The Sierpinski Space As An Open-Set Classifier

Open Sets Encoded by Maps

In topology, continuity is tested by asking whether inverse images of open sets are open. A carefully chosen target space can make that test record a particular subset of the domain. The two-point Sierpinski space does this for open subsets: each such subset determines a continuous map to the space, and each continuous map to it determines an open subset.

Let \(X\) be a set equipped with a topology \(\mathcal{T}\). For a subset \(A\subseteq X\), define its characteristic map \(\chi_A:X\to S\) by sending points in \(A\) to \(1\) and points outside \(A\) to \(0\). The key is the topology on \(S\), not merely the fact that it has two points.

Definition: The Sierpinski space is the set \(S=\{0,1\}\) with topology \(\mathcal{T}_S=\{\varnothing,\{1\},S\}\). For \(A\subseteq X\), its characteristic map is \(\chi_A:X\to S\), where \(\chi_A(x)=1\) if \(x\in A\) and \(\chi_A(x)=0\) if \(x\notin A\). A map \(f:X\to Y\) between topological spaces is continuous if \(f^{-1}(V)\) is open in \(X\) for every open set \(V\) in \(Y\).

The listed collection really is a topology: it contains \(\varnothing\) and \(S\); any union of its sets is one of \(\varnothing,\{1\},S\); and the intersection of any finite collection of its sets is also one of those three sets. In particular, \(\{1\}\) is open but \(\{0\}\) is not. That asymmetry is what allows the space to detect openness rather than closedness.

The Classifier Theorem

Theorem: Let \(X\) be any topological space and \(A\subseteq X\). The characteristic map \(\chi_A:X\to S\) is continuous if and only if \(A\) is open in \(X\).

Proof. Suppose first that \(\chi_A\) is continuous. The set \(\{1\}\) is open in \(S\), so its inverse image under \(\chi_A\) must be open in \(X\). By the definition of \(\chi_A\), \(\chi_A^{-1}(\{1\})=A\). Therefore \(A\) is open.

Conversely, suppose \(A\) is open in \(X\). To prove continuity, check the inverse image of every open subset of \(S\). There are exactly three: \(\varnothing,\{1\},S\). Their inverse images are, respectively, \(\varnothing,A,X\). Each is open in \(X\): the empty set and \(X\) are open by the topology axioms, and \(A\) is open by assumption. Thus \(\chi_A\) is continuous. \(\square\)

The proof depends on checking all target open sets. The decisive one is \(\{1\}\), whose inverse image is exactly \(A\); the other two inverse images are automatically open. This also explains the role of the topology on \(S\): it makes the condition on \(A\) precisely openness.

Corollary: For every topological space \(X\), assigning \(A\mapsto\chi_A\) gives a one-to-one correspondence between the open subsets of \(X\) and the continuous maps from \(X\) to \(S\).

Proof. The theorem shows that every open subset \(A\) gives a continuous map \(\chi_A\). Conversely, let \(g:X\to S\) be continuous and set \(A=g^{-1}(\{1\})\). This set is open because \(g\) is continuous and \(\{1\}\) is open. For each \(x\in X\), if \(x\in A\), then \(g(x)=1=\chi_A(x)\); if \(x\notin A\), then \(g(x)\ne1\), and since \(S=\{0,1\}\), \(g(x)=0=\chi_A(x)\). Hence \(g=\chi_A\). Finally, the subset is uniquely determined by the map, since it equals the inverse image of \(\{1\}\). \(\square\)

Worked Examples

Worked Example: An Open Interval in the Real Line

Give \(\mathbb{R}\) its standard topology and take \(A=(1,3)\). Determine whether \(\chi_A:\mathbb{R}\to S\) is continuous, and compute the inverse images of all open sets in \(S\).

The interval \(A=(1,3)\) is open in the standard topology on \(\mathbb{R}\). The classifier theorem therefore says that \(\chi_A\) is continuous. Directly, the three computations are \(\chi_A^{-1}(\varnothing)=\varnothing\), \(\chi_A^{-1}(\{1\})=(1,3)\), and \(\chi_A^{-1}(S)=\mathbb{R}\). All three sets are open in \(\mathbb{R}\), which verifies continuity from its definition.

