Continuity into a Countable Product
A point of \(\mathbb{R}^{\mathbb{N}}\) is a sequence of real numbers, so a map into this space can be specified by giving one real-valued function for each coordinate. The product topology makes this description useful: each basic open neighborhood restricts only finitely many coordinates. Consequently, continuity can be tested one coordinate at a time, even though the output has infinitely many entries.
Throughout, \(\mathbb{R}\) has its standard topology, \(\mathbb{N}=\{1,2,\dots\}\), and \(\mathbb{R}^{\mathbb{N}}=\prod_{n\in\mathbb{N}}\mathbb{R}\) has the product topology. For each \(n\in\mathbb{N}\), the coordinate projection is \(\pi_n:\mathbb{R}^{\mathbb{N}}\to\mathbb{R}\), defined by \(\pi_n((y_k)_{k\in\mathbb{N}})=y_n\). A basic open set in the product topology restricts a finite set of coordinates and leaves every other coordinate unrestricted.
In particular, a basic open neighborhood of \(f(x)\) may require its first coordinate to lie in one open interval, its fifth coordinate to lie in another, and its twelfth coordinate to lie in a third, while placing no restrictions on all other coordinates. There is no requirement in this topology that infinitely many coordinate conditions be met by one basic neighborhood.
The Coordinate Continuity Criterion
Proof. First suppose \(f\) is continuous. For any open set \(U\subseteq\mathbb{R}\), the set \(\pi_n^{-1}(U)\) is open in the product topology: it restricts coordinate \(n\) to \(U\) and leaves all other coordinates unrestricted. Therefore \((\pi_n\circ f)^{-1}(U)=f^{-1}(\pi_n^{-1}(U))\) is open in \(X\), by continuity of \(f\). This proves that every coordinate function is continuous.
Conversely, suppose every coordinate function \(f_n=\pi_n\circ f\) is continuous. A basic open set in the product has the form \(V=\bigcap_{j\in F}\pi_j^{-1}(U_j)\), where \(F\subseteq\mathbb{N}\) is finite and each \(U_j\subseteq\mathbb{R}\) is open. Its preimage is \(f^{-1}(V)=\bigcap_{j\in F}f_j^{-1}(U_j)\). Each set in this finite intersection is open in \(X\), since \(f_j\) is continuous, so \(f^{-1}(V)\) is open. If \(F\) is empty, \(V=\mathbb{R}^{\mathbb{N}}\) and its preimage is \(X\), also open. Every open set in the product is a union of basic open sets, and the preimage of a union is the union of the preimages. Thus the preimage of every open set is open, which proves that \(f\) is continuous. \(\square\)
The finite set \(F\) is the key in the reverse direction. Even though the target has infinitely many coordinates, an individual basic neighborhood tests only finitely many of them. If \(X=\mathbb{R}\), this can also be phrased using the epsilon-delta test at a point \(x\): for each of the finitely many restricted coordinates, continuity supplies a suitable positive \(\delta\); taking the minimum of those finitely many values gives one \(\delta\) that works for that product neighborhood. No minimum over infinitely many coordinates is called for.
Worked Example: A Power Sequence
Define \(f:\mathbb{R}\to\mathbb{R}^{\mathbb{N}}\) by \(f(t)=(t^n)_{n\in\mathbb{N}}\). Its \(n\)-th coordinate is the polynomial \(f_n(t)=t^n\), which is continuous on \(\mathbb{R}\). By the coordinate criterion, \(f\) is continuous.
For a concrete value, take \(t=\frac12\). Substituting into the coordinate formula gives \(f_1(\frac12)=(\frac12)^1=\frac12\), \(f_2(\frac12)=(\frac12)^2=\frac14\), and \(f_3(\frac12)=(\frac12)^3=\frac18\). Thus \(f(\frac12)=(\frac12,\frac14,\frac18,\dots)\), with the \(n\)-th entry \((\frac12)^n\).
To check the neighborhood reasoning directly, consider any basic open neighborhood of \(f(\frac12)\) that restricts coordinates in a finite set \(F\). For each \(j\in F\), continuity of \(t\mapsto t^j\) gives an open neighborhood of \(\frac12\) whose image in coordinate \(j\) stays in the prescribed open set. The intersection of these finitely many neighborhoods is open, contains \(\frac12\), and maps into the chosen product neighborhood. If \(F\) is empty, the whole domain works. This agrees with the theorem, including the case of no coordinate restrictions.
