Baire Does Not Mean Complete
Completeness is a property of a metric: every Cauchy sequence must converge to a point of the space. The Baire property, by contrast, is a property of a topology. An open interval with its usual metric gives a useful example of the distinction. It is Baire, even though some Cauchy sequences in that metric converge only to an endpoint that is not in the interval.
A set is dense when its closure is the whole space, or, equivalently, when it meets every nonempty open set. Thus the Baire property says that countably many density conditions imposed by open dense sets can hold simultaneously on a dense set of points. It does not say that Cauchy sequences converge.
Why an Open Interval Is Baire
We use the locally compact Hausdorff Baire theorem, established in the earlier material on locally compact Hausdorff spaces: every locally compact Hausdorff space is Baire. Here “locally compact” means that every point has a compact neighborhood, and a neighborhood of \(x\) is a set containing an open set that contains \(x\). We verify these hypotheses for an open interval rather than treating the endpoints as if they belonged to it.
Proof. First, \(\mathbb{R}\) is Hausdorff: if \(x\ne y\), the open intervals of radius \(|x-y|/3\) about \(x\) and \(y\) are disjoint. Intersecting these separating open sets with \((a,b)\) shows that the subspace \((a,b)\) is Hausdorff.
Now fix \(x\in(a,b)\), and put \(r=\tfrac12\min\{x-a,b-x\}\). Both \(x-a\) and \(b-x\) are positive, so \(r>0\). Also \(r\le (x-a)/2\) and \(r\le (b-x)/2\), hence \(a<x-r<x+r<b\). The closed interval \(K=[x-r,x+r]\) is therefore a subset of \((a,b)\). It is compact by the Heine–Borel theorem, and its subspace topology as a subset of \((a,b)\) is the same as its subspace topology as a subset of \(\mathbb{R}\). The open interval \((x-r,x+r)\), which contains \(x\), is contained in \(K\). Thus \(K\) is a compact neighborhood of \(x\), and \((a,b)\) is locally compact. Applying the locally compact Hausdorff Baire theorem gives the conclusion. \(\square\)
The compact neighborhood need not be the whole space. In fact, the open interval is not compact: for example, the open sets \((a,b-1/n)\), for sufficiently large \(n\) that \(b-1/n>a\), cover \((a,b)\), but no finite subcollection covers points arbitrarily close to \(b\). Local compactness asks for compact neighborhoods around individual points, not compactness of the entire space.
Worked Example: A Compact Neighborhood in \((0,1)\)
Take \(X=(0,1)\) with its usual subspace topology and \(x=1/3\). Choose \(r=1/6\). The proposed compact neighborhood is $$ K=[x-r,x+r]=[1/3-1/6,1/3+1/6]=[1/6,1/2]. $$
Both endpoints lie strictly between \(0\) and \(1\), since \(0<1/6\) and \(1/2<1\). The open interval \((1/6,1/2)\) contains \(1/3\) and is contained in \(K\), so \(K\) is a neighborhood of \(x\) in \(X\). The interval \(K\) is closed and bounded in \(\mathbb{R}\), hence compact by Heine–Borel. Therefore \(x\) has a compact neighborhood in \(X\). The same construction works at every \(x\in(0,1)\) by choosing \(r=\tfrac12\min\{x,1-x\}\).
Check: the chosen radius gives \(1/3-1/6=1/6\) and \(1/3+1/6=1/2\), and \(0<1/6<1/3<1/2<1\). Thus the compact neighborhood does not accidentally include either missing endpoint.
Worked Example: What the Baire Conclusion Gives
In \(X=(0,1)\), for each \(n\in\mathbb{N}\) let \(p_n=1/(n+1)\) and \(U_n=X\setminus\{p_n\}\). Each singleton \(\{p_n\}\) is closed in \(X\), so \(U_n\) is open. It is also dense: every nonempty open subset of \(X\) contains a nonempty interval and therefore has a point other than \(p_n\). Thus each \(U_n\) is open and dense in \(X\).
Since \(X\) is Baire, \(\bigcap_{n=1}^{\infty}U_n\) is dense in \(X\). In particular, it meets every nonempty open subset of \(X\). This conclusion uses the Baire property; it does not require the intersection itself to be open. The example illustrates the role of the theorem: countably many open dense requirements still leave a dense set of points satisfying all of them.
Check: \(p_n=1/(n+1)\) satisfies \(0<p_n<1\) for every \(n\ge1\), so every removed point belongs to \(X\). Each \(U_n\) is open because its complement in \(X\) is exactly the closed singleton \(\{p_n\}\).
The Usual Metric Is Incomplete
For a metric \(d\), a sequence \((x_n)\) is Cauchy if for every \(\varepsilon>0\) there is \(N\in\mathbb{N}\) such that \(d(x_m,x_n)<\varepsilon\) whenever \(m,n\ge N\). A metric space is complete if every Cauchy sequence converges to a point in that space. On \((0,1)\), use the usual metric \(d(x,y)=|x-y|\).
