Why Two Coordinates Are Enough
A map into the plane has two real-valued coordinates. The product topology makes those coordinates a practical continuity test: a neighborhood of a point in the plane contains a product of neighborhoods in the two coordinate axes. Thus, to control the image near a point, it is enough to control each coordinate there. This test applies to maps from any topological space, not just from an interval or another Euclidean space.
Throughout, \(\mathbb{R}\) has its standard topology, and \(\mathbb{R}^2=\mathbb{R}\times\mathbb{R}\) has the product topology. Its basis consists of open rectangles \(O_1\times O_2\), where \(O_1,O_2\) are open in \(\mathbb{R}\). A map \(F:X\to Y\) is continuous if the preimage of every open subset of \(Y\) is open in \(X\). If \(F:X\to\mathbb{R}^2\), write its coordinates as \(F(x)=(f(x),g(x))\), where \(f,g:X\to\mathbb{R}\).
Proof. Suppose first that \(F\) is continuous. For any open \(O\subseteq\mathbb{R}\), the sets \(O\times\mathbb{R}\) and \(\mathbb{R}\times O\) are open in the product topology. Their preimages satisfy \(f^{-1}(O)=F^{-1}(O\times\mathbb{R})\) and \(g^{-1}(O)=F^{-1}(\mathbb{R}\times O)\). These preimages are open in \(X\), so both coordinates are continuous.
Conversely, suppose \(f\) and \(g\) are continuous. For any open rectangle \(O_1\times O_2\), \(F^{-1}(O_1\times O_2)=f^{-1}(O_1)\cap g^{-1}(O_2)\). Both sets on the right are open in \(X\), so their intersection is open. Every open subset \(W\) of \(\mathbb{R}^2\) is a union of basis rectangles. Its preimage is the union of the preimages of those rectangles, and a union of open sets is open. Therefore \(F\) is continuous. \(\square\)
The same argument gives a test at a single point. Here, continuity of a coordinate at \(x_0\) means that for every open neighborhood \(O\) of its value, there is an open neighborhood \(U\) of \(x_0\) whose image under that coordinate lies in \(O\).
Proof. Suppose \(F\) is continuous at \(x_0\). Given an open neighborhood \(O\) of \(f(x_0)\), the set \(O\times\mathbb{R}\) is an open neighborhood of \(F(x_0)\). Continuity at \(x_0\) gives an open neighborhood \(U\) of \(x_0\) with \(F(U)\subseteq O\times\mathbb{R}\). Hence \(f(U)\subseteq O\), proving continuity of \(f\) at \(x_0\). Using \(\mathbb{R}\times O\) proves continuity of \(g\) there.
Now suppose both coordinates are continuous at \(x_0\), and let \(W\) be any open neighborhood of \(F(x_0)=(f(x_0),g(x_0))\). By the rectangle basis, there are open sets \(O_1,O_2\subseteq\mathbb{R}\) such that \(F(x_0)\in O_1\times O_2\subseteq W\). Coordinate continuity gives open neighborhoods \(U_1,U_2\) of \(x_0\) with \(f(U_1)\subseteq O_1\) and \(g(U_2)\subseteq O_2\). The open neighborhood \(U_1\cap U_2\) is mapped into \(O_1\times O_2\), hence into \(W\). This is continuity of \(F\) at \(x_0\). \(\square\)
Polynomial Coordinate Maps
For a map from \(\mathbb{R}^2\) to \(\mathbb{R}^2\), the test reduces the problem to two real-valued functions on \(\mathbb{R}^2\). We use the standard real-analysis facts that sums and products of continuous real-valued functions are continuous. The coordinate functions \(s\) and \(t\) on \(\mathbb{R}^2\) are continuous: they are the coordinate projections, whose preimages of open sets are open in the product topology. Therefore polynomial expressions in \(s\) and \(t\) are continuous.
Worked Example: A Polynomial Map of the Plane
Consider \(F:\mathbb{R}^2\to\mathbb{R}^2\) given by \(F(s,t)=(s^2+3t,\,st-2)\). Its coordinate functions are \(f(s,t)=s^2+3t\) and \(g(s,t)=st-2\). The coordinate projections \(s,t\) are continuous; taking products, constant multiples, and sums preserves continuity. Thus \(f\) and \(g\) are continuous, and the coordinatewise test proves that \(F\) is continuous.
Check at \((1,2)\): Substitution gives \(f(1,2)=1^2+3(2)=1+6=7\) and \(g(1,2)=(1)(2)-2=2-2=0\). Hence \(F(1,2)=(7,0)\). Substituting these values into the defining ordered pair gives \((1^2+3\cdot2,\,1\cdot2-2)=(7,0)\), as required. This value check does not replace the continuity proof; the proof applies at every point.
Piecewise Coordinates and Their Joining Points
When coordinates are defined by different formulas on different parts of the domain, check the joining points as well as the interiors of those parts. Away from a joining point, a sufficiently small neighborhood stays within one formula. At a joining point, the values and nearby behavior from each side must be controlled.
Worked Example: A Piecewise-Defined Map
Define \(F:\mathbb{R}\to\mathbb{R}^2\) by \(F(x)=(f(x),g(x))\), where \(f(x)=x^2\) for \(x\leq 0\), \(f(x)=x\) for \(x>0\), and \(g(x)=2x+1\) for \(x<1\), \(g(x)=x^2+2\) for \(x\geq1\). On each open interval that does not contain \(0\) or \(1\), the coordinate formulas are polynomials, so they are continuous there.
