Continuity as a Test on Open Sets
The neighborhood definition of continuity asks whether the preimage of each neighborhood of an output is a neighborhood of the input. For a function defined on an entire space, this point-by-point test has a concise global form: preimages of open sets must be open. This characterization is useful because it turns a local condition into a statement about collections of sets.
Let \(f:X\to Y\) be a function between topological spaces. For a subset \(V\subseteq Y\), its preimage is \(f^{-1}(V)=\{x\in X:f(x)\in V\}\). The preimage is a subset of the domain, whether or not \(f\) is one-to-one or onto. The following theorem connects open preimages with the neighborhood definition of continuity from the previous tutorial.
Proof. Suppose first that \(f\) is continuous, and let \(V\) be open in \(Y\). We show that \(f^{-1}(V)\) is open in \(X\). If \(x\in f^{-1}(V)\), then \(f(x)\in V\). Since \(V\) is open and contains \(f(x)\), it is a neighborhood of \(f(x)\). Continuity at \(x\) implies that \(f^{-1}(V)\) is a neighborhood of \(x\). By the definition of neighborhood, there is an open set \(U_x\) in \(X\) such that \(x\in U_x\subseteq f^{-1}(V)\). This holds for every \(x\in f^{-1}(V)\). Consequently, $$ f^{-1}(V)=\bigcup_{x\in f^{-1}(V)} U_x, $$ so \(f^{-1}(V)\) is open, since a union of open sets is open. If \(f^{-1}(V)\) is empty, it is open as well.
Conversely, suppose that the preimage of every open subset of \(Y\) is open in \(X\). Fix \(x\in X\), and let \(W\) be any neighborhood of \(f(x)\) in \(Y\). There is an open set \(V\) in \(Y\) such that \(f(x)\in V\subseteq W\). By the supposition, \(f^{-1}(V)\) is open in \(X\), and \(x\in f^{-1}(V)\). Thus \(f^{-1}(V)\) is a neighborhood of \(x\), and \(f^{-1}(V)\subseteq f^{-1}(W)\). Hence \(f^{-1}(W)\) is a neighborhood of \(x\). The neighborhood definition shows that \(f\) is continuous at \(x\). Since \(x\) was arbitrary, \(f\) is continuous on \(X\). \(\square\)
The proof also identifies a pointwise version. At \(x\), it is enough to test open sets \(V\) containing \(f(x)\): continuity at \(x\) holds exactly when \(f^{-1}(V)\) is a neighborhood of \(x\) for every such \(V\). Globally, the open-set theorem says more compactly that all those preimages are open sets in the domain.
Worked Examples with Open Preimages
Worked Example: A Map into the Sierpiński Space
Let \(Y=\{0,1\}\) have the Sierpiński topology \(\{\varnothing,\{1\},Y\}\), and give \(\mathbb{R}\) its usual topology. Define \(f:\mathbb{R}\to Y\) by \(f(x)=1\) if \(x>2\), and \(f(x)=0\) if \(x\leq 2\). To test continuity, we check the preimage of each open set of \(Y\).
The preimages of the three open sets are $$ f^{-1}(\varnothing)=\varnothing,\qquad f^{-1}(\{1\})=(2,\infty),\qquad f^{-1}(Y)=\mathbb{R}. $$ Each is open in \(\mathbb{R}\). The open-set characterization therefore proves that \(f\) is continuous.
The topology of the target matters to this calculation: \(\{0\}\) is not open in the Sierpiński space, so it is not one of the sets whose preimage must be checked. Continuity is tested against the open sets the target topology actually supplies.
Worked Example: A Slight Change That Breaks Continuity
Keep the same spaces and define \(g:\mathbb{R}\to Y\) by \(g(x)=1\) if \(x\geq 2\), and \(g(x)=0\) if \(x<2\). The target set \(\{1\}\) is open, but $$ g^{-1}(\{1\})=[2,\infty). $$ This is not open in the usual topology on \(\mathbb{R}\): every open interval containing \(2\) contains points less than \(2\), and those points are not in \([2,\infty)\). Thus \(g\) is not continuous.
The two definitions differ only in what happens at the boundary point \(2\), but that difference changes the relevant preimage from an open ray to a nonopen ray. Checking open preimages makes the source of the failure explicit.
