Extending a Map One Coordinate at a Time
A real-valued function has one output coordinate, so the bounded Tietze extension theorem applies to it directly. A map into \(\mathbb{R}^n\) has finitely many real-valued coordinate functions. Extending each of these separately gives a map on the whole space, and finite-dimensional continuity ensures that the assembled map is continuous. Its coordinate bounds are controlled, although its Euclidean norm bound may be larger than the bound of the original map.
Let \((X,\mathcal{T})\) be a topological space and \(A\subseteq X\) have the subspace topology. Throughout, normal means \(T_1\) and that every pair of disjoint closed sets has disjoint open neighborhoods. The space \(\mathbb{R}^n\) has its usual Euclidean topology. A map \(f:A\to\mathbb{R}^n\) is bounded if some \(M\geq 0\) satisfies \(\|f(a)\|\leq M\) for every \(a\in A\).
The scalar result used below is the bounded Tietze extension theorem: if \(X\) is normal, \(A\) is closed in \(X\), and \(g:A\to\mathbb{R}\) is continuous with \(|g(a)|\leq c\) for all \(a\in A\), where \(c\geq0\), then there is a continuous \(G:X\to\mathbb{R}\) such that \(G|_A=g\) and \(|G(x)|\leq c\) for all \(x\in X\). This standard result was established in the earlier material on the bounded Tietze extension theorem; we use it here, rather than reproving the scalar construction.
Why Coordinate Continuity Is Enough
Proof. Suppose first that \(f\) is continuous. Each coordinate projection \(\pi_i:\mathbb{R}^n\to\mathbb{R}\) is continuous in the Euclidean topology, so \(f_i=\pi_i\circ f\) is continuous as a composition of continuous maps.
Conversely, suppose every \(f_i\) is continuous. The usual topology on \(\mathbb{R}^n\) is also the finite product topology. A basis for that topology consists of sets \(V_1\times\cdots\times V_n\), where each \(V_i\) is open in \(\mathbb{R}\). For any such basis set, \[ f^{-1}(V_1\times\cdots\times V_n) =\bigcap_{i=1}^n f_i^{-1}(V_i). \] Each \(f_i^{-1}(V_i)\) is open in \(A\), and a finite intersection of open sets is open. Thus the preimage of every basis set is open. Every open set in \(\mathbb{R}^n\) is a union of basis sets, and preimages preserve unions, so the preimage of every open set is open. Therefore \(f\) is continuous. \(\square\)
The Coordinatewise Extension Theorem
Proof. If \(X=\varnothing\), then \(A=\varnothing\) and the unique map from \(X\) to \(\mathbb{R}^n\) is the required extension. If \(A=\varnothing\) and \(X\ne\varnothing\), take \(F\) to be the constant zero map; it is continuous and bounded.
Now suppose \(A\ne\varnothing\). Boundedness of \(f\) gives an \(M\geq0\) such that \(\|f(a)\|\leq M\) for every \(a\in A\). Since \(|f_i(a)|\leq\|f(a)\|\), each coordinate function is bounded and \(0\leq M_i\leq M\). Each \(f_i\) is continuous by the coordinate-continuity proposition. Apply the bounded Tietze extension theorem to each \(f_i\): there is a continuous \(F_i:X\to\mathbb{R}\) extending \(f_i\), with \(|F_i(x)|\leq M_i\) for all \(x\in X\). Define \[ F(x)=(F_1(x),\ldots,F_n(x)). \] Its restriction to \(A\) is \(f\), because each coordinate agrees there. The coordinate-continuity proposition shows that \(F\) is continuous. Finally, for every \(x\in X\), \[ \|F(x)\|^2=\sum_{i=1}^n |F_i(x)|^2 \leq\sum_{i=1}^n M_i^2. \] Taking nonnegative square roots gives the asserted bound, so \(F\) is bounded. \(\square\)
The theorem gives separate control of each coordinate. If \(\|f(a)\|\leq M\), then each \(M_i\leq M\), so this argument guarantees \(\|F(x)\|\leq\sqrt{n}\,M\). It does not, in general, guarantee \(\|F(x)\|\leq M\), nor that \(F\) takes values in a particular subset such as a sphere.
Explicit Extensions in the Plane
Worked Example: A Bounded Map on a Closed Line
Give an explicit extension to \(\mathbb{R}^2\) of \[ f(t,0)=\left(\frac{t}{\sqrt{1+t^2}},\frac{1}{1+t^2}\right), \qquad (t,0)\in A=\mathbb{R}\times\{0\}. \] All plane spaces here have the standard Euclidean topology, and \(A\) has the subspace topology.
First, \(A\) is closed: it is the zero set of the continuous second-coordinate projection \((x,y)\mapsto y\), since \(\{0\}\) is closed in \(\mathbb{R}\). The displayed coordinate functions are continuous because \(1+t^2>0\) for all \(t\). Also, \[ \left|\frac{t}{\sqrt{1+t^2}}\right|\leq1 \quad\text{and}\quad 0<\frac{1}{1+t^2}\leq1, \] using \(t^2\leq1+t^2\) for the first inequality. Thus \(f\) is bounded.
