relative open ball about \(q\) misses \(q_n\). Also, \(F_n\) has empty interior in \(X\): every relative open ball about \(q_n\) contains another rational, for example \(q_n+1/k\) for any sufficiently large positive integer \(k\). The sets cover \(X\), since every rational is one of the \(q_n\).
This does not contradict the theorem because \(\mathbb{Q}\) is not complete in the inherited metric. For example, the rational numbers \(a_m=\lfloor 10^m\sqrt{2}\rfloor/10^m\) satisfy \(0\le\sqrt{2}-a_m<10^{-m}\), so they converge in \(\mathbb{R}\) to \(\sqrt{2}\) and form a Cauchy sequence in \(\mathbb{Q}\). Their limit is not rational: if \(\sqrt{2}=p/q\) in lowest terms, then \(p^2=2q^2\), so \(p\) is even, and substituting \(p=2r\) shows \(q\) is even as well, contradicting that \(p/q\) was in lowest terms.
Check: the estimate \(0\le\sqrt{2}-a_m<10^{-m}\) follows from the defining property of the floor function, and \(10^{-m}\) tends to zero. The sequence therefore has no limit in \(\mathbb{Q}\), although it is Cauchy there.
Common pitfall. One might argue that the singleton cover of \(\mathbb{Q}\) contradicts the closed-cover theorem because every singleton is closed and has empty interior. The error is applying the theorem without checking completeness: the theorem requires a complete metric space, and \(\mathbb{Q}\) with its inherited metric is not complete. Another necessary hypothesis is nonemptiness. If \(X=\varnothing\), then the sequence \(F_n=\varnothing\) consists of closed sets and covers \(X\), while every \(F_n\) has empty interior. Thus the correct statement is about a nonempty complete metric space.
The closed-cover consequence is often useful in contrapositive form: if a nonempty complete metric space is covered by countably many closed sets, it is impossible that all those sets have empty interior. In particular, a countable collection of closed sets cannot cover such a space while each member is nowhere dense (that is, while each has empty interior). The conclusion is local and concrete: some member contains an open ball. It does not say that this member is the whole space, nor that every member has nonempty interior.
Check Your Understanding
Use the theorem and the definitions to decide each question, paying attention to which space supplies the topology.
- Why does the theorem require that the metric space be nonempty? Give a closed cover that shows the issue when the space is empty.
- For \(X=\mathbb{R}\) and \(F_n=[-n,n]\), verify directly that \(B(0,1/2)\subseteq F_1\).
- In \(I=[0,1]\), why is \([0,1-1/n)\) open in the relative topology for \(n\ge2\), even though it is not open in \(\mathbb{R}\)?
- Which hypothesis of the closed-cover theorem fails for the singleton cover of \(\mathbb{Q}\), and how does the sequence of decimal truncations of \(\sqrt{2}\) demonstrate the failure?
- State the contrapositive of the theorem using the phrase “empty interior.”