Worked Example: An Incomplete Cauchy Sequence in \((0,1)\)
Let \(x_n=1/(n+1)\) for \(n\ge1\). Each term is in \((0,1)\). If \(m,n\ge N\), then both \(x_m\) and \(x_n\) lie in \((0,1/(N+1)]\), so $$ |x_m-x_n|<\frac{1}{N+1}. $$ Given \(\varepsilon>0\), choose \(N\) with \(1/(N+1)<\varepsilon\). Then \(|x_m-x_n|<\varepsilon\) for all \(m,n\ge N\), proving that \((x_n)\) is Cauchy.
As a sequence of real numbers, \(x_n\to0\). For example, \(0<x_n=1/(n+1)<\varepsilon\) whenever \(n+1>1/\varepsilon\). But \(0\notin(0,1)\), so this limit is not in \(X\). The sequence cannot converge to a different \(x\in(0,1)\): convergence to \(x\) in the usual metric would also be convergence to \(x\) in \(\mathbb{R}\), and limits in \(\mathbb{R}\) are unique. Since its real limit is \(0\), that would force \(x=0\), a contradiction. Thus the usual metric on \((0,1)\) is incomplete.
Check: the terms tend to the excluded endpoint \(0\), not to a point of the interval. The sequence is Cauchy but has no limit in \(X\), which is exactly the failure of completeness.
Completeness Depends on the Metric
The example proves that Baire does not imply completeness of the particular metric under consideration. It does not prove that \((0,1)\) cannot carry any complete metric inducing its topology. In fact, it can. This is a useful distinction: the Baire property belongs to the topology, whereas completeness is being tested for a specified metric.
Worked Example: A Complete Compatible Metric on \((0,1)\)
Define \(\phi:(0,1)\to\mathbb{R}\) by $$ \phi(x)=\ln\!\left(\frac{x}{1-x}\right), \qquad d'(x,y)=|\phi(x)-\phi(y)|. $$ The function \(\phi\) is a homeomorphism: its inverse is \(\phi^{-1}(t)=e^t/(1+e^t)\), and both formulas define continuous functions on their domains. Therefore \(d'\) is a metric, and it induces the usual topology on \((0,1)\): its open balls are the preimages under \(\phi\) of open intervals in \(\mathbb{R}\).
To verify completeness, let \((x_n)\) be Cauchy for \(d'\). By the definition of \(d'\), the real sequence \((\phi(x_n))\) is Cauchy. Completeness of \(\mathbb{R}\) gives a real number \(t\) with \(\phi(x_n)\to t\). Set \(x=\phi^{-1}(t)=e^t/(1+e^t)\). Since \(e^t>0\), we have \(0<x<1\), and $$ d'(x_n,x)=|\phi(x_n)-\phi(x)|=|\phi(x_n)-t|\longrightarrow0. $$ Thus every \(d'\)-Cauchy sequence converges in \((0,1)\), so \(d'\) is complete. This does not change the earlier calculation: the usual metric is incomplete, while this different, compatible metric is complete.
Check: \(e^t/(1+e^t)\) is strictly positive, and it is less than \(1\) because \(e^t<1+e^t\). Hence the constructed limit really is in the interval.
A Common Mistake
Common pitfall: “\(1/(n+1)\to0\), so the sequence converges in \((0,1)\).” The calculation establishes convergence in \(\mathbb{R}\), but convergence in a metric space requires the limit to be a point of that space. Here \(0\notin(0,1)\), and the uniqueness-of-limits argument above rules out every other possible limit in the interval. The correct conclusion is that the sequence is Cauchy in the usual metric on \((0,1)\) but does not converge there.
The logical conclusions should therefore be kept separate. Local compactness and the Hausdorff property give the Baire property through the locally compact Hausdorff Baire theorem. The sequence \(1/(n+1)\) shows incompleteness of the usual metric. Neither conclusion contradicts the other, and the compatible metric \(d'\) further shows that this interval even admits a complete metric for the same topology.
Check Your Understanding
Use the definitions and the hypotheses of the locally compact Hausdorff Baire theorem in your answers.
- For \(x\in(a,b)\), explain why \(r=\tfrac12\min\{x-a,b-x\}\) is positive and why \([x-r,x+r]\subset(a,b)\).
- Why does a compact neighborhood at every point not imply that the whole space \((0,1)\) is compact?
- Verify directly that \(1/(n+1)\) is Cauchy in the usual metric, and identify why its real limit does not make it convergent in \((0,1)\).
- State the Baire conclusion for any countable collection of open dense subsets of \((0,1)\).
- In one sentence, distinguish the Baire property from completeness of a specified metric.