Why a Basis Can Test Continuity
Continuity is defined by asking whether the preimage of every open set is open. A codomain can have infinitely many open sets, so checking them one at a time is usually impractical. A basis gives a smaller collection of open sets from which every open set is formed by unions. Since preimages preserve unions, checking the basis elements is enough.
Let \((X,\mathcal{T}_X)\) and \((Y,\mathcal{T}_Y)\) be topological spaces, and let \(f:X\to Y\). The map \(f\) is continuous if \(f^{-1}(V)\) is open in \(X\) for every open set \(V\) in \(Y\). For the criterion below, only the codomain needs a chosen basis; the domain may have any topology.
For example, the open intervals \((a,b)\), where \(a<b\) are real numbers, form a basis for the standard topology on \(\mathbb{R}\). The open balls \(B(p,r)=\{y\in\mathbb{R}^n:\|y-p\|<r\}\), with \(p\in\mathbb{R}^n\) and \(r>0\), form a basis for the standard topology on \(\mathbb{R}^n\). These bases let us express familiar continuity tests in terms of interval or ball preimages.
The Basis Criterion
Proof. Suppose first that \(f\) is continuous. Every \(B\in\mathcal{B}\) is open in \(Y\), so the definition of continuity gives that \(f^{-1}(B)\) is open in \(X\).
Conversely, suppose \(f^{-1}(B)\) is open in \(X\) for every \(B\in\mathcal{B}\). Let \(V\) be any open set in \(Y\). By the basis property, \(V\) is a union of basis elements; write \(V=\bigcup_{\alpha\in J}B_\alpha\), where each \(B_\alpha\in\mathcal{B}\). The inverse-image identity for unions gives \(f^{-1}(V)=\bigcup_{\alpha\in J}f^{-1}(B_\alpha)\). Each set in this union is open in \(X\) by assumption, and an arbitrary union of open sets is open. Thus \(f^{-1}(V)\) is open for every open \(V\) in \(Y\), which is exactly continuity. This argument also covers \(V=\varnothing\), whose preimage is the empty open set. \(\square\)
The theorem requires checking all members of the chosen basis, not just some convenient examples from it. It does not require a basis for the domain. It also does not say that a preimage must itself be a basis element: it only has to be open in the domain.
Testing Maps with Interval Bases
Worked Example: An Affine Map of the Real Line
Let \(f:\mathbb{R}\to\mathbb{R}\) be \(f(x)=2x+1\), with the standard topology on both copies of \(\mathbb{R}\). We test continuity using the basis of open intervals in the codomain.
Take any basis interval \((a,b)\), where \(a<b\). A real number \(x\) lies in \(f^{-1}((a,b))\) exactly when \(a<2x+1<b\). Subtracting \(1\) and dividing by the positive number \(2\) preserves both strict inequalities, so \(f^{-1}((a,b))=((a-1)/2,(b-1)/2)\). Since \(a<b\), we have \((a-1)/2<(b-1)/2\), and this preimage is an open interval in the domain. The calculation applies to every basis interval, so the basis criterion proves that \(f\) is continuous.
Check: for \((a,b)=(1,5)\), the preimage is \((0,2)\). If \(0<x<2\), then \(1<2x+1<5\); at the endpoints \(x=0\) and \(x=2\), the function values are \(1\) and \(5\), which are excluded. This confirms the strict endpoints in the computed preimage.
Worked Example: A Map from a Closed Interval
Give \(I=[0,1]\) the subspace topology inherited from the standard topology on \(\mathbb{R}\), and give the codomain \(\mathbb{R}\) its standard topology. Define \(g:I\to\mathbb{R}\) by \(g(t)=3t-1\). We again use the open-interval basis in the codomain.
For any \(a<b\), membership in the preimage means \(a<3t-1<b\). Adding \(1\), dividing by \(3>0\), and imposing the domain restriction \(0\le t\le1\) gives \(g^{-1}((a,b))=I\cap((a+1)/3,(b+1)/3)\). This is open in \(I\): by the definition of the subspace topology, the intersection of \(I\) with an open subset of \(\mathbb{R}\) is open in \(I\). Every basis interval therefore has an open preimage in the domain, so \(g\) is continuous.