For example, \(\chi_A(2)=1\), because \(1<2<3\), while \(\chi_A(1)=0\), because \(1\notin(1,3)\). These values agree with the inverse-image computation: the endpoints are excluded from the preimage of \(\{1\}\).

Check: the only nontrivial inverse image is \((1,3)\), which is open; the other two are \(\varnothing\) and \(\mathbb{R}\), also open. Thus the direct test confirms the theorem.

Worked Example: A Subset That Is Not Open

Again give \(\mathbb{R}\) its standard topology, and let \(A=[0,1]\). Determine whether \(\chi_A:\mathbb{R}\to S\) is continuous.

The inverse image of the open target set \(\{1\}\) is \(\chi_A^{-1}(\{1\})=[0,1]\). This subset is not open in \(\mathbb{R}\): it contains \(0\), but every open neighborhood of \(0\) contains an interval \((-\varepsilon,\varepsilon)\) for some \(\varepsilon>0\), and that interval contains \(-\varepsilon/2<0\), which is not in \([0,1]\). Hence no open neighborhood of \(0\) is contained in \([0,1]\). Since a continuous map must have an open inverse image for every open target set, \(\chi_A\) is not continuous.

The other inverse images are \(\chi_A^{-1}(\varnothing)=\varnothing\) and \(\chi_A^{-1}(S)=\mathbb{R}\), both open. The failure occurs exactly at the open target set \(\{1\}\), as the theorem predicts.

Check: \(0\in[0,1]\), but for every \(\varepsilon>0\), \(-\varepsilon/2\in(-\varepsilon,\varepsilon)\) and \(-\varepsilon/2\notin[0,1]\). This verifies that \([0,1]\) is not open and that the characteristic map fails the continuity test.

Worked Example: Classifying Maps from a Two-Point Space

Let \(X=\{p,q\}\) have topology \(\mathcal{T}_X=\{\varnothing,\{p\},X\}\), and use the Sierpinski space \(S\) as target. For \(A=\{p\}\) and \(B=\{q\}\), determine whether the characteristic maps are continuous.

The open subsets of \(X\) are exactly \(\varnothing,\{p\},X\). Thus \(A=\{p\}\) is open, while \(B=\{q\}\) is not. For \(A\), the map sends \(p\) to \(1\) and \(q\) to \(0\). Its inverse images of \(\varnothing,\{1\},S\) are \(\varnothing,\{p\},X\), respectively, so all are open and \(\chi_A\) is continuous.

For \(B\), the map sends \(q\) to \(1\) and \(p\) to \(0\). In this case \(\chi_B^{-1}(\{1\})=\{q\}\), which is not open in \(X\). Therefore \(\chi_B\) is not continuous.

Check: the two subsets of \(X\) that give continuous characteristic maps are precisely the open subsets \(\varnothing\) and \(\{p\}\) and \(X\). Their maps are, respectively, the constant map with value \(0\), the map just checked for \(A\), and the constant map with value \(1\). For each, the inverse image of \(\{1\}\) is its corresponding subset.

Pulling Back the Classified Set

The classifier also interacts directly with continuous maps between spaces. If \(f:Y\to X\) is continuous, then the subset of \(Y\) detected by the composite \(\chi_A\circ f\) is \(f^{-1}(A)\). This is useful because it translates continuity of maps into a statement about inverse images of open subsets.

Proposition: Let \(f:Y\to X\) be a continuous map and \(A\subseteq X\). Then \(\chi_A\circ f=\chi_{f^{-1}(A)}\). In particular, \(\chi_A\circ f:Y\to S\) is continuous if and only if \(f^{-1}(A)\) is open in \(Y\).