One explicit extension is \[ F(x,y)=\left(\frac{x}{\sqrt{1+x^2}},\frac{1}{1+x^2}\right). \] The denominator is positive at every \((x,y)\), so both coordinates, and hence \(F\), are continuous. For \((t,0)\in A\), substitution gives \(F(t,0)=f(t,0)\). For example, \(F(0,0)=(0,1)\), while \(F(1,0)=(1/\sqrt{2},1/2)\), agreeing with the prescribed values. The same inequalities with \(x\) in place of \(t\) show that the extension is bounded. This verifies the extension directly, without needing to construct it by the scalar theorem.
Worked Example: Coordinate Extensions from the Unit Circle
Let \(A=S^1=\{(x,y)\in\mathbb{R}^2:x^2+y^2=1\}\), with its subspace topology, and let \(f:A\to\mathbb{R}^2\) be \(f(x,y)=(x,y)\). The set \(A\) is closed because it is the zero set of the continuous function \((x,y)\mapsto x^2+y^2-1\). On \(A\), \(|x|\leq1\) and \(|y|\leq1\), so \(f\) is bounded and continuous.
For \(s\in\mathbb{R}\), define \(c(s)=-1\) if \(s<-1\), \(c(s)=s\) if \(-1\leq s\leq1\), and \(c(s)=1\) if \(s>1\). This function is continuous: its formulas are continuous on the three intervals and agree at the joining points \(-1\) and \(1\). Set \[ F(x,y)=(c(x),c(y)). \] It is continuous by coordinate continuity. Every point of \(S^1\) has both coordinates in \([-1,1]\), so substitution yields \(F(x,y)=(x,y)=f(x,y)\) on \(A\). Furthermore, \(|F_1|\leq1\) and \(|F_2|\leq1\) throughout the plane. At the point \((2,2)\), the calculation gives \(F(2,2)=(c(2),c(2))=(1,1)\), whose norm is \(\sqrt{1^2+1^2}=\sqrt{2}\). This confirms that the separate coordinate bounds do not imply a Euclidean norm bound of \(1\).
Worked Example: A Map on a Closed Segment
Let \(A=[-1,1]\times\{0\}\subseteq\mathbb{R}^2\), with the subspace topology, and prescribe \[ f(t,0)=\left(t^2,\frac{t+1}{2}\right). \] The segment \(A\) is closed: it is the intersection of the closed sets \(\mathbb{R}\times\{0\}\), \(\{(x,y):x\geq-1\}\), and \(\{(x,y):x\leq1\}\). On \(A\), \(0\leq t^2\leq1\) and \(0\leq(t+1)/2\leq1\), so both coordinates are bounded and continuous.
Define \(c_0(s)=\min(1,\max(0,s))\). The functions \(\min\) and \(\max\) of two real variables are continuous, so \(c_0\) is continuous; also \(c_0(s)=s\) whenever \(0\leq s\leq1\). An extension is \[ F(x,y)=\left(c_0(x^2),c_0\left(\frac{x+1}{2}\right)\right). \] It is continuous because its coordinates are compositions of continuous functions. For \(t\in[-1,1]\), both inputs to \(c_0\) lie in \([0,1]\), so \(F(t,0)=(t^2,(t+1)/2)=f(t,0)\). Direct checks at three points give \(F(-1,0)=(1,0)\), \(F(0,0)=(0,1/2)\), and \(F(1,0)=(1,1)\); each is the prescribed value. Since \(c_0\) always lies in \([0,1]\), the extension has both coordinates bounded by \(1\) everywhere.
What the Bounds Do—and Do Not—Say
The coordinatewise argument is useful because it reduces a finite-dimensional extension problem to scalar extension problems, each with a known bound. If the image of \(f\) lies in a box \(\prod_{i=1}^n[-M_i,M_i]\), the constructed extension also lies in that box. But a box is not usually a ball, and coordinatewise membership in intervals does not preserve relationships between coordinates.
Common pitfall. From \(|F_1(x)|\leq1\) and \(|F_2(x)|\leq1\), it is incorrect to conclude that \(\|F(x)\|\leq1\). At \((2,2)\), the circle example gives \(F(2,2)=(1,1)\), so \[ \|F(2,2)\|=\sqrt{1^2+1^2}=\sqrt{2}>1. \] The correct conclusion from coordinate bounds \(M_1,M_2\) is \(\|F(x)\|\leq\sqrt{M_1^2+M_2^2}\). Likewise, the extension of the identity map on \(S^1\) just constructed is not \(S^1\)-valued: \(F(0,0)=(0,0)\), and \(0^2+0^2=0\ne1\). The theorem guarantees an extension into \(\mathbb{R}^2\), not into the sphere.
There is a second distinction worth keeping: boundedness of the original map is part of the bounded extension theorem being applied. It lets us choose finite coordinate bounds \(M_i\) and use the bound-preserving scalar theorem. The conclusion proved here is an extension that is itself bounded, with explicit coordinate controls; it is not a claim that an unbounded map can have a bounded extension agreeing with it.
Check Your Understanding
In each question, use the stated topology and distinguish coordinate bounds from a Euclidean norm bound.
- Why is each coordinate function of a bounded map \(f:A\to\mathbb{R}^n\) bounded?
- If \(|F_1(x)|\leq2\) and \(|F_2(x)|\leq3\) for every \(x\), what Euclidean norm bound follows from these inequalities?
- For the segment example, calculate \(F(1/2,0)\) and verify that it agrees with \(f(1/2,0)\).
- Why is \(S^1\) closed in \(\mathbb{R}^2\) with the standard topology?
- Does coordinatewise extension by itself guarantee that a map originally taking values in the unit circle extends as a unit-circle-valued map? Explain using the circle example.