Continuity Through Local Tests
Continuity is defined by checking inverse images of open sets, but a target topology may have many open sets, and a product may have many coordinates. The practical strategy is to use a smaller collection that generates the topology. For a subbasis, it is enough to check its members; for a product, it is enough to check the coordinate maps. These reductions work because inverse images preserve unions and finite intersections.
Let \((X,\mathcal{T}_X)\) and \((Y,\mathcal{T}_Y)\) be topological spaces. A map \(f:X\to Y\) is continuous if \(f^{-1}(V)\in\mathcal{T}_X\) for every \(V\in\mathcal{T}_Y\). A subbasis \(\mathcal{S}\) for \(\mathcal{T}_Y\) is a collection of open sets whose finite intersections form a basis for \(\mathcal{T}_Y\). The empty intersection is \(Y\) itself. Thus every open set in \(Y\) is a union of finite intersections of members of \(\mathcal{S}\).
Proof. If \(f\) is continuous, then \(f^{-1}(S)\) is open for every subbasic open set \(S\), since \(S\) is open in \(Y\). Conversely, suppose that \(f^{-1}(S)\) is open for every \(S\in\mathcal{S}\). For a finite intersection \(B=S_1\cap\cdots\cap S_k\), the inverse-image identity gives \(f^{-1}(B)=f^{-1}(S_1)\cap\cdots\cap f^{-1}(S_k)\), which is open because a finite intersection of open sets is open. If \(k=0\), then \(B=Y\) and \(f^{-1}(B)=X\), also open. Every open set \(V\) in \(Y\) is a union of such finite intersections, and \(f^{-1}(\bigcup_\lambda B_\lambda)=\bigcup_\lambda f^{-1}(B_\lambda)\). This is open because arbitrary unions of open sets are open. Hence the inverse image of every open set is open, so \(f\) is continuous. \(\square\)
A Subbasis Test on the Real Line
In \(\mathbb{R}\) with its standard topology, the collection \(\mathcal{S}=\{(-\infty,a),(a,\infty):a\in\mathbb{R}\}\) is a subbasis. Each member is open. Intersecting one left ray and one right ray gives an open interval \((a,b)\) when \(a<b\); such intervals form a basis for the standard topology. Thus the subbasis test lets us check a function using only the inverse images of these rays.
Worked Example: Testing the Square Function
Let \(f:\mathbb{R}\to\mathbb{R}\) be \(f(x)=x^2\), with the standard topology on both copies of \(\mathbb{R}\). Test continuity using the ray subbasis.
For a subbasic ray \((-\infty,a)\), its inverse image is \(f^{-1}((-\infty,a))=\{x\in\mathbb{R}:x^2<a\}\). If \(a\leq 0\), this set is empty: squares are nonnegative, and even when \(a=0\), the strict inequality \(x^2<0\) has no solutions. If \(a>0\), the inequality is equivalent to \(-\sqrt a<x<\sqrt a\), so the inverse image is \((-\sqrt a,\sqrt a)\). In both cases it is open.
For a subbasic ray \((a,\infty)\), the inverse image is \(\{x:x^2>a\}\). If \(a<0\), every real square is greater than \(a\), so the inverse image is \(\mathbb{R}\). If \(a=0\), the inequality excludes exactly \(x=0\), giving \((-\infty,0)\cup(0,\infty)\). If \(a>0\), solving the strict inequality gives \((-\infty,-\sqrt a)\cup(\sqrt a,\infty)\). Each result is open in the standard topology. The subbasis test now proves that \(f\) is continuous.
Check: at \(a=0\), the preimage of \((-\infty,0)\) is empty, while the preimage of \((0,\infty)\) is \(\mathbb{R}\setminus\{0\}\). These agree with the formulas and respect the strict endpoints.
