Why a Subbasis Can Be Enough
Continuity is defined by testing the preimages of all open sets in the codomain. That can be a large collection. A subbasis reduces the task: if every subbasic set has an open preimage, then the same is true of the finite intersections and arbitrary unions that make up the codomain topology. The key is to check that both operations preserve openness under preimages.
Throughout, \((X,\mathcal{T})\) and \((Y,\mathcal{U})\) are topological spaces. A map \(f:X\to Y\) is continuous if \(f^{-1}(V)\in\mathcal{T}\) for every \(V\in\mathcal{U}\).
When \(\mathcal{S}=\varnothing\), its finite intersections consist of the empty intersection \(Y\). Arbitrary unions of this family give \(\{\varnothing,Y\}\), the indiscrete topology on \(Y\). If \(Y=\varnothing\), then \(\{\varnothing,Y\}=\{\varnothing\}\), which is the topology on the empty set. In particular, the empty subbasis does not generate only \(\{\varnothing\}\) when \(Y\) is nonempty.
The subbasis condition lets us test only the sets in \(\mathcal{S}\), rather than every open set. The reason is that preimages commute with unions and intersections: for any family \((V_i)_{i\in J}\) of subsets of \(Y\), \(f^{-1}(\bigcup_{i\in J}V_i)=\bigcup_{i\in J}f^{-1}(V_i)\) and \(f^{-1}(\bigcap_{i\in J}V_i)=\bigcap_{i\in J}f^{-1}(V_i)\). These equalities follow by testing whether an arbitrary \(x\in X\) has \(f(x)\) in the set on each side. They also hold for empty families: the preimage of \(\varnothing\) is \(\varnothing\), and the preimage of \(Y\) is \(X\).
The Subbasis Criterion
Proof. Suppose first that \(f\) is continuous. Each \(S\in\mathcal{S}\) is open in \(Y\), so the definition of continuity gives \(f^{-1}(S)\in\mathcal{T}\).
Conversely, suppose \(f^{-1}(S)\) is open in \(X\) for every \(S\in\mathcal{S}\). Every open set \(V\in\mathcal{U}\) is a union of finite intersections of members of \(\mathcal{S}\). The preimage of each such finite intersection is the corresponding finite intersection of sets \(f^{-1}(S)\), so it is open in \(X\), since a topology is closed under finite intersections. This includes the empty intersection: its preimage is \(f^{-1}(Y)=X\), which is open. The preimage of \(V\), being a union of those open preimages, is open in \(X\), since a topology is closed under arbitrary unions. If the union is empty, its preimage is \(\varnothing\), also open. Thus the preimage of every open set in \(Y\) is open in \(X\), and \(f\) is continuous. The argument also covers \(\mathcal{S}=\varnothing\): the condition on its members is vacuous, and every open set of the generated topology is either \(\varnothing\) or \(Y\), both of whose preimages are open. \(\square\)
The proof shows the exact division of work: the subbasis provides the sets to check, while the closure of open sets under finite intersections and arbitrary unions extends the check to the whole topology. The codomain is the space where the subbasis is used; the domain can be any topological space.
Worked Example: The Square Map on the Standard Real Line
Let \(f:\mathbb{R}\to\mathbb{R}\) be \(f(x)=x^2\), with the standard topology on both copies of \(\mathbb{R}\). Use the subbasis \(\mathcal{S}=\{(-\infty,a),(a,\infty):a\in\mathbb{R}\}\) for the codomain topology.
These rays are open in the standard topology. Their finite intersections include the open intervals \((a,b)=(-\infty,b)\cap(a,\infty)\) when \(a<b\). Every standard open set is a union of open intervals, so these rays generate the standard topology and form a subbasis.
For a left ray \((-\infty,a)\), its preimage is \(\{x\in\mathbb{R}:x^2<a\}\). If \(a\leq 0\), this set is empty, because squares are nonnegative. If \(a>0\), the inequality is equivalent to \(-\sqrt{a}<x<\sqrt{a}\), so the preimage is the open interval \((-\sqrt{a},\sqrt{a})\).
For a right ray \((a,\infty)\), the preimage is \(\{x\in\mathbb{R}:x^2>a\}\). If \(a<0\), every real square is at least \(0\), which is greater than \(a\), so the preimage is \(\mathbb{R}\). If \(a=0\), the preimage is \(\mathbb{R}\setminus\{0\}=(-\infty,0)\cup(0,\infty)\). If \(a>0\), the inequality holds exactly when \(x<-\sqrt{a}\) or \(x>\sqrt{a}\), giving \((-\infty,-\sqrt{a})\cup(\sqrt{a},\infty)\). In every case the preimage is standard-open.
Check at \(a=4\): \(\sqrt{4}=2\), so \(f^{-1}((-\infty,4))=(-2,2)\) and \(f^{-1}((4,\infty))=(-\infty,-2)\cup(2,\infty)\). These agree with the inequalities \(x^2<4\) and \(x^2>4\), respectively. Since all subbasic preimages are open, the subbasis criterion proves that \(f\) is continuous.
