Why Openness Matters
A Baire space has a useful stability property: each of its open subspaces is again Baire. The key point is that a dense open subset of an open subspace need not itself be dense in the whole space. It can, however, be enlarged outside the subspace to make an ambient dense open set. We prove this extension step first, then use it to establish the result and to relate nowhere-dense sets in the two spaces.
Here and throughout, a set is nowhere dense when its closure has empty interior. We use the dense-open-intersection definition of Baire directly. In particular, it also handles the empty space: the only subset of the empty space is empty, which is dense in it, so the Baire condition holds vacuously.
Extending Dense Open Sets
Let \(U\) be open in \(X\), and let \(G\) be open and dense in the subspace \(U\). Since \(U\) is open, \(G\) is open in \(X\) as well: write \(G=U\cap H\) for some open \(H\subseteq X\), and then \(G\) is the intersection of two ambient open sets. The set \(G\) need not be dense in \(X\). To extend it, add the region outside the closure of \(U\).
Proof. Since \(U\) is open, \(G\) is open in \(X\), as noted above. Also \(X\setminus\overline{U}\) is open because \(\overline{U}\) is closed. Their union \(W\) is therefore open. Density of \(G\) in \(U\) means that every point of \(U\) lies in \(\overline{G}\), where this closure is taken in \(X\) for the present argument: equivalently, \(\overline{G}\cap U=U\) in the subspace topology. Thus \(U\subseteq\overline{G}\), and taking ambient closures gives \(\overline{U}\subseteq\overline{G}\subseteq\overline{W}\). Also \(X\setminus\overline{U}\subseteq W\subseteq\overline{W}\). These two inclusions show that \(\overline{W}=X\), so \(W\) is dense. Finally, \(X\setminus\overline{U}\) is disjoint from \(U\), and \(G\subseteq U\), so \(W\cap U=G\). \(\square\)
The equality \(W\cap U=G\) is important: the enlargement adds points outside \(U\), without changing the part of the set inside \(U\). Applying this construction separately to each member of a sequence will let us use the Baire property of \(X\) while preserving the original sequence inside \(U\).
The Open-Subspace Theorem
Proof. If \(U=\varnothing\), it is Baire by the vacuous case of the definition. Now suppose \(U\ne\varnothing\). Let \((G_n)_{n\in\mathbb{N}}\) be any sequence of dense open subsets of \(U\). For each \(n\), define $$ W_n=G_n\cup(X\setminus\overline{U}). $$ The lemma shows that \(W_n\) is open and dense in \(X\), and \(W_n\cap U=G_n\). Since \(X\) is Baire, \(D=\bigcap_{n\in\mathbb{N}}W_n\) is dense in \(X\). To prove that \(\bigcap_{n\in\mathbb{N}}G_n\) is dense in \(U\), take any nonempty open subset \(O\) of \(U\). Because \(U\) is open in \(X\), \(O\) is open in \(X\): it is the intersection of \(U\) with an ambient open set. Density of \(D\) gives a point \(x\in O\cap D\). Since \(x\in O\subseteq U\), membership \(x\in W_n\) forces \(x\in G_n\) for every \(n\), because \(W_n\cap U=G_n\). Hence \(O\) meets \(\bigcap_{n\in\mathbb{N}}G_n\). Every nonempty open subset of \(U\) meets this intersection, so the intersection is dense in \(U\). This proves that \(U\) is Baire. \(\square\)
Worked Example: The Open Interval in the Real Line
Use the theorem to check a sequence of dense open sets in \(U=(0,1)\), where \(X=\mathbb{R}\) has its standard topology. The real line is Baire by the complete-metric Baire theorem, since its usual metric is complete. For each \(n\in\mathbb{N}\), set $$ G_n=(0,1)\setminus\left\{\frac{1}{n+2}\right\}. $$
Each \(G_n\) is open in \(U\): the removed singleton is closed in \(U\). It is dense in \(U\), because every nonempty open interval in \(U\) contains points other than the single removed point. The closure of \(U\) in \(\mathbb{R}\) is \([0,1]\), so the ambient extension is $$ W_n=\left((0,1)\setminus\left\{\frac{1}{n+2}\right\}\right)\cup(-\infty,0)\cup(1,\infty). $$ It is open. It is dense because its closure contains \([0,1]\), as \(G_n\) is dense in \(U\), and also contains \((-\infty,0)\cup(1,\infty)\); together these sets cover \(\mathbb{R}\).
The point \(1/2\) belongs to every \(G_n\): for \(n\geq1\), \(n+2\geq3\), so \(1/(n+2)\leq1/3<1/2\). This checks that the intersection is nonempty, while the theorem proves the stronger fact that \(\bigcap_{n\in\mathbb{N}}G_n\) is dense in \(U\). The example also shows why extending the sets is useful: each \(G_n\) lies inside \((0,1)\), so none is dense in \(\mathbb{R}\).
Translating Nowhere-Dense Sets
The same openness assumption controls how closure behaves when a set \(A\) is contained in \(U\). Its closure in the subspace is $$ \overline{A}^{\,U}=U\cap\overline{A}^{\,X}, $$ where the superscripts specify the space in which closure is taken. Indeed, a point of \(U\) belongs to the relative closure exactly when each of its neighborhoods in \(U\) meets \(A\), which is equivalent to each ambient open neighborhood meeting \(A\). This identity lets us compare nowhere-denseness in \(U\) and \(X\).
