Piecewise Maps on Real Intervals

Why Breakpoints Matter

A formula may change from one part of an interval to another, but continuity is a property of the resulting map, not of the notation used to describe it. On each piece, continuity can be checked using the formula there. At a shared endpoint, the formulas must also give the same value. The pasting lemmas explain why these local checks are enough, and why it matters whether the pieces are open or closed.

Unless stated otherwise, every real interval below has the subspace topology inherited from \(\mathbb{R}\) with its standard topology, and every codomain \(\mathbb{R}\) has its standard topology. A map \(f:X\to Y\) is continuous if the preimage \(f^{-1}(V)\) of every open set \(V\subseteq Y\) is open in \(X\). An open or closed subset of an interval is understood relative to that interval.

Proof. The agreement condition makes the definition \(f(x)=f_i(x)\) for any \(i\) with \(x\in U_i\) independent of the choice of \(i\). It also gives uniqueness, since every point belongs to at least one \(U_i\). Let \(V\subseteq Y\) be open. Then \(f^{-1}(V)=\bigcup_{i\in J} f_i^{-1}(V)\), where each preimage on the right is taken within \(U_i\). Since \(f_i\) is continuous, \(f_i^{-1}(V)\) is open in \(U_i\); since \(U_i\) is open in \(X\), a set open in \(U_i\) is open in \(X\). Thus their union is open in \(X\). The open-preimage definition proves that \(f\) is continuous. \(\square\)

Proof. Agreement on intersections defines a single map \(f\) on all of \(X\). Let \(F\) be closed in \(Y\). By continuity on each \(C_i\), the set \(f_i^{-1}(F)\) is closed in \(C_i\). A subset closed in a closed subspace \(C_i\) is closed in \(X\): it has the form \(C_i\cap D_i\) for some closed \(D_i\subseteq X\), and this intersection is closed in \(X\). Therefore each \(f_i^{-1}(F)\), viewed as a subset of \(X\), is closed. Because the \(C_i\) cover \(X\), \(f^{-1}(F)=\bigcup_{i=1}^{n} f_i^{-1}(F)\). This is a finite union of closed sets, hence closed in \(X\). The closed-set test for continuity now gives continuity of \(f\). \(\square\)

A Criterion for Piecewise Maps on an Interval

Let \(a=x_0<x_1<\cdots<x_n=b\) be a finite partition of a nondegenerate closed interval \([a,b]\). The subintervals \(C_i=[x_{i-1},x_i]\) are closed in \([a,b]\), and their union is the whole interval. Consecutive pieces meet at one breakpoint; nonconsecutive pieces do not meet. This gives a direct test for piecewise formulas.

Proof. The only points assigned by two consecutive branches are their common endpoint \(x_i\). The stated equality gives the same value there, so the definition is well-defined. The pieces form a finite closed cover of \([a,b]\), and each restriction is continuous by assumption. The closed-cover pasting lemma therefore gives continuity of \(f\). Conversely, a single-valued map has only one value at \(x_i\); its restrictions to the two pieces must both equal that value. Hence \(f_i(x_i)=f_{i+1}(x_i)\). \(\square\)

The theorem includes the endpoints \(a\) and \(b\): each belongs to just one piece, so there is no matching condition at an outer endpoint. If the domain consists of a single point, continuity is automatic; the partition criterion above is stated for \(a<b\).

Computing Preimages Branch by Branch

Once continuity has been established, the same piecewise description also helps compute preimages. For a target set \(V\), first solve \(f_i(x)\in V\) on each branch, keeping that branch’s domain restrictions. Then take the union of the solutions. Strict inequalities in an open target set remain strict when solving the branch conditions.

When an Open Cover Is More Useful

The closed-cover lemma is natural for a finite partition into closed subintervals. The open-cover lemma is useful when local definitions are given on overlapping open neighborhoods instead. A finite partition such as \([-1,0]\) and \([0,1]\) is not an open cover: its pieces are not open in \([-1,1]\). For instance, every relative neighborhood of \(0\) in \([-1,1]\) contains positive points, so \([-1,0]\) is not relatively open. Agreement at a breakpoint alone does not turn closed pieces into open pieces.

A Common Mistake and a Practical Check

Common pitfall: applying the open-cover lemma to the closed halves of an interval just because they cover it. For the preceding piecewise polynomial-linear example, \([-1,0]\) and \([0,1]\) cover \([-1,1]\), but neither is open in \([-1,1]\). The open-cover lemma does not apply to those sets. The correction is to use the closed-cover lemma, check that both branches are continuous on their closed pieces, and verify agreement at \(0\).

A second common error is to solve a branch inequality without retaining that branch’s domain restriction. For example, in the polynomial-linear example the condition \(x^2<1/2\) gives \(-\sqrt{2}/2<x<\sqrt{2}/2\) on the real line. But on the negative branch the domain also requires \(x\le0\), so its contribution is only \((-\sqrt{2}/2,0]\), not the entire interval \((-\sqrt{2}/2,\sqrt{2}/2)\). The linear branch must be solved separately. Intersecting each solution with its branch domain before taking the union prevents this error.

For a finite piecewise definition on a closed interval, the reliable procedure is to identify the topology and the pieces, confirm that the pieces cover the domain and are closed, check continuity on each piece, and compare values at every shared breakpoint. For a local definition on open neighborhoods, instead confirm that the sets are open and cover the domain, then check agreement on every overlap. The hypotheses determine which pasting argument is valid.