Check: for the interval \((-2,2)\), the inequalities \(-2<3t-1<2\) give \(-1/3<t<1\). Intersecting with \(I=[0,1]\) gives \([0,1)\). This set is not open in \(\mathbb{R}\), but it is open in \(I\), since \([0,1)=I\cap(-1,1)\). Using the domain's actual topology matters.
Testing Maps with Euclidean Ball Bases
Worked Example: Projection onto One Coordinate
Give both the domain and codomain \(\mathbb{R}^2\) their standard Euclidean topologies. Define \(P:\mathbb{R}^2\to\mathbb{R}^2\) by \(P(x,y)=(x,0)\). We test \(P\) against every open ball in the codomain.
Fix a ball with centre \((a,b)\) and radius \(r>0\). A point \((x,y)\) belongs to its preimage exactly when its image lies in the ball: \((x-a)^2+b^2<r^2\). Notice that this condition does not depend on \(y\). If \(|b|<r\), then \(r^2-b^2>0\), and the condition is equivalent to \(a-\sqrt{r^2-b^2}<x<a+\sqrt{r^2-b^2}\). Thus the preimage is \((a-\sqrt{r^2-b^2},a+\sqrt{r^2-b^2})\times\mathbb{R}\), which is open in \(\mathbb{R}^2\). Indeed, around any point of this set one can choose a positive-radius ball small enough that its first coordinate stays inside the displayed interval.
If \(|b|\ge r\), then \(b^2\ge r^2\), so \((x-a)^2+b^2\ge r^2\) for every \(x\). The strict ball inequality has no solutions, including when \(|b|=r\), and the preimage is empty, hence open. We have checked every ball, so the basis criterion proves that \(P\) is continuous.
Check: for the ball centred at \((0,3)\) with radius \(5\), the condition is \(x^2+9<25\), or \(x^2<16\). Its preimage is \((-4,4)\times\mathbb{R}\). For \(x=3\), the image is \((3,0)\), whose squared distance from \((0,3)\) is \(3^2+(-3)^2=18<25\). At \(x=4\) or \(x=-4\), that squared distance is \(16+9=25\), so those boundary points are correctly excluded.
A Common Pitfall: Open Does Not Mean a Basis Element
A basis is a collection that generates all the open sets by unions. An open set can therefore be a union of several basis elements and need not be one basis element itself. For example, on \(\mathbb{R}\) with its standard topology, take \(h(x)=x^2\) and the basis interval \((1,4)\). The condition \(1<x^2<4\) is equivalent to \(1<|x|<2\), so \(h^{-1}((1,4))=(-2,-1)\cup(1,2)\). This is open, even though it is not a single open interval.
Common mistake. One might incorrectly reject this preimage because it is not one basis interval, or incorrectly claim that checking just this one preimage proves \(h\) continuous. The correction is that this particular preimage is open, as it is a union of open intervals, but the basis criterion proves continuity only after the preimage of every basis interval has been shown open.
The criterion can also expose a failure. Define \(s:\mathbb{R}\to\mathbb{R}\) by \(s(x)=0\) for \(x<0\) and \(s(x)=1\) for \(x\ge0\). For the basis interval \((1/2,3/2)\), the preimage is \([0,\infty)\): the value \(1\) lies in the interval, while \(0\) does not. This preimage is not open in \(\mathbb{R}\), because every open interval around \(0\) contains negative numbers and hence is not contained in \([0,\infty)\). The basis criterion therefore shows that \(s\) is not continuous.
Check Your Understanding
For each question, identify the topology on the domain and use the basis criterion where appropriate.
- For \(f(x)=5x-2\) from standard \(\mathbb{R}\) to standard \(\mathbb{R}\), find the preimage of an arbitrary basis interval \((a,b)\).
- For \(q:I\to\mathbb{R}\), \(q(t)=2t\), compute the preimage of \((1,3)\) and say whether it is open in \(I\).
- For \(R(x,y)=(0,y)\) on standard \(\mathbb{R}^2\), describe the preimage of a ball with centre \((a,b)\) and radius \(r>0\), distinguishing the cases \(|a|<r\) and \(|a|\ge r\).
- Explain why a preimage that is a union of two open intervals can still satisfy the basis criterion.
- For the step map \(s\) in the tutorial, find a basis interval whose preimage is not open and justify the failure of openness.