Encoding a Space by Continuous Coordinates
A continuous function from a space into the unit interval records one numerical feature of each point. A family of such functions records many features at once: each point becomes a sequence of numbers. The essential question is whether those coordinates preserve all the topological information of the original space. For a regular second-countable space, the countable separating-family construction supplies functions with enough local control to do this.
Write \(I=[0,1]\), with its usual subspace topology from \(\mathbb{R}\), and let \(Q=I^{\mathbb{N}}\) have the product topology. A basic open set in \(Q\) restricts finitely many coordinates to open subsets of \(I\), leaving all other coordinates unrestricted. The space \(Q\) is the Hilbert cube.
Here is the precise input from the construction of countably many separating functions. Suppose \((X,\mathcal{T})\) is regular and second-countable, and let \((f_n)_{n\in\mathbb{N}}\) be continuous maps \(f_n:X\to I\) with the following local property:
This property says more than that the functions distinguish points: near each point, at least one coordinate can detect a set lying inside any prescribed open neighborhood. The countable separating-family construction provides this local property. We now prove that it is exactly what the coordinate-map argument needs.
The Coordinate Map and the Embedding Theorem
Define \(e:X\to Q\) by collecting all the function values:
Proof. First we prove injectivity. Take distinct points \(x,y\in X\). Regular spaces are \(T_1\), so \(\{y\}\) is closed and \(U=X\setminus\{y\}\) is open. Apply the local coordinate property at \(x\in U\). It gives \(n\) and an open \(V\subseteq I\) with \(f_n(x)\in V\) and \(f_n^{-1}(V)\subseteq U\). Since \(x\in f_n^{-1}(V)\), but \(y\notin U\), we have \(f_n(x)\ne f_n(y)\). Thus the \(n\)th coordinates of \(e(x)\) and \(e(y)\) differ, and \(e(x)\ne e(y)\).
Next we prove continuity. A basis for the product topology on \(Q\) consists of finite intersections of sets of the form \(\pi_n^{-1}(V)\), where \(V\) is open in \(I\) and \(\pi_n:Q\to I\) is the \(n\)th projection. For each such set, \(e^{-1}(\pi_n^{-1}(V))=f_n^{-1}(V)\), which is open because \(f_n\) is continuous. The preimage of a finite intersection is the corresponding finite intersection of these open preimages, so is open. Since every open set in \(Q\) is a union of basis sets, its preimage under \(e\) is open. Hence \(e\) is continuous.
Finally, we show that \(e^{-1}:e(X)\to X\) is continuous. It suffices to prove that for each open \(U\subseteq X\), the set \(e(U)\) is open in the subspace \(e(X)\). For each \(x\in U\), choose \(n_x\) and an open \(V_x\subseteq I\) given by the local coordinate property, and put \(W_x=\pi_{n_x}^{-1}(V_x)\). Each \(W_x\) is open in \(Q\), and \(e(x)\in W_x\). Moreover, if \(e(y)\in W_x\), then \(f_{n_x}(y)\in V_x\), so \(y\in f_{n_x}^{-1}(V_x)\subseteq U\). Consequently, every point of \(e(X)\cap\bigcup_{x\in U}W_x\) belongs to \(e(U)\). Conversely, each point \(e(x)\in e(U)\) lies in \(W_x\). Therefore
The right-hand side is relatively open in \(e(X)\), so \(e^{-1}\) is continuous. This proves that \(e\) is an embedding. If \(X=\varnothing\), injectivity is vacuous and the same openness statement holds with the empty union; thus the conclusion includes the empty space. \(\square\)
Worked Examples
Worked Example: Coordinates on the Real Line
Take \(X=\mathbb{R}\) with its standard topology. For each rational pair \(a<b\), define \(f_{a,b}(x)=\max(0,\min(1,x-a,b-x))\). There are countably many rational pairs, so enumerate these functions as a sequence \((f_n)\), repeating functions if needed.
Each function takes values in \(I\). Indeed, the outer maximum makes the value nonnegative, and the inner minimum with \(1\) makes it at most \(1\). The functions \(x\mapsto x-a\) and \(x\mapsto b-x\) are continuous; the minimum and maximum of two continuous real-valued functions are continuous, so \(f_{a,b}\) is continuous. Directly from the formula, \(f_{a,b}(x)>0\) exactly when \(a<x<b\). The set \((0,1]\) is open in \(I\), and hence \(f_{a,b}^{-1}((0,1])=(a,b)\).
To check the local coordinate property, let \(x\in U\), where \(U\) is open in \(\mathbb{R}\). There is \(\delta>0\) such that \((x-\delta,x+\delta)\subseteq U\). Choose rational numbers \(a,b\) with \(x-\delta<a<x<b<x+\delta\). Then \((a,b)\subseteq U\), and \(f_{a,b}(x)>0\). Taking \(V=(0,1]\), which is open in \(I\), gives \(x\in f_{a,b}^{-1}(V)=(a,b)\subseteq U\).
