Residuality and Dense Intersections
Meager sets describe sets that can be covered by countably many nowhere-dense sets. Residuality reverses this viewpoint: a set is residual when its complement is meager. Countable intersections of open dense sets provide a useful way to build residual sets, but the word dense matters. A residual set need not be dense in every topological space, and a dense set need not be residual.
Throughout, \(X\) denotes a topological space. A subset \(N\subseteq X\) is nowhere dense if the interior of its closure is empty, so \(\operatorname{int}\overline{N}=\varnothing\). A set is meager if it is a countable union of nowhere-dense sets. These definitions do not require \(X\) to be a metric space.
Residuality is defined by the size, in the category sense, of the complement; it does not include a density requirement. A G-delta description, on the other hand, specifies how a set is built from open sets. The following result connects these ideas in one direction without any extra assumptions on the space.
Proof. Let \(G=\bigcap_{n\in\mathbb{N}}U_n\), where each \(U_n\) is open and dense. By the set-complement laws, \(X\setminus G=\bigcup_{n\in\mathbb{N}}(X\setminus U_n)\). For each \(n\), \(X\setminus U_n\) is closed because \(U_n\) is open. It has empty interior: if it contained a nonempty open set \(V\), then \(V\cap U_n=\varnothing\), contradicting the density of \(U_n\). Thus \(X\setminus U_n\) is nowhere dense. Its countable union \(X\setminus G\) is meager, so \(G\) is residual by definition. \(\square\)
The converse needs care. If \(A\) is residual, write \(X\setminus A=\bigcup_{n\in\mathbb{N}}N_n\), where each \(N_n\) is nowhere dense. Replacing each \(N_n\) by its closure gives closed nowhere-dense sets that still cover \(X\setminus A\). Taking complements then produces open dense sets. Their intersection is contained in \(A\), but need not equal \(A\), and the definitions alone do not guarantee that this intersection is dense.
Proof. Since \(A\) is residual, there are nowhere-dense sets \(N_n\) with \(X\setminus A=\bigcup_{n\in\mathbb{N}}N_n\). Define \(V_n=X\setminus\overline{N_n}\). The set \(V_n\) is open because \(\overline{N_n}\) is closed. It is dense because \(\operatorname{int}\overline{N_n}=\varnothing\): no nonempty open set can be contained in \(\overline{N_n}\), so every nonempty open set meets \(V_n\). If \(x\in\bigcap_nV_n\), then \(x\notin\overline{N_n}\), and hence \(x\notin N_n\), for every \(n\). Therefore \(x\notin X\setminus A\), which means \(x\in A\). This proves the inclusion. The argument establishes no density claim about the intersection itself. \(\square\)
So a residual set contains an intersection of open dense sets, while a dense G-delta set is itself such an intersection and is residual. The distinction between “contains” and “is” is significant. It also prevents an unjustified inference that a countable intersection of dense open sets must be dense in an arbitrary space.
Worked Example: A Residual Dense G-Delta Set in the Real Line
Give a dense G-delta description of the irrational numbers in \(\mathbb{R}\) with its standard topology, and verify that the set is residual.
Write \(\mathbb{Q}=\{q_n:n\in\mathbb{N}\}\), using an enumeration of the rationals. Each \(\{q_n\}\) is closed in \(\mathbb{R}\). Its interior is empty: every open interval containing \(q_n\) also contains a distinct real number. Thus \(\{q_n\}\) is nowhere dense, and \(\mathbb{Q}=\bigcup_{n\in\mathbb{N}}\{q_n\}\) is meager.
For each \(n\), \(U_n=\mathbb{R}\setminus\{q_n\}\) is open and dense. It is dense because every nonempty open interval contains points other than the single point \(q_n\). Their intersection is exactly the set of irrationals: a real number belongs to every \(U_n\) precisely when it is not equal to any rational \(q_n\). Therefore \[ \mathbb{R}\setminus\mathbb{Q}=\bigcap_{n\in\mathbb{N}}U_n. \] This is a G-delta set, and it is residual because its complement \(\mathbb{Q}\) is meager.
It is also dense. Given a nonempty open interval \((a,b)\), choose a rational \(r\) with \(a<r<b\). Choose a positive rational \(t\) with \(t\sqrt{2}<b-r\). Then \(r+t\sqrt{2}\) lies in \((a,b)\). It is irrational: if it were rational, subtracting the rational \(r\) and dividing by the nonzero rational \(t\) would make \(\sqrt{2}\) rational. Thus every nonempty open interval contains an irrational.
Check: the complement of the displayed intersection is \(\bigcup_n\{q_n\}=\mathbb{Q}\), a countable union of the nowhere-dense singletons just verified. This directly confirms residuality.
Density Alone Does Not Imply Residuality
The next example uses a topology different from the standard topology on the real line. It makes it possible to check density, meagerness, and residuality directly, without relying on a category theorem.
Worked Example: A Dense Set That Is Not Residual
Let the underlying set be \(X=\mathbb{R}\), equipped with the finite-complement topology: the open sets are \(\varnothing\) and all subsets whose complement in \(X\) is finite. Let \(D=\mathbb{N}\). We will show that \(D\) is dense but not residual.