Worked Example: Testing a Translation with a Basis
Consider \(h:\mathbb{R}\to\mathbb{R}\), \(h(x)=x+4\), with the usual topology on both spaces. Open intervals form a basis for the usual topology. For any open interval \((a,b)\), where \(a<b\), $$ h^{-1}((a,b))=\{x\in\mathbb{R}:a<x+4<b\}=(a-4,b-4). $$ The last equality follows by subtracting \(4\) from each part of the double inequality. The resulting interval is open. The basis test proved below then shows that \(h\) is continuous.
This calculation illustrates a practical strategy: rather than begin with every open subset of the target, check a sufficiently rich collection of simple open sets and then use the way open sets are built from that collection.
Checking Only a Basis
A basis for a topology is a collection of open sets such that every open set is a union of basis elements. Bases are useful for continuity because preimages commute with unions. In particular, if the preimage of every basis element is open, then the preimage of every open set is open.
Proof. If \(f\) is continuous, the open-set characterization applies to each \(B\in\mathcal{B}\), since every basis element is open in \(Y\). Therefore \(f^{-1}(B)\) is open in \(X\).
For the converse, suppose that \(f^{-1}(B)\) is open in \(X\) for every \(B\in\mathcal{B}\). Let \(V\) be any open set in \(Y\). Since \(\mathcal{B}\) is a basis, there is a subcollection \(\mathcal{B}_V\subseteq\mathcal{B}\) such that \(V=\bigcup_{B\in\mathcal{B}_V}B\). For every \(x\in X\), \(x\in f^{-1}(V)\) exactly when \(f(x)\in B\) for some \(B\in\mathcal{B}_V\). Hence $$ f^{-1}(V)=\bigcup_{B\in\mathcal{B}_V} f^{-1}(B). $$ Each set in this union is open in \(X\) by hypothesis, so the union is open. Thus preimages of all open sets in \(Y\) are open, and the open-set characterization proves that \(f\) is continuous. \(\square\)
The basis test does not say that checking an arbitrary selection of open sets is enough. The collection must generate every open set by unions. For the usual topology on \(\mathbb{R}\), for example, open intervals form a basis; checking their preimages suffices because every usual open set is a union of open intervals.
Preimages, Images, and a Common Pitfall
The open-set characterization concerns preimages of open sets in the target. It does not require a continuous function to send every open set in its domain to an open set in its target. These are different properties, and confusing them leads to incorrect continuity tests.
For example, let \(c:\mathbb{R}\to\mathbb{R}\) be the constant function \(c(x)=5\), with the usual topology on both spaces. If \(V\subseteq\mathbb{R}\) is open, then \(c^{-1}(V)=\mathbb{R}\) when \(5\in V\), and \(c^{-1}(V)=\varnothing\) when \(5\notin V\). Both possibilities are open, so \(c\) is continuous. But the open interval \((0,1)\) has image \(c[(0,1)]=\{5\}\), which is not open in \(\mathbb{R}\). Thus a continuous function need not send open sets to open sets.
There is a structural reason preimages are the right sets to use. For any function \(f:X\to Y\) and any family \(\{V_\lambda\}\) of subsets of \(Y\), $$ f^{-1}\left(\bigcup_\lambda V_\lambda\right)=\bigcup_\lambda f^{-1}(V_\lambda), \qquad f^{-1}\left(\bigcap_\lambda V_\lambda\right)=\bigcap_\lambda f^{-1}(V_\lambda). $$ These identities follow directly from membership: an input maps into a union exactly when it maps into at least one member of the union, and maps into an intersection exactly when it maps into every member. In particular, the union identity is what makes the basis criterion work. Open sets in the target are assembled by unions, and their preimages are assembled by the same unions.
Continuity is therefore a compatibility condition between the two topologies: every open region in the target pulls back to an open region in the domain. It does not require distances, nor does it require the function to preserve openness of sets in the forward direction. When applying the criterion, identify the target's open sets (or a basis for them), compute their preimages, and check openness in the domain topology.
Check Your Understanding
Use the open-set characterization and the basis theorem to answer the following questions.
- State the open-set condition equivalent to continuity of \(f:X\to Y\).
- Why does continuity imply that the preimage of an open target set is open, rather than merely a neighborhood at each of its points?
- If \(\mathcal{B}\) is a basis for \(Y\), why is it enough to check preimages of its members?
- For the map into the Sierpiński space defined by the condition \(x>2\), what is the preimage of the open set \(\{1\}\), and why is it open?
- Why does the constant function from \(\mathbb{R}\) to \(\mathbb{R}\) show that continuity does not require images of open sets to be open?