Check: for the first three coordinates at \(t=\frac12\), the formula gives \((\frac12,\frac14,\frac18)\); raising each displayed input \(\frac12\) to the corresponding exponent \(1,2,3\) reproduces those entries.
Coordinate Continuity Does Not Require Uniform Bounds
For a real-valued function, an epsilon-delta estimate may use a \(\delta\) depending on the function and on the chosen tolerance. When there are infinitely many coordinate functions, the coordinate criterion asks for continuity of each one; it does not ask for a single estimate that works simultaneously for every coordinate and every tolerance. The next example makes that distinction explicit.
Worked Example: Coordinates with Increasing Slopes
Define \(g:\mathbb{R}\to\mathbb{R}^{\mathbb{N}}\) by \(g(t)=(nt)_{n\in\mathbb{N}}\), so \(g_n(t)=nt\). Each coordinate is a linear function and is continuous. Therefore \(g\) is continuous into the product.
Here is the finite-coordinate estimate at any \(t_0\in\mathbb{R}\). Suppose a basic neighborhood restricts the coordinates in a nonempty finite set \(F\), with coordinate \(j\) allowed to vary by less than \(\varepsilon_j>0\) from \(g_j(t_0)\). Since \(|g_j(t)-g_j(t_0)|=j|t-t_0|\), it is enough to take \(\delta=\min_{j\in F}(\varepsilon_j/j)>0\). If \(|t-t_0|<\delta\), then for each \(j\in F\), \(|g_j(t)-g_j(t_0)|=j|t-t_0|<j\delta\leq\varepsilon_j\). If \(F\) is empty, any \(\delta>0\) works.
For example, take \(F=\{1,3,5\}\) and \(\varepsilon_j=\frac1{10}\) for each \(j\in F\). The three candidate bounds are \(\frac{1}{10}\), \(\frac{1}{30}\), and \(\frac{1}{50}\), so \(\delta=\frac{1}{50}\). If \(|t-t_0|<\frac1{50}\), the coordinate changes are respectively less than \(\frac1{50}\), \(\frac3{50}\), and \(\frac5{50}=\frac1{10}\). Each is strictly less than \(\frac1{10}\), as required.
There cannot, however, be a positive \(\delta\) that keeps every coordinate change below \(1\) whenever \(|t-0|<\delta\). Given any \(\delta>0\), let \(t=\delta/2\), so \(0<t<\delta\). Choose \(n\in\mathbb{N}\) with \(n>2/\delta\). Then \(|g_n(t)-g_n(0)|=n\delta/2>1\). Thus no such uniform estimate holds across all coordinates, even though \(g\) is continuous into the product.
Check: at \(t=\frac1{10}\), the first three entries are \(1\cdot\frac1{10}=\frac1{10}\), \(2\cdot\frac1{10}=\frac2{10}=\frac15\), and \(3\cdot\frac1{10}=\frac3{10}\). Their differences from the corresponding entries at \(0\) are exactly those values, in agreement with \(|g_n(t)-g_n(0)|=n|t|\).
A uniform estimate can still be a convenient sufficient tool. The point is that it is stronger than the coordinate criterion requires: a common bound can prove continuity of every coordinate at once, but a map may be continuous even when no such common bound is available.
Worked Example: A Common Estimate That Is Available
Define \(h:\mathbb{R}\to\mathbb{R}^{\mathbb{N}}\) by \(h_n(t)=t/(n+1)\). For any \(s,t\in\mathbb{R}\) and \(n\in\mathbb{N}\), \[ |h_n(t)-h_n(s)| =\frac{|t-s|}{n+1} \leq\frac{|t-s|}{2}, \] because \(n\geq1\) implies \(n+1\geq2\). This single estimate works for every coordinate. Given any tolerance \(\varepsilon>0\), choosing \(\delta=2\varepsilon\) ensures that \(|t-s|<\delta\) implies \(|h_n(t)-h_n(s)|<\varepsilon\) for every \(n\). In particular each coordinate is continuous, so the coordinate criterion proves that \(h\) is continuous into the product.
