From Bounded Functions to a Compact Product
A compactification can be built by placing a space inside a compact Hausdorff space and then taking the closure of its image. The challenge is to choose an ambient compact space that retains enough information to distinguish points and recover the original topology. Bounded continuous real-valued functions provide the coordinates: together, they record how every such function behaves at each point.
Let \((X,\mathcal{T})\) be a completely regular Hausdorff space. Here, completely regular means that whenever \(F\subseteq X\) is closed and \(x\notin F\), there is a continuous function \(u:X\to[0,1]\) such that \(u(x)=0\) and \(u(y)=1\) for every \(y\in F\). Write \(C_b(X,\mathbb{R})\) for the set of all bounded continuous functions \(f:X\to\mathbb{R}\).
Each coordinate interval is closed and bounded in \(\mathbb{R}\), hence compact by the Heine–Borel theorem; it is also Hausdorff. The definition of \(M_f\) ensures that \(f(x)\in J_f\) for every \(x\), so the map really does take values in the stated product. Define \(\beta X=\overline{e(X)}\), where the closure is taken in \(P\). We will prove that \(e\) embeds \(X\) into \(P\), and that \(\beta X\) is a compact Hausdorff space in which \(e(X)\) is dense.
The Evaluation Map Is an Embedding
Proof. First suppose \(x,y\in X\) and \(x\ne y\). Since \(X\) is Hausdorff, it is \(T_1\), so \(\{y\}\) is closed. Complete regularity gives a continuous \(u:X\to[0,1]\) with \(u(x)=0\) and \(u(y)=1\). This \(u\) is bounded and is one of the coordinates in the product. Thus the \(u\)-coordinates of \(e(x)\) and \(e(y)\) differ, so \(e(x)\ne e(y)\). Hence \(e\) is injective.
For continuity, a basic open set in the product restricts finitely many coordinates to open subsets of their intervals. Its preimage under \(e\) is the finite intersection of the corresponding preimages under those coordinate functions \(f\). Each such preimage is open because \(f\) is continuous. Basic open sets generate the product topology, so \(e\) is continuous.
It remains to show that the inverse on \(e(X)\) is continuous. It is enough to prove that for every open \(U\subseteq X\), the set \(e(U)\) is open in \(e(X)\). Fix \(x\in U\). The set \(F=X\setminus U\) is closed, and \(x\notin F\). Choose, by complete regularity, a continuous \(u:X\to[0,1]\) with \(u(x)=0\) and \(u(y)=1\) for all \(y\in F\). Since \(u\) is bounded, it is a coordinate in \(P\), and its coordinate interval is \(J_u=[-1,1]\). The set of product points whose \(u\)-coordinate is less than \(1/2\) is open in \(P\). It contains \(e(x)\), and its intersection with \(e(X)\) is contained in \(e(U)\): if \(y\notin U\), then \(u(y)=1\), so \(e(y)\) is not in that intersection. Taking these neighborhoods for all \(x\in U\) proves that \(e(U)\) is open in \(e(X)\). If \(U=\varnothing\), its image is open as well. Thus the inverse is continuous. \(\square\)
This argument identifies the separate roles of the hypotheses. Hausdorffness makes distinct points separable by a continuous coordinate, while complete regularity supplies coordinates that detect the topology near a point and a closed set. Merely having many continuous functions does not guarantee either conclusion.
Worked Example: The Two-Point Discrete Space
Let \(X=\{a,b\}\) with the discrete topology, in which every subset is open. Define \(f:X\to\mathbb{R}\) by \(f(a)=0\) and \(f(b)=1\). This function is continuous because every subset of \(X\) is open, and it is bounded with \(M_f=\max(1,1)=1\). Its coordinate interval is \(J_f=[-1,1]\).
The \(f\)-coordinates of the evaluation points are \(0\) and \(1\), so \(e(a)\ne e(b)\). In fact, the product-open condition that the \(f\)-coordinate be less than \(1/2\) contains \(e(a)\) and not \(e(b)\); the condition that it be greater than \(1/2\) contains \(e(b)\) and not \(e(a)\). These conditions show directly that the subspace \(e(X)\) has the discrete topology. Since \(e(X)\) is a finite subset of a Hausdorff space, it is closed. Therefore \(\overline{e(X)}=e(X)\), and this construction adds no new points for this compact space.
Check: the two listed coordinate values are \(f(a)=0\in[-1,1]\) and \(f(b)=1\in[-1,1]\), as required by the definition of the product.
Worked Example: A Local Coordinate on the Real Line
Give \(X=\mathbb{R}\) its standard topology. The function \(g(x)=x/(1+|x|)\) is continuous and bounded. For \(x\ge0\), \(g(x)=x/(1+x)\), which lies in \([0,1)\); for \(x<0\), \(g(x)=x/(1-x)\), which lies in \((-1,0)\). Thus \(|g(x)|<1\) for every \(x\), and \(\sup_{x\in\mathbb{R}}|g(x)|=1\), so \(M_g=1\) and \(J_g=[-1,1]\).