At \(x=0\), for \(f\): The defined value is \(f(0)=0^2=0\). If \(|x|<1\), then for \(x\leq0\), \(|f(x)-f(0)|=x^2\leq|x|\), while for \(x>0\), \(|f(x)-f(0)|=x=|x|\). Thus \(|f(x)-f(0)|\leq|x|\) whenever \(|x|<1\). Given \(\varepsilon>0\), choosing \(\delta=\min(1,\varepsilon)\) ensures that \(|x-0|<\delta\) implies \(|f(x)-f(0)|<\varepsilon\). The coordinate \(g\) is given by \(2x+1\) on a neighborhood of \(0\), so it is continuous there.
At \(x=1\), for \(g\): The defined value is \(g(1)=1^2+2=3\). If \(|x-1|<1/2\), then \(1/2<x<3/2\). For \(x<1\), \(|g(x)-3|=|2x+1-3|=2|x-1|\). For \(x\geq1\), \(|g(x)-3|=|x^2+2-3|=|x-1||x+1|<3|x-1|\), because \(x<3/2\) implies \(x+1<5/2<3\). Therefore, on this neighborhood, \(|g(x)-3|\leq3|x-1|\). Given \(\varepsilon>0\), choose \(\delta=\min(1/2,\varepsilon/3)\); then \(|x-1|<\delta\) implies \(|g(x)-g(1)|<\varepsilon\). The coordinate \(f\) is given by \(x\) on a neighborhood of \(1\), so it is continuous there.
Both coordinates are continuous at their joining points and on the intervals between them. Hence they are continuous on \(\mathbb{R}\), and the coordinatewise test proves \(F\) is continuous. Check: At \(x=-1\), \(f(-1)=(-1)^2=1\) and \(g(-1)=2(-1)+1=-1\), so \(F(-1)=(1,-1)\). At \(x=1\), \(f(1)=1\) and \(g(1)=1^2+2=3\), so \(F(1)=(1,3)\). Each value uses the formula whose condition includes that input.
One Discontinuous Coordinate Is Enough
The test is an if-and-only-if statement: continuity of the whole map requires continuity of every coordinate. A continuous first coordinate cannot compensate for a discontinuous second one.
Worked Example: A Step in One Coordinate
Let \(F:\mathbb{R}\to\mathbb{R}^2\) be \(F(x)=(x,h(x))\), where \(h(x)=0\) for \(x<0\) and \(h(x)=1\) for \(x\geq0\). The first coordinate \(x\mapsto x\) is continuous. For the second coordinate, the open interval \((1/2,3/2)\) has preimage \(h^{-1}((1/2,3/2))=[0,\infty)\), which is not open in \(\mathbb{R}\): every open interval containing \(0\) contains negative points. Thus \(h\) is not continuous, and the coordinatewise test shows that \(F\) is not continuous.
One can also see the failure directly. The open rectangle \(W=(-1,1)\times(1/2,3/2)\) contains \(F(0)=(0,1)\). Its preimage is \(F^{-1}(W)=[0,1)\), which is not open in \(\mathbb{R}\) because it contains \(0\) but contains no open interval around \(0\). This agrees with the definition of continuity.
Common pitfall. A wrong argument is: “The first coordinate \(x\mapsto x\) is continuous, so \(F(x)=(x,h(x))\) is continuous.” This checks only one coordinate and does not establish the required condition. The correction is to check both: \(x\mapsto x\) is continuous, but \(h^{-1}((1/2,3/2))=[0,\infty)\) is not open, so \(h\) is discontinuous and \(F\) is discontinuous. In the product topology, a neighborhood in the plane can restrict either coordinate, so neither coordinate may be ignored.
The practical value of the coordinatewise test is that it replaces a potentially complicated collection of open subsets of the plane with two ordinary real-valued continuity checks. For piecewise maps, this means identifying the breakpoints and verifying each coordinate there; for polynomial maps, it means recognizing the coordinate formulas as continuous real functions. The test depends on the product topology, whose basic neighborhoods are rectangles. It does not say that checking just one coordinate suffices, nor does it require that the two coordinates have the same formula or share the same joining points.
Check Your Understanding
For each question, identify the two coordinates before applying the continuity test.
- Use the coordinatewise test to decide whether \(F(s,t)=(s+t,\,s^2t)\) is continuous from \(\mathbb{R}^2\) to \(\mathbb{R}^2\).
- For the piecewise function \(u(x)=x\) when \(x<2\) and \(u(x)=x^2-2\) when \(x\geq2\), check the two formulas at \(x=2\) and decide whether \(u\) is continuous there.
- If \(f:X\to\mathbb{R}\) is continuous and \(g:X\to\mathbb{R}\) is discontinuous, can \(x\mapsto(f(x),g(x))\) be continuous? Explain using the theorem.
- For \(F(x)=(|x|,x^2)\) from \(\mathbb{R}\) to \(\mathbb{R}^2\), identify the coordinates and state why the coordinatewise test applies.
- Give an open rectangle containing \((0,1)\) whose preimage under the step map in the worked example is not open.