First, every infinite subset \(E\subseteq X\) is dense. Indeed, a nonempty open set has the form \(X\setminus F\) with \(F\) finite. An infinite \(E\) cannot be contained in \(F\), so \(E\cap(X\setminus F)\ne\varnothing\). In particular, \(D=\mathbb{N}\) is dense.
Next, the nowhere-dense subsets of this space are exactly the finite subsets. A finite set \(F\) is closed because \(X\setminus F\) is open. It has empty interior: every nonempty open set is cofinite and therefore infinite, so it cannot be contained in \(F\). Hence \(F\) is nowhere dense. Conversely, if \(E\) is infinite, every nonempty open set \(X\setminus F\) meets \(E\), since \(F\) is finite. Thus \(E\) is dense, its closure is \(X\), and \(\operatorname{int}\overline{E}=X\ne\varnothing\); it is not nowhere dense.
It follows that every meager set is countable, since it is a countable union of finite sets. Conversely, every countable subset is meager because it is a countable union of singletons, each of which is finite and nowhere dense. The complement \(X\setminus D\) is uncountable, so it is not meager. Therefore \(D\) is not residual, despite being dense.
Check: \(D\) is infinite, so it intersects every nonempty cofinite open set; but \(X\setminus D\) is uncountable, so it cannot be a countable union of finite nowhere-dense sets. These are separate tests for density and residuality.
This topology also supplies a dense G-delta set. For each \(n\in\mathbb{N}\), let \(V_n=X\setminus\{n\}\). Each \(V_n\) is open and dense, and \[ \bigcap_{n\in\mathbb{N}}V_n=X\setminus\mathbb{N}. \] The intersection is dense because it is infinite, and it is residual because its complement \(\mathbb{N}\) is countable and hence meager in this topology. This illustrates Proposition 1 concretely; the preceding dense set \(D=\mathbb{N}\), by contrast, is not residual.
A Residual Set Need Not Be Dense
There is a separate issue in the other direction. The definition of residuality does not force the set itself to be dense. The rational numbers with their usual subspace topology give a direct example.
Worked Example: A Residual Set That Is Not Dense
Let \(X=\mathbb{Q}\) with the subspace topology inherited from the standard topology on \(\mathbb{R}\). Every singleton \(\{q\}\) is closed in \(X\). It has empty interior: for every open neighborhood of \(q\) in \(X\), there is a different rational in that neighborhood. For example, if the neighborhood contains \((q-\varepsilon,q+\varepsilon)\cap\mathbb{Q}\) for \(\varepsilon>0\), it contains \(q+\min(\varepsilon/2,1/2)\), which is rational and differs from \(q\). Thus every singleton in \(X\) is nowhere dense.
The set \(X=\mathbb{Q}\) is countable, so it is a countable union of nowhere-dense singletons and is meager in itself. Take \(A=\varnothing\). Its complement in \(X\) is \(X\), which is meager, so \(A\) is residual. But \(A\) is not dense in the nonempty space \(X\), since its closure is empty.
In fact, if \((q_n)\) enumerates \(\mathbb{Q}\), each \(X\setminus\{q_n\}\) is open and dense in \(X\), and \[ \bigcap_{n\in\mathbb{N}}(X\setminus\{q_n\})=\varnothing. \] Each factor is dense, but the intersection is not. This verifies directly why density of the individual open sets cannot, by itself, be used to conclude density of their countable intersection.
Common Pitfalls and a Practical Test
Common pitfall: “Dense means residual.” In the finite-complement topology on \(\mathbb{R}\), the incorrect calculation would be: “\(\mathbb{N}\) is infinite, so it is dense; therefore it is residual.” The first claim is correct, but the conclusion does not follow. Residuality requires the complement \(\mathbb{R}\setminus\mathbb{N}\) to be meager. In this topology meager sets are exactly the countable sets, and \(\mathbb{R}\setminus\mathbb{N}\) is uncountable. Hence \(\mathbb{N}\) is not residual.
Correction: test density by asking whether the set meets every nonempty open set. Test residuality by asking whether its complement is a countable union of nowhere-dense sets. Neither test replaces the other. Likewise, to show that a set is a dense G-delta, one must verify both its countable-open-intersection description and its density; being a G-delta or being dense alone is insufficient.
Check Your Understanding
For each question, distinguish the definition being tested from any additional density claim.
- Prove directly that a countable union of closed nowhere-dense sets is meager.
- In \(\mathbb{R}\) with its standard topology, explain why \(\mathbb{R}\setminus\mathbb{Z}\) is a dense G-delta set and residual.
- In the finite-complement topology on an infinite set \(X\), prove that a subset is dense exactly when it is infinite.
- Can a residual set fail to be a G-delta set based only on the definitions here? Which proposition gives a G-delta subset, rather than an equality?
- Why does the example \(X=\mathbb{Q}\) with its subspace topology show that a countable intersection of dense open sets need not be dense?