Proof. For \(y\in Y\), the value \((\chi_A\circ f)(y)\) is \(1\) exactly when \(f(y)\in A\). This condition is equivalent to \(y\in f^{-1}(A)\), which in turn is equivalent to \(\chi_{f^{-1}(A)}(y)=1\). If \(f(y)\notin A\), both maps have value \(0\). Thus the maps agree at every \(y\), proving the equality. Applying the classifier theorem to the subset \(f^{-1}(A)\subseteq Y\) shows that the composite is continuous exactly when \(f^{-1}(A)\) is open. Since \(f\) is continuous and \(A\) is not assumed open here, this last condition is a specific test for this particular subset; when \(A\) is open, it holds by the definition of continuity of \(f\). \(\square\)

Worked Example: Pulling an Open Set Back

Let \(f:\mathbb{R}\to\mathbb{R}\) be \(f(x)=2x\), with the standard topology on both copies of \(\mathbb{R}\), and let \(A=(1,3)\). Compute \(f^{-1}(A)\) and identify the characteristic map of the composite.

By definition, \(x\in f^{-1}(A)\) exactly when \(1<2x<3\). Dividing all parts of the inequality by \(2>0\) preserves the inequalities and gives \(1/2<x<3/2\). Therefore \(f^{-1}(A)=(1/2,3/2)\), which is open in \(\mathbb{R}\).

At any \(x\), \((\chi_A\circ f)(x)=1\) exactly when \(f(x)\in(1,3)\), which is exactly when \(x\in(1/2,3/2)\). Hence \(\chi_A\circ f=\chi_{(1/2,3/2)}\), and this composite is continuous by the classifier theorem.

Check: \(f(1)=2\in(1,3)\), so the composite has value \(1\) at \(1\); also \(1\in(1/2,3/2)\). At \(x=2\), \(f(2)=4\notin(1,3)\), so the composite has value \(0\); also \(2\notin(1/2,3/2)\). The endpoint tests are strict: \(f(1/2)=1\) and \(f(3/2)=3\), neither of which belongs to \((1,3)\).

Why the Target Topology Matters

A common mistake is to treat the two-point target as if its topology were irrelevant. For example, if one uses the opposite characteristic convention—sending points of \(A\) to \(0\) and points outside \(A\) to \(1\)—then the inverse image of the open set \(\{1\}\) is \(X\setminus A\), not \(A\). The resulting map into \(S\) is continuous exactly when \(X\setminus A\) is open, or equivalently when \(A\) is closed. The convention for which value represents membership matters.

Another incorrect argument is to say that \(\chi_A\) is continuous whenever \(A\) is open because the target is finite. Finiteness alone says nothing about the inverse images being open. For instance, in the second example, \(A=[0,1]\) is a subset of the real line and the target has only two points, yet \(\chi_A^{-1}(\{1\})=[0,1]\) is not open.

The distinction from the discrete two-point topology makes the role of the chosen target especially clear. If \(\{0,1\}\) is instead given the discrete topology, both \(\{1\}\) and \(\{0\}\) are open. The characteristic map \(\chi_A\) is then continuous exactly when both \(A\) and \(X\setminus A\) are open: continuity requires both of these inverse images to be open, and those are the inverse images of the two singleton open sets. Thus the Sierpinski topology detects open subsets, whereas the discrete topology requires the subset and its complement both to be open.

The classifier theorem is a small but useful instance of a general topological strategy: encode a property of subsets as continuity into a selected space, then use inverse images to test that property. Here the encoding is exact: no extra condition is imposed, and every continuous map into \(S\) arises from one and only one open subset.

Check Your Understanding

Use inverse images of open sets in the target to justify each answer.

For a topological space \(X\), find the inverse images of \(\varnothing,\{1\},S\) under \(\chi_A\).
Give an example of a subset \(A\subseteq\mathbb{R}\) for which \(\chi_A:\mathbb{R}\to S\) is not continuous, and identify the failed inverse-image test.
Why does the characteristic map with membership sent to \(0\), rather than \(1\), test whether the complement of \(A\) is open?
For the discrete topology on \(\{0,1\}\), state the necessary and sufficient condition on \(A\subseteq X\) for \(\chi_A\) to be continuous.
If \(f:Y\to X\) is continuous and \(A\) is open in \(X\), explain why the subset classified by \(\chi_A\circ f\) is open in \(Y\).