Using Different Topologies on the Real Line
The same underlying set can carry different topologies, so the direction of a map matters. Write \(\mathbb{R}_\ell\) for the real line with the lower-limit topology, generated by the basis of intervals \([a,b)\), where \(a<b\). These intervals are half-open: their left endpoints belong to them and their right endpoints do not. Their finite intersections are either empty or another interval of this form: when two intervals overlap, the intersection is \([\max(a,c),\min(b,d))\), provided \(\max(a,c)<\min(b,d)\). Thus this basis is also a subbasis.
Worked Example: From the Lower-Limit Line to the Standard Line
Let \(f:\mathbb{R}_\ell\to\mathbb{R}\) be the identity map, where the codomain has the standard topology. Test continuity using the standard subbasis of left and right rays.
For each \(a\in\mathbb{R}\), the preimage of the left ray is \((-\infty,a)\). In the lower-limit topology, \((-\infty,a)=\bigcup_{t<a}[t,a)\). Each interval in this union is a basis element, and every \(x<a\) belongs to the particular interval \([x,a)\). No point \(x\geq a\) belongs to any interval in the union. Therefore \((-\infty,a)\) is open in \(\mathbb{R}_\ell\).
The preimage of the right ray is \((a,\infty)\). For each \(t\in\mathbb{R}\), the ray \([t,\infty)\) is open in \(\mathbb{R}_\ell\), because \([t,\infty)=\bigcup_{b>t}[t,b)\). Indeed, each point \(x\geq t\) lies in \([t,b)\) for some \(b>t\) with \(b>x\), while no point below \(t\) lies in any interval in the union. It follows that \((a,\infty)=\bigcup_{t>a}[t,\infty)\) is open: each \(x>a\) lies in the term with \(t=x\), and every point of every term is greater than \(a\). Thus the preimages of all subbasic rays are open, and the criterion proves that the identity map \(\mathbb{R}_\ell\to\mathbb{R}\) is continuous.
Check: for \(a=2\), \((-\infty,2)=\bigcup_{t<2}[t,2)\) and \((2,\infty)=\bigcup_{t>2}[t,\infty)\). The first union contains exactly the points less than \(2\), and the second contains exactly the points greater than \(2\). The strict endpoint exclusions match the two rays.
Worked Example: A Translation of the Lower-Limit Line
Let \(g:\mathbb{R}_\ell\to\mathbb{R}_\ell\) be \(g(x)=x+3\). Use the subbasis consisting of all basis intervals \([a,b)\), where \(a<b\). For any such interval, \(g^{-1}([a,b))=\{x:a\leq x+3<b\}=[a-3,b-3)\). The inequalities are equivalent after subtracting \(3\) from each side, and \(a-3<b-3\), so this preimage is a basis interval and is open in \(\mathbb{R}_\ell\). The criterion proves that \(g\) is continuous.
Check with specific endpoints: for \([1,5)\), the preimage is \(\{x:1\leq x+3<5\}\). Subtracting \(3\) gives \(-2\leq x<2\), hence \(g^{-1}([1,5))=[-2,2)\). Applying \(g\) to the included endpoint gives \(-2+3=1\), and to the excluded endpoint gives \(2+3=5\), confirming the half-open endpoints.
A Common Pitfall: Reversing the Map
It is tempting to assume that the identity map must be continuous whichever topologies are placed on its domain and codomain. Continuity is defined by preimages of open sets in the codomain, so changing the direction changes the test.
Common mistake. The identity \(\mathbb{R}\to\mathbb{R}_\ell\) is not continuous. A wrong argument might say that \([0,1)\) is open in \(\mathbb{R}_\ell\), and therefore its preimage is open in the standard real line. That conclusion does not follow: openness of the set in the codomain does not imply openness of its preimage.
Correction. The preimage of the lower-limit open set \([0,1)\) under this identity is \([0,1)\) itself. It is not open in the standard topology: it contains \(0\), but every standard open interval about \(0\) contains negative numbers, which are not in \([0,1)\). More explicitly, for every \(\varepsilon>0\), the point \(-\varepsilon/2\) lies in \((-\varepsilon,\varepsilon)\) but not in \([0,1)\). Thus a codomain-open set has a non-open preimage, and the identity map in this direction is not continuous.
The subbasis criterion is useful because it turns a condition on every codomain-open set into a check on a generating family. It does not remove the need to identify the topologies correctly: the same subset may be open in one topology and not another, and a map’s preimages must be tested for openness in its domain.
Check Your Understanding
For each question, keep track of which topology is on the domain and which is on the codomain.
- Explain why the empty subbasis on a nonempty set \(Y\) generates \(\{\varnothing,Y\}\), not just \(\{\varnothing\}\).
- For \(f(x)=x^2\) with standard topologies, find the preimages of \((-\infty,0)\), \((0,\infty)\), and \((-\!4,\infty)\).
- Show directly that \([t,\infty)\) is open in \(\mathbb{R}_\ell\) by expressing it as a union of basis intervals.
- For \(g(x)=x+3\) on \(\mathbb{R}_\ell\), compute \(g^{-1}([4,7))\), including its endpoint brackets.
- Why does the fact that \([0,1)\) is open in \(\mathbb{R}_\ell\) not establish continuity of the identity \(\mathbb{R}\to\mathbb{R}_\ell\)?