This one coordinate distinguishes points. If \(x,y\ge0\) and \(g(x)=g(y)\), then \(x/(1+x)=y/(1+y)\), so \(x(1+y)=y(1+x)\), giving \(x=y\). If \(x,y<0\), equality gives \(x(1-y)=y(1-x)\), again giving \(x=y\). Values on opposite sides of zero have opposite signs, and only zero maps to zero. For instance, \(g(0)=0\), \(g(2)=2/3\), and \(g(-2)=-2/3\).
To see how a coordinate detects a neighborhood, let \(U=(x-r,x+r)\) with \(r>0\). Define \(h(y)=\min(1,|y-x|/r)\). This is continuous, takes values in \([0,1]\), and therefore is a bounded coordinate. It satisfies \(h(x)=0\). If \(y\notin U\), then \(|y-x|\ge r\), so \(h(y)=1\). Consequently, the product-open condition that the \(h\)-coordinate be less than \(1/2\) contains \(e(x)\), and its preimage is \((x-r/2,x+r/2)\), which is contained in \(U\).
Check: at \(y=x+r/2\), the formula gives \(h(y)=\min(1,(r/2)/r)=1/2\). Since the coordinate condition is strict, this endpoint is excluded, in agreement with the preimage \((x-r/2,x+r/2)\).
Taking the Closure Produces a Compactification
Proof. Every factor \(J_f\) is a nonempty compact Hausdorff space. By Tychonoff’s theorem, their product \(P\), with the product topology, is compact; this theorem is used in its standard form, with the axiom of choice. The product is Hausdorff: if two product points differ, they differ in some coordinate \(f\), and disjoint neighborhoods of their distinct coordinate values in the Hausdorff space \(J_f\) pull back under the projection to disjoint product neighborhoods. The closure \(\beta X\) is closed in \(P\), so it is compact as a closed subset of a compact space, and it is Hausdorff as a subspace of a Hausdorff space. By the first theorem, \(e\) is an embedding into \(P\); since \(e(X)\subseteq\beta X\), it is also an embedding into \(\beta X\). Finally, \(\beta X=\overline{e(X)}\) means precisely that \(e(X)\) is dense in \(\beta X\). These are the required properties of a compactification. \(\square\)
When \(X\) is empty, the evaluation image and its closure are both empty, and the same conclusion holds under the definition of compactification as a dense embedding into a compact Hausdorff space. For nonempty \(X\), the construction also explains why the product has many coordinates: a single function usually cannot preserve all the topological information of \(X\).
Worked Example: Why the Hausdorff Hypothesis Matters
Let \(X=\{a,b\}\) have the topology \(\mathcal{T}=\{\varnothing,\{a\},X\}\). This space is not Hausdorff: every open neighborhood of \(b\) is \(X\), so no neighborhood of \(b\) can be disjoint from a neighborhood of \(a\).
Every continuous real-valued function on \(X\) is constant. Indeed, suppose \(f(a)\ne f(b)\). Choose an open interval in \(\mathbb{R}\) containing \(f(b)\) but not \(f(a)\). Its preimage under \(f\) is then exactly \(\{b\}\), which is not open in \(X\), contradicting continuity. Hence all coordinates in the evaluation map have the same value at \(a\) and \(b\), and \(e(a)=e(b)\). The evaluation map is not injective, so it is not an embedding.
Check: the obstruction is not a failure to choose large enough intervals: each constant coordinate has a valid compact interval \(J_f\). The failure is that continuous coordinates cannot distinguish the two points.
A Common Pitfall: The Coordinates Must Be Bounded
A tempting but invalid construction on \(\mathbb{R}\) is to use the identity function \(f(x)=x\) as a coordinate in \([-1,1]\). This is not a coordinate of the stated product: \(f\) is unbounded, and, for example, \(f(2)=2\notin[-1,1]\). The correction is to use only bounded continuous functions and to assign each its own interval \(J_f\). For the bounded function \(g(x)=x/(1+|x|)\), the values at \(2\) and \(-2\) are \(2/3\) and \(-2/3\), both in its correctly chosen interval \([-1,1]\).
Another distinction is important: the theorem makes \(\beta X\) compact, not necessarily the embedded image \(e(X)\). Compactness belongs to the closure in the product. The construction here establishes the compactification as a dense embedding; properties describing how maps extend over that compactification require additional arguments.
Check Your Understanding
Use the definitions and the proofs above to answer the following questions.
- For \(f\in C_b(X,\mathbb{R})\), why is \(f(x)\in J_f\) for every \(x\in X\)?
- Which hypothesis is used to obtain a bounded continuous function that separates two distinct points?
- For an open set \(U\) containing \(x\), explain why the coordinate condition \(u<1/2\) can be chosen to have preimage contained in \(U\).
- Why does Tychonoff’s theorem apply to \(P\), and what role does the axiom of choice have in the stated theorem?
- In the two-point topology \(\{\varnothing,\{a\},X\}\), why can no continuous real-valued function take different values at \(a\) and \(b\)?