Maps into Product Spaces
Let \(\{X_i:i\in J\}\) be a family of topological spaces, where \(J\) is an index set. The product set is \(\prod_{i\in J}X_i\). Its product topology has a basis consisting of sets that restrict only finitely many coordinates: choose a finite set \(F\subseteq J\), choose an open set \(U_i\subseteq X_i\) for each \(i\in F\), and take \(\{(x_i)_{i\in J}:x_i\in U_i\text{ for every }i\in F\}\). All other coordinates are unrestricted. The projection \(\pi_i:\prod_{j\in J}X_j\to X_i\) sends a tuple to its \(i\)-th coordinate. The sets \(\pi_i^{-1}(U_i)\), for \(U_i\) open in \(X_i\), form a subbasis: their finite intersections are exactly the basic sets just described.
Proof. If \(f\) is continuous, then each projection \(\pi_i\) is continuous: for open \(U_i\subseteq X_i\), the set \(\pi_i^{-1}(U_i)\) is subbasic, hence open in the product. Therefore \(f_i=\pi_i\circ f\) is continuous, since \(f_i^{-1}(U_i)=f^{-1}(\pi_i^{-1}(U_i))\), an inverse image of an open set under \(f\).
Conversely, suppose every \(f_i\) is continuous. For each subbasic set \(\pi_i^{-1}(U_i)\), its inverse image under \(f\) is \(f^{-1}(\pi_i^{-1}(U_i))=f_i^{-1}(U_i)\), which is open in \(X\). The subbasis test proves that \(f\) is continuous. If \(J\) is empty, the product is a one-point space, every map into it is continuous, and there are no coordinate conditions to check; the equivalence still holds. \(\square\)
The finite-restriction feature matters: a basic product neighborhood does not require all coordinates to lie in prescribed open sets. It imposes conditions on only finitely many coordinates at a time. The following examples use the coordinate criterion and then check particular product neighborhoods directly.
Worked Example: A Map into a Finite Product
Define \(f:\mathbb{R}\to\mathbb{R}\times\mathbb{R}\) by \(f(t)=(t,t^2)\), where each factor has the standard topology and the product has the product topology. The coordinate functions are \(f_1(t)=t\) and \(f_2(t)=t^2\).
The first coordinate is continuous: for \(a\in\mathbb{R}\) and \(\varepsilon>0\), choosing \(\delta=\varepsilon\) gives \(|t-a|<\delta\Rightarrow|f_1(t)-f_1(a)|<\varepsilon\). For the second coordinate, fix \(a\in\mathbb{R}\) and \(\varepsilon>0\). Choose \(\delta=\min(1,\varepsilon/(2|a|+1))\), which is positive. If \(|t-a|<\delta\), then \(|t-a|<1\), so \(|t+a|\leq |t-a|+2|a|<1+2|a|\). Consequently, \[ |f_2(t)-f_2(a)|=|t^2-a^2|=|t-a||t+a|<\delta(2|a|+1)\leq\varepsilon. \] Thus both coordinates are continuous, and the coordinatewise theorem proves that \(f\) is continuous.
For a direct check at \(t=2\), consider the basic open rectangle \(U=(1,3)\times(3,5)\), which contains \(f(2)=(2,4)\). Its inverse image is \[ f^{-1}(U)=\{t:1<t<3\text{ and }3<t^2<5\}=(\sqrt{3},\sqrt{5}). \] Indeed, \(3<t^2<5\) gives two intervals, \((-\sqrt5,-\sqrt3)\cup(\sqrt3,\sqrt5)\), and intersecting with \((1,3)\) leaves \((\sqrt3,\sqrt5)\). This is open and contains \(2\), as required by the local continuity test.
Check: substituting \(t=2\) into the original map gives \((2,4)\), and \(1<2<3\) and \(3<4<5\), so the selected point lies in the rectangle. The inverse-image interval includes \(2\) because \(3<4<5\).
Worked Example: A Map into an Infinite Product
Let \(\mathbb{R}^{\mathbb{N}}=\prod_{n\in\mathbb{N}}\mathbb{R}\) have the product topology, and define \(g:\mathbb{R}\to\mathbb{R}^{\mathbb{N}}\) by \(g(t)=(t,t/2,t/3,\ldots)\). Its \(n\)-th coordinate is \(g_n(t)=t/n\). For any \(a\in\mathbb{R}\) and \(\varepsilon>0\), if \(|t-a|<n\varepsilon\), then \(|g_n(t)-g_n(a)|=|t-a|/n<\varepsilon\). So each coordinate is continuous, and the theorem proves that \(g\) is continuous.
