Why Nested Balls Give a Dense Intersection
Suppose a metric space has a sequence of dense open sets. Density guarantees that each open region contains points of each set, but it does not by itself identify a point that belongs to all the sets at once. Completeness supplies the missing step: we can choose smaller and smaller closed balls, placing each new ball inside the next dense open set and inside the preceding ball. Their centers become a Cauchy sequence, and its limit belongs to every ball constructed.
Throughout, \((X,d)\) is a metric space with its metric topology. A set is dense if its closure is \(X\). We use the Baire-space criterion: \(X\) is a Baire space if every countable intersection of dense open subsets of \(X\) is dense in \(X\). The result below proves this criterion for complete metric spaces.
For \(x\in X\) and \(r>0\), write \(B(x,r)=\{y\in X:d(x,y)<r\}\) and \(\overline{B}(x,r)=\{y\in X:d(x,y)\le r\}\). Open balls are open and closed balls are closed in the metric topology. In particular, if \(B(x,\varepsilon)\subseteq O\), then \(\overline{B}(x,r)\subseteq O\) whenever \(0<r<\varepsilon\): every point of that closed ball has distance at most \(r<\varepsilon\) from \(x\).
The Complete-Metric Baire Theorem
Proof. If \(X=\varnothing\), its only subset is \(\varnothing=X\), which is dense, so the statement holds. Assume \(X\ne\varnothing\). Let \((G_n)_{n\in\mathbb{N}}\) be dense open subsets of \(X\). To prove that \(\bigcap_{n=1}^{\infty}G_n\) is dense, take an arbitrary nonempty open set \(W\subseteq X\). We will construct closed balls such that \[ \overline{B}(x_1,r_1)\subseteq W\cap G_1,\qquad \overline{B}(x_{n+1},r_{n+1})\subseteq B(x_n,r_n)\cap G_{n+1}, \] with \(0<r_n\le 2^{-n}\).
Since \(G_1\) is dense and \(W\) is nonempty and open, \(W\cap G_1\ne\varnothing\). Choose \(x_1\in W\cap G_1\). This intersection is open, so for some \(\varepsilon_1>0\), \(B(x_1,\varepsilon_1)\subseteq W\cap G_1\). Choose \(r_1=\min(\varepsilon_1/2,2^{-1})\). Then \(r_1>0\) and \(\overline{B}(x_1,r_1)\subseteq B(x_1,\varepsilon_1)\subseteq W\cap G_1\).
Now suppose \(x_n,r_n\) have been chosen. The set \(B(x_n,r_n)\) is nonempty and open, so density of \(G_{n+1}\) gives a point \(x_{n+1}\in B(x_n,r_n)\cap G_{n+1}\). This intersection is open. Choose \(\varepsilon_{n+1}>0\) such that \(B(x_{n+1},\varepsilon_{n+1})\subseteq B(x_n,r_n)\cap G_{n+1}\), and set \(r_{n+1}=\min(\varepsilon_{n+1}/2,2^{-(n+1)})\). Then \(0<r_{n+1}\le 2^{-(n+1)}\), and \(\overline{B}(x_{n+1},r_{n+1})\subseteq B(x_n,r_n)\cap G_{n+1}\). This completes the recursive construction.
For \(m\ge n\), nesting implies \(x_m\in\overline{B}(x_n,r_n)\), hence \(d(x_m,x_n)\le r_n\le 2^{-n}\). Given \(\varepsilon>0\), choose \(N\) so that \(2^{-N}<\varepsilon\). If \(m,n\ge N\), let \(k=\min(m,n)\). Then \(k\ge N\), and nesting gives \(d(x_m,x_n)\le r_k\le2^{-k}\le2^{-N}<\varepsilon\). Thus \((x_n)\) is Cauchy. Completeness gives a limit \(x\in X\).
For each fixed \(n\), every \(x_m\) with \(m\ge n\) lies in the closed set \(\overline{B}(x_n,r_n)\). A closed set contains the limit of every convergent sequence of its points, so \(x\in\overline{B}(x_n,r_n)\). The construction placed this ball inside \(G_n\), so \(x\in G_n\) for every \(n\). It also placed the first ball inside \(W\), so \(x\in W\). Therefore every nonempty open \(W\) meets \(\bigcap_{n=1}^{\infty}G_n\), which proves that this intersection is dense. \(\square\)
The construction uses all three key conditions. Openness lets us fit a sufficiently small closed ball around a chosen point; density ensures that the next open set meets the current open ball; completeness puts the limit of the centers back in the space. The bounds \(r_n\le2^{-n}\) ensure that the radii tend to zero, which is what makes the centers Cauchy.
Worked Applications
Worked Example: Dense Open Sets in the Real Line
In \(\mathbb{R}\) with its standard topology, let \(G_n=\mathbb{R}\setminus\{n\}\) for each \(n\in\mathbb{N}\). Check that these sets are dense and open, then exhibit nested closed balls lying in them.
A singleton is closed in \(\mathbb{R}\), so each \(G_n\) is open. Every nonempty open interval contains a point other than \(n\), so every nonempty open set meets \(G_n\); thus \(G_n\) is dense. For every \(n\), choose \(x_n=0\) and \(r_n=2^{-(n+2)}\). These radii are positive and decrease with \(n\), so the closed intervals are nested. Also, \(r_n\le 2^{-3}=1/8<1\le n\), so \(n\notin[-r_n,r_n]\). Consequently, \[ \overline{B}(0,r_n)=[-2^{-(n+2)},2^{-(n+2)}]\subseteq G_n. \]
Check: \(0\in[-2^{-(n+2)},2^{-(n+2)}]\) for every \(n\), and \(0\ne n\) for every positive integer \(n\). Thus \(0\in\bigcap_{n=1}^{\infty}G_n\). This explicit choice verifies the theorem’s conclusion for this particular sequence of dense open sets.