At \(t=2\), direct substitution gives \(h_1(2)=2/2=1\), \(h_2(2)=2/3\), and \(h_3(2)=2/4=1/2\). At \(s=0\), each coordinate is \(0\); the differences in these three coordinates are \(1\), \(2/3\), and \(1/2\). The general estimate gives \(|h_n(2)-h_n(0)|=2/(n+1)\leq1\), which these three values satisfy. This stronger estimate is useful here, but the theorem needs only continuity of each coordinate.
Check: the displayed values at \(t=2\) are obtained from \(2/(n+1)\) for \(n=1,2,3\), and substituting \(s=0\) gives \(0/(n+1)=0\) in each case.
One Discontinuous Coordinate Is Enough
The criterion is an if and only if: every coordinate must be continuous, not merely most of them. A failure in even one coordinate prevents continuity of the full map. The projection formulation also explains why: if \(f\) were continuous, composing it with the continuous projection onto that coordinate would give a continuous real-valued function.
Worked Example: A Step in the First Coordinate
Define \(q:\mathbb{R}\to\mathbb{R}^{\mathbb{N}}\) by \(q_1(t)=0\) for \(t\leq0\), \(q_1(t)=1\) for \(t>0\), and \(q_n(t)=0\) for every \(n\geq2\). The first coordinate is discontinuous at \(0\). This also gives a direct open-set test for the map.
Let \(V=\pi_1^{-1}((-1/2,1/2))\), a basic open set in the product. The value \(q_1(t)\) lies in \((-1/2,1/2)\) exactly when \(t\leq0\): the value \(0\) belongs to that interval, while \(1\) does not. Hence \(q^{-1}(V)=(-\infty,0]\), which is not open in \(\mathbb{R}\). Indeed, every open interval around \(0\) contains a positive \(t\), and that \(t\) is not in \((-\infty,0]\). Therefore \(q\) is not continuous.
Check: \(q_1(-1)=0\in(-1/2,1/2)\), \(q_1(0)=0\in(-1/2,1/2)\), and \(q_1(1)=1\notin(-1/2,1/2)\). These substitutions confirm that the preimage includes \(0\) and all negative inputs, but excludes every positive input.
A Common Mistake
Common pitfall. For \(g_n(t)=nt\), one might try to prove continuity by requiring a single positive \(\delta\) such that \(|t|<\delta\) implies \(|nt|<1\) for every \(n\). This demand is too strong for continuity into the product. For example, if the proposed \(\delta\) is \(1/10\), take \(t=1/20\) and \(n=21\). Then \(|t|=1/20<1/10\), but \(|nt|=21/20>1\). The calculation rules out that proposed uniform bound, not continuity of \(g\).
The correction is to use the actual neighborhood being tested. It restricts only a finite set \(F\) of coordinates, so one takes the minimum of the finitely many required \(\delta\)-values. The coordinate criterion is not a claim that all coordinates can be controlled by one estimate at once; it is a statement about continuity of each coordinate function and the finite restrictions in product neighborhoods.
Check Your Understanding
Use the coordinate criterion and the finite-coordinate description of product neighborhoods to answer the following.
- Define \(u:\mathbb{R}\to\mathbb{R}^{\mathbb{N}}\) by \(u_n(t)=t^2+n\). Is \(u\) continuous? State the coordinate function used in your check.
- For \(v_n(t)=t/n\), find a bound for \(|v_n(t)-v_n(s)|\) that is valid for every \(n\in\mathbb{N}\).
- Define \(w_n(t)=0\) for every \(n\) when \(t\leq1\), and \(w_n(t)=1\) for every \(n\) when \(t>1\). Give a basic open set whose preimage shows that \(w\) is not continuous.
- Explain why, in checking continuity at a point for a given basic product neighborhood, only finitely many coordinate estimates need to be combined.
- For the map \(g_n(t)=nt\), show that no positive \(\delta\) guarantees \(|g_n(t)-g_n(0)|<1\) for every \(n\) whenever \(|t|<\delta\), and explain why this does not contradict continuity into the product.