Here is an explicit product-neighborhood calculation. Let \(U\) restrict coordinates \(1,3,5\) to \((1,3)\), \((1/2,5/6)\), and \((0,1/2)\), respectively, leaving every other coordinate unrestricted. This is a basic open set in the product topology. Since \(g(2)=(2,1,2/3,1/2,2/5,\ldots)\), the point \(g(2)\) belongs to \(U\). The three restrictions pull back to \(1<t<3\), \(1/2<t/3<5/6\), and \(0<t/5<1/2\). Multiplying the second inequality by \(3\) gives \(3/2<t<5/2\); multiplying the third by \(5\) gives \(0<t<5/2\). Their intersection with \((1,3)\) is therefore \(g^{-1}(U)=(3/2,5/2)\), an open neighborhood of \(2\).
Check: for \(t=2\), the restricted coordinates are \(2\), \(2/3\), and \(2/5\). They satisfy \(1<2<3\), \(1/2<2/3<5/6\), and \(0<2/5<1/2\), confirming the membership and the strict endpoints.
A Common Pitfall: Replacing Product by Box
The coordinatewise theorem depends on the product topology. In the box topology on \(\mathbb{R}^{\mathbb{N}}\), a basic open set may restrict every coordinate at once. It is a mistake to reason that coordinatewise continuity automatically makes inverse images of such sets open: the resulting inverse image may be an infinite intersection of open sets, and an infinite intersection need not be open.
For a concrete failure, let \(h:\mathbb{R}\to\mathbb{R}^{\mathbb{N}}\) be the diagonal map \(h(t)=(t,t,t,\ldots)\). Every coordinate map is the identity on \(\mathbb{R}\), hence continuous. The set \(V=\prod_{n\in\mathbb{N}}(-1/n,1/n)\) is open in the box topology and contains \(h(0)\). Its inverse image is \[ h^{-1}(V)=\{t\in\mathbb{R}: |t|<1/n\text{ for every }n\in\mathbb{N}\}=\{0\}. \] To verify the equality, \(t=0\) satisfies every inequality; if \(t\ne0\), choose \(n\in\mathbb{N}\) with \(n>1/|t|\), so \(1/n<|t|\) and the \(n\)-th inequality fails. The singleton \(\{0\}\) is not open in the standard real topology: every open interval containing \(0\) contains nonzero real numbers. Thus \(h\) is not continuous into the box topology.
The tempting but incorrect calculation is “each coordinate condition has an open inverse image, so their intersection is open.” The correction is that topology guarantees openness of arbitrary unions and only finite intersections. In the product topology, a basic set restricts finitely many coordinates, which is precisely why the coordinatewise test works. In the box topology, the infinite family of restrictions can destroy openness.
Check Your Understanding
For each question, identify the topology being used and keep track of whether restrictions are finite or unrestricted.
- For \(f(x)=x^2\), compute \(f^{-1}((a,\infty))\) separately when \(a<0\), \(a=0\), and \(a>0\).
- Let \(f:X\to Y\), and suppose inverse images of every member of a subbasis for \(Y\) are open. Explain why this implies that inverse images of arbitrary open sets are open.
- For \(q:\mathbb{R}\to\mathbb{R}^2\) defined by \(q(t)=(t,3t)\), state its coordinate maps and use the coordinatewise theorem to decide continuity.
- In \(\mathbb{R}^{\mathbb{N}}\) with the product topology, describe the form of a basic open neighborhood and explain why only finitely many coordinates are restricted.
- For the diagonal map \(h(t)=(t,t,\ldots)\), explain why the box-open set \(\prod_{n\in\mathbb{N}}(-1/n,1/n)\) has a non-open inverse image.