Worked Example: A Finite Discrete Metric Space
Let \(X=\{a,b,c\}\) and define \(d(x,y)=0\) when \(x=y\) and \(d(x,y)=1\) when \(x\ne y\). This metric gives \(X\) the discrete topology, in which every subset is open. Show directly that \(X\) is complete and satisfies the theorem’s conclusion.
Take a Cauchy sequence \((x_n)\) in \(X\). Apply the Cauchy condition with \(\varepsilon=1/2\). There is \(N\) such that \(d(x_m,x_n)<1/2\) whenever \(m,n\ge N\). Distances in this metric are either \(0\) or \(1\), so this strict inequality forces \(x_m=x_n\) for all \(m,n\ge N\). The sequence is eventually constant and therefore converges in \(X\); hence \(X\) is complete.
If \(G\subseteq X\) is dense, then \(\overline{G}=G\), because every subset of a discrete space is closed as well as open. Thus \(\overline{G}=X\) forces \(G=X\). Every dense open set in \(X\) is therefore \(X\), and the intersection of any countable sequence of them is \(X\), which is dense.
Check: For \(0<r<1\), \(\overline{B}(a,r)=\{a\}\), since \(d(a,a)=0\) and \(d(a,b)=d(a,c)=1>r\). These singleton closed balls are allowed by the construction; a complete metric space need not have balls containing more than one point.
Worked Example: What Can Fail Without Completeness
Let \(X=\mathbb{Q}\) with the metric inherited from \(\mathbb{R}\), and enumerate its elements as \((a_n)_{n\in\mathbb{N}}\). Define \(G_n=\mathbb{Q}\setminus\{a_n\}\). Each singleton is closed in a metric space, so \(G_n\) is open in \(\mathbb{Q}\). It is dense there: every nonempty relatively open subset of \(\mathbb{Q}\) contains infinitely many rationals, and therefore contains a rational other than \(a_n\). But \[ \bigcap_{n=1}^{\infty}G_n=\varnothing, \] because each rational \(a_n\) is omitted from \(G_n\). The empty set is not dense in the nonempty space \(\mathbb{Q}\).
For completeness, consider \(s_n=\lfloor10^n\sqrt{2}\rfloor/10^n\), where \(\lfloor t\rfloor\) is the greatest integer at most \(t\). Each \(s_n\) is rational, and the defining property of the floor gives \(0\le\sqrt{2}-s_n<10^{-n}\). If \(m,n\ge N\), then \[ |s_m-s_n|\le |s_m-\sqrt{2}|+|s_n-\sqrt{2}|<2\cdot10^{-N}. \] For every \(\varepsilon>0\), we can choose \(N\) with \(2\cdot10^{-N}<\varepsilon\), so \((s_n)\) is Cauchy in \(\mathbb{Q}\). It converges to \(\sqrt{2}\) in \(\mathbb{R}\), which is irrational: if \(\sqrt{2}=p/q\) in lowest terms, then \(p^2=2q^2\) makes \(p\) even, and substituting \(p=2k\) makes \(q\) even, contradicting lowest terms. It cannot converge to a rational number as well, since a sequence in a metric space has at most one limit. Therefore \(\mathbb{Q}\) is not complete. This example shows that the theorem’s completeness hypothesis cannot simply be dropped.
A Common Pitfall: Shrinking Is Not Enough
Small radii alone do not ensure that a closed ball lies inside the dense open set required at that stage. For instance, let \(G=\mathbb{R}\setminus\{0\}\), which is dense and open. The point \(1/10\) belongs to \(G\), but the proposed closed ball \(\overline{B}(1/10,1/5)=[-1/10,3/10]\) contains \(0\), so it is not a subset of \(G\). The calculation \(1/10-1/5=-1/10<0<3/10=1/10+1/5\) shows exactly where containment fails.
A corrected choice is \(\overline{B}(1/10,1/20)=[1/20,3/20]\). Its left endpoint \(1/20\) is positive, so the interval does not contain \(0\), and it lies in \(G\). In the nested-ball proof, the radius is chosen only after finding a point in the required open set; it is then made small enough that the entire closed ball, not just its center, stays inside that set.
The theorem is useful whenever a problem can be expressed through countably many dense open conditions. Its conclusion is stronger than finding one point that meets the conditions: the points meeting all of them are dense, so every nonempty open region contains such a point. The proof also explains precisely why completeness matters: it guarantees that the limiting point produced by the nested construction remains in the space.
Check Your Understanding
Use the nested-ball argument and its hypotheses to work through these questions.
- Why must the construction place each closed ball inside an open ball at the preceding stage, rather than only require the centers to lie in the preceding balls?
- In the real-line example, verify directly that \([-2^{-(n+2)},2^{-(n+2)}]\) does not contain \(n\) for every \(n\in\mathbb{N}\).
- For the discrete metric on \(\{a,b,c\}\), what is \(\overline{B}(a,1/2)\), and why does this not obstruct the construction?
- Which step in the proof can fail if a metric space is not complete?
- In \(\mathbb{Q}\), why is \(\mathbb{Q}\setminus\{a_n\}\) dense even though it omits \(a_n\)?