For a concrete coordinate check, use \(a=0\), \(b=2\), and \(x=1\). The formula gives \(f_{0,2}(1)=\max(0,\min(1,1,1))=1\). At \(y=3\), it gives \(f_{0,2}(3)=\max(0,\min(1,3,-1))=0\). Thus this coordinate distinguishes \(1\) from \(3\). The general theorem proves that the full sequence of rational-interval coordinates embeds \(\mathbb{R}\) in \(Q\).
Check: for \(a=0,b=2\), the positive-coordinate set is \((0,2)\), not \([0,2]\): at \(x=0\), \(f_{0,2}(0)=\max(0,\min(1,0,2))=0\), and at \(x=2\), \(f_{0,2}(2)=\max(0,\min(1,2,0))=0\). Thus the preimage calculation agrees with the open interval used above.
Worked Example: A Finite Discrete Space
Let \(X=\{p,q\}\) with the discrete topology, in which every subset is open. Define \(f_1(p)=1\) and \(f_1(q)=0\), and let \(f_n\) be the constant-zero function for every \(n\ge2\). All these functions are continuous: the preimage of any open subset of \(I\) is a subset of \(X\), hence open.
The coordinate map has values \(e(p)=(1,0,0,\ldots)\) and \(e(q)=(0,0,0,\ldots)\), so it is injective. The local property can be checked for every open set. For \(U=\{p\}\), take \(V=(1/2,1]\); then \(f_1^{-1}(V)=\{p\}\). For \(U=\{q\}\), take \(V=[0,1/2)\); then \(f_1^{-1}(V)=\{q\}\). Both \(V\)'s are open in \(I\). For \(U=X\), take \(V=I\); for \(U=\varnothing\), there are no points at which to check the property.
The image is discrete in its subspace topology: the open coordinate neighborhoods \(\pi_1^{-1}((1/2,1])\) and \(\pi_1^{-1}([0,1/2))\) meet \(e(X)\) in the respective singletons \(\{e(p)\}\) and \(\{e(q)\}\). This directly verifies that the map is an embedding.
Why Injectivity and Continuity Alone Are Not Enough
Worked Example: A Continuous Injection That Is Not an Embedding
Let \(X=\mathbb{N}\) with the discrete topology, and define \(g:X\to Q\) by \(g(1)=(0,0,\ldots)\) and \(g(n+1)=(1/n,0,0,\ldots)\) for \(n\in\mathbb{N}\). These values are distinct, so \(g\) is injective. It is continuous because every subset of \(X\) is open, and therefore the preimage of every open subset of \(Q\) is open.
However, \(\{1\}\) is open in \(X\), while \(g(\{1\})\) is not open in \(g(X)\). To verify the latter claim, consider any open neighborhood of \(g(1)\) in \(Q\). It contains a basic product neighborhood restricting finitely many coordinates. All coordinates after the first are zero for every point of \(g(X)\); the first-coordinate restriction contains an open neighborhood of \(0\) in \(I\), and therefore contains \([0,\eta)\) for some \(\eta>0\). Choose \(n\in\mathbb{N}\) with \(n>1/\eta\). Then \(0<1/n<\eta\), so \(g(n+1)\) lies in that neighborhood as well as \(g(1)\). No neighborhood of \(g(1)\) in the image isolates \(g(1)\). Thus \(g^{-1}:g(X)\to X\) is not continuous, and \(g\) is not an embedding.
This example pinpoints the role of the local coordinate property. Continuity controls preimages of open sets in the target, but it does not guarantee that open sets of the domain appear as relatively open subsets of the image.
A Common Pitfall and the Topological Payoff
Common pitfall. In the real-line construction, it is unnecessary to choose rational \(a<x<b\) with \([a,b]\subseteq U\). For example, if \(U=(0,1)\), \(x=1/2\), \(a=0\), and \(b=1\), then \((a,b)\subseteq U\) but \([a,b]\not\subseteq U\). The coordinate neighborhood is \(f_{a,b}^{-1}((0,1])=(a,b)\), so the open-interval containment is sufficient. The correct selection is to find rational \(a<x<b\) with \((a,b)\subseteq U\); openness of \(U\) and density of \(\mathbb{Q}\) guarantee such a choice.
The embedding result turns a question about an abstract topology into a question about coordinates in a familiar product space. Injectivity says that different points receive different sequences; continuity says that the coordinate assignment respects the topology; and the local property ensures that the image does not lose any open-set information. The theorem therefore realizes every regular second-countable space as a subspace of the Hilbert cube, not merely as a set of distinct points in it.
Check Your Understanding
Use the coordinate map and its local property to organize your reasoning.
- Why does the \(T_1\) property allow the open set \(X\setminus\{y\}\) to be used when proving that \(e(x)\ne e(y)\)?
- Explain why checking preimages of finite-coordinate basic open sets suffices to prove continuity into \(Q\).
- For \(f_{a,b}(x)=\max(0,\min(1,x-a,b-x))\), compute \(f_{1,4}(2)\) and \(f_{1,4}(5)\), and identify its positive-coordinate set.
- In the finite-space example, give an open coordinate neighborhood whose intersection with the image is exactly \(\{e(q)\}\).
- In the non-embedding example, explain why every neighborhood of \(g(1)\) in the image contains some \(g(n+1)\).