Topological Proof Workshop I

A Workshop on Following Closures Through Maps

In “Sequential Characterizations in Topology,” sequences provided useful tests for closedness and continuity under appropriate hypotheses. In general topology, a proof often works better directly with neighborhoods: to show that a point lies in a closure, take an arbitrary open neighborhood and prove that it meets the set. This workshop uses that method to examine how closure and density behave under continuous maps.

The direction of a map matters. A continuous map sends points in the closure of a set to points in the closure of its image. For inverse images, continuity gives a different inclusion involving the closure of a preimage. These statements are related, but neither should be replaced by an unexamined claim that images and closures commute.

Two Closure Inclusions

We use the topological definition of closure from “Interior and Closure Topologically”: \(x\in\overline{A}\) exactly when every open neighborhood of \(x\) meets \(A\). The first proof pattern applies this test in the target space, then pulls a neighborhood back to the domain.

Proof. Take \(x\in\overline{A}\). We show that every open neighborhood of \(f(x)\) meets \(f[A]\). Let \(V\) be any open set in \(Y\) containing \(f(x)\). By continuity, \(f^{-1}(V)\) is open in \(X\); it contains \(x\), so it is a neighborhood of \(x\). Since \(x\in\overline{A}\), there is a point \(a\in f^{-1}(V)\cap A\). Then \(f(a)\in V\), and \(f(a)\in f[A]\). Thus \(V\cap f[A]\ne\varnothing\). Every open neighborhood of \(f(x)\) meets \(f[A]\), so \(f(x)\in\overline{f[A]}\). This proves the inclusion. \(\square\)

The proof uses only continuity and the neighborhood test for closure. It does not require the map to be injective, surjective, or open. It also explains why the inclusion points in this direction: points near \(A\) have images that cannot be separated from \(f[A]\) by an open neighborhood.

Proof. Take \(x\in\overline{f^{-1}(B)}\). We prove that \(f(x)\in\overline{B}\). Let \(V\) be any open neighborhood of \(f(x)\) in \(Y\). The set \(f^{-1}(V)\) is open by continuity and contains \(x\). Because \(x\) is in the closure of \(f^{-1}(B)\), the open neighborhood \(f^{-1}(V)\) meets \(f^{-1}(B)\). Thus there is some \(z\in f^{-1}(V)\cap f^{-1}(B)\). In particular, \(f(z)\in V\) and \(f(z)\in B\), so \(V\cap B\ne\varnothing\). Every open neighborhood \(V\) of \(f(x)\) meets \(B\), which means \(f(x)\in\overline{B}\). Equivalently, \(x\in f^{-1}(\overline{B})\), as required. \(\square\)

The second result can also be understood through the Closed-Set Characterization of Continuity from “Continuity Through Closed Sets.” The set \(\overline{B}\) is closed and contains \(B\), so its preimage is closed and contains \(f^{-1}(B)\). The closure of \(f^{-1}(B)\), being the smallest closed set containing that preimage, must lie inside \(f^{-1}(\overline{B})\). The neighborhood proof makes the mechanism explicit and is useful when unpacking an inclusion directly.

Dense Sets and Continuous Surjections

A subset \(D\) of a space \(X\) is dense when \(\overline{D}=X\). Equivalently, by the density characterization from “Separable Spaces,” every nonempty open subset of \(X\) meets \(D\). For continuous surjections, this open-set formulation gives a short and reliable proof that density passes to the image.

Proof. Let \(V\) be any nonempty open subset of \(Y\). Since \(f\) is surjective, choose \(x\in X\) with \(f(x)\in V\). Then \(f^{-1}(V)\) is open by continuity, and it is nonempty because it contains \(x\). Density of \(D\) implies that \(f^{-1}(V)\cap D\ne\varnothing\). Choose \(d\) in this intersection. Since \(d\in f^{-1}(V)\), we have \(f(d)\in V\); since \(d\in D\), we also have \(f(d)\in f[D]\). Therefore \(V\cap f[D]\ne\varnothing\). Every nonempty open subset of \(Y\) meets \(f[D]\), so \(f[D]\) is dense in \(Y\). \(\square\)

Surjectivity is used to make the preimage of each nonempty open set nonempty. Without it, continuity alone does not imply that the image of a dense set is dense in the whole target. The conclusion does hold in the image \(f[X]\), with its subspace topology, when \(D\) is dense in \(X\): the restriction \(f:X\to f[X]\) is continuous and surjective, so the theorem applies with \(f[X]\) as the target.

Proof Choices and Common Pitfalls

These arguments illustrate a useful choice of proof language. To prove an inclusion involving closures, begin with a point in the left-hand set and test an arbitrary open neighborhood of the point that should lie in the closure. To prove density, begin with an arbitrary nonempty open set and show that it meets the proposed dense subset. In each case, writing down the arbitrary neighborhood prevents the proof from relying on an unstated approximation argument.

A common mistake is to treat continuous maps as if they preserved every set operation involving closure. The two inclusions above are generally not equalities, as the worked examples demonstrate. A second mistake is to forget the target when saying that an image is dense. A subset can be dense in \(f[X]\) without being dense in a larger space \(Y\). For instance, the constant map \(f:\mathbb{R}\to\mathbb{R}\), \(f(x)=0\), sends the dense set \(\mathbb{Q}\) to \(\{0\}\). This image is dense in \(f[\mathbb{R}]=\{0\}\), but it is not dense in \(\mathbb{R}\).

The central habit is to match each definition to its quantifiers. Closure asks about every open neighborhood of one fixed point; density asks about every nonempty open set. Continuity connects these tests by ensuring that the preimage of an open set is open. Once those steps are explicit, the proof’s hypotheses and conclusion are easier to track.