Proof. First suppose \(A\) is nowhere dense in \(X\). If \(\overline{A}^{\,U}\) had nonempty interior in \(U\), there would be a nonempty open set \(O\) in \(X\) with \(O\subseteq U\cap\overline{A}^{\,X}\). This would be a nonempty ambient open subset of \(\overline{A}^{\,X}\), contrary to the assumption that \(A\) is nowhere dense in \(X\). Thus \(A\) is nowhere dense in \(U\).
Conversely, suppose \(A\) is nowhere dense in \(U\). Assume for contradiction that \(\overline{A}^{\,X}\) has nonempty interior, and choose a nonempty ambient open set \(O\subseteq\overline{A}^{\,X}\). Since \(A\subseteq U\), we have \(\overline{A}^{\,X}\subseteq\overline{U}^{\,X}\). If \(O\cap U\ne\varnothing\), then \(O\cap U\) is a nonempty open subset of \(U\), and $$ O\cap U\subseteq U\cap\overline{A}^{\,X}=\overline{A}^{\,U}. $$ This contradicts the empty interior of \(\overline{A}^{\,U}\). If \(O\cap U=\varnothing\), choose \(x\in O\). Since \(x\in\overline{U}^{\,X}\) and \(O\) is an open neighborhood of \(x\), the definition of closure implies \(O\cap U\ne\varnothing\), again a contradiction. Therefore \(\overline{A}^{\,X}\) has empty interior, and \(A\) is nowhere dense in \(X\). \(\square\)
Worked Example: A Nowhere-Dense Set Near an Endpoint
Let \(X=\mathbb{R}\), \(U=(0,1)\), and \(A=\{1/k:k\in\mathbb{N},\,k\geq2\}\). The only possible accumulation point of these numbers outside \(U\) is \(0\), because \(1/k\to0\). More explicitly, if \(y\in(0,1)\setminus A\), choose \(N\geq2\) so large that \(1/N<y/2\). For \(k\geq N\), \(1/k\leq1/N<y/2\), so these terms stay at least \(y/2\) below \(y\). The remaining terms \(1/2,\ldots,1/(N-1)\) form a finite set not containing \(y\); a sufficiently small open interval around \(y\) avoids them as well. Thus \(y\notin\overline{A}^{\,U}\), and \(\overline{A}^{\,U}=A\).
No point of \(A\) is an interior point of \(A\): around \(1/k\), there are points strictly between \(1/k\) and the next larger member \(1/(k-1)\) when \(k>2\); for \(k=2\), there are points between \(1/2\) and \(1\). These points are in \(U\) but not in \(A\). Hence \(A\) is nowhere dense in \(U\). In \(\mathbb{R}\), its closure is \(A\cup\{0\}\), which also has empty interior, in agreement with the proposition. In particular, \(U\setminus A\) is dense and open in \(U\), and its ambient dense-open extension is $$ \bigl((0,1)\setminus A\bigr)\cup\bigl((-\infty,0)\cup(1,\infty)\bigr). $$
Worked Example: A Baire Space Without Separation Assumptions
Let \(X=\{a,b\}\) have the Sierpinski topology \(\{\varnothing,\{a\},X\}\), and let \(U=\{a\}\), which is open. The set \(\{a\}\) is dense in \(X\), since its only closed superset is \(X\). The dense open subsets of \(X\) are therefore \(\{a\}\) and \(X\): the empty set is not dense, and these are the two nonempty open sets. Any countable intersection of sets chosen from \(\{\{a\},X\}\) is either \(\{a\}\) or \(X\), both of which are dense in \(X\). Thus \(X\) is Baire.
The subspace \(U\) is a one-point space, whose only open sets are \(\varnothing\) and \(U\). Its only dense open set is \(U\), so every countable intersection of dense open sets is \(U\). Hence \(U\) is Baire as well. This example does not rely on a metric or a separation axiom; the open-subspace proof uses only the stated topological definitions.
A Common Mistake
Pitfall: It is incorrect to infer that a dense open subset of \(U\) is dense in \(X\). In the interval example, \(G=(0,1)\setminus\{1/2\}\) is dense in \(U=(0,1)\), but its closure in \(\mathbb{R}\) is \([0,1]\), not \(\mathbb{R}\). The wrong calculation is to claim \(\overline{G}^{\,\mathbb{R}}=\mathbb{R}\) merely because \(\overline{G}^{\,U}=U\). The correction is to add \(X\setminus\overline{U}\): then \(G\cup(\mathbb{R}\setminus[0,1])\) is open and dense in \(\mathbb{R}\), while its intersection with \(U\) remains exactly \(G\). The endpoints are not added to this extension, but they lie in its closure, which is enough for density.
Openness is doing two jobs: it ensures that a relatively open set in \(U\) is open in \(X\), and it makes every open subset of \(U\) available as an ambient open set when testing density. Without those steps, the extension and the final density argument do not follow.
Check Your Understanding
Use the dense-open definition of Baire and distinguish closures taken in the subspace from closures taken in the ambient space.
- For an open \(U\subseteq X\) and dense open \(G\subseteq U\), explain why \(G\cup(X\setminus\overline{U})\) is open.
- In \(\mathbb{R}\), find the closure of \((0,1)\setminus\{1/2\}\), and state in which space this set is dense.
- Why does the proof of the open-subspace theorem treat \(U=\varnothing\) separately?
- Show that if \(U\) is open and \(A\subseteq U\) is nowhere dense in \(U\), then \(A\) is nowhere dense in \(X\).
- In the Sierpinski example, list the dense open subsets of \(X\) and describe their possible countable intersections.