Why Bounded Functions Matter
A continuous real-valued function can record information about points of a space: if two points receive different values, the function distinguishes them. The central question here is whether there are enough such functions to distinguish points and closed sets, and whether requiring the functions to be bounded loses any separating power. Complete regularity gives a precise answer. It is also the reason bounded functions are central to the Stone–Čech compactification.
Throughout, a continuous function means a continuous map between topological spaces. A real-valued function \(f:X\to\mathbb{R}\) is bounded if there is a real number \(M\ge 0\) such that \(|f(x)|\le M\) for every \(x\in X\). We use the following convention for complete regularity.
The definition makes one particular separation requirement: a single point is separated from a closed set not containing it. It does not, by itself, state that any two disjoint closed sets can be separated by one function. Keeping these requirements distinct is important.
The Separation Theorem
Proof. If \(C\) is closed and \(x\notin C\), complete regularity gives a continuous \(f:X\to[0,1]\) with \(f(x)=0\) and \(f(c)=1\) for every \(c\in C\). Since \(0\le f(z)\le1\) for every \(z\in X\), this function is bounded and has the required values.
Now take distinct points \(x,y\in X\). Because \(X\) is Hausdorff, \(\{y\}\) is closed: for each \(z\ne y\), there are disjoint open neighborhoods of \(z\) and \(y\), so the neighborhood of \(z\) misses \(y\); the union of these neighborhoods over \(z\ne y\) is \(X\setminus\{y\}\), which is open. Apply the first part with \(C=\{y\}\). The resulting function satisfies \(f(x)=0\) and \(f(y)=1\), so it distinguishes the points. \(\square\)
The second conclusion means that the family of bounded continuous real-valued functions distinguishes every pair of points. The function can depend on the pair. The conclusion does not say that one function must distinguish every pair in \(X\).
Bounding a Function Without Losing Its Values
There is also a general way to make any continuous real-valued function bounded while preserving whether two of its values are different. Define \(\phi:\mathbb{R}\to(-1,1)\) by \(\phi(t)=t/(1+|t|)\). The following proposition explains why boundedness is not an obstacle when using real-valued functions to distinguish points.
Proof. For \(t\ge0\), \(\phi(t)=t/(1+t)\), and for \(t<0\), \(\phi(t)=t/(1-t)\). These formulas agree at \(t=0\), so \(\phi\) is continuous on \(\mathbb{R}\). For every \(t\), \(|\phi(t)|=|t|/(1+|t|)<1\), so \(\phi\) is bounded. Also, for \(u\in(-1,1)\), solving \(u=t/(1+|t|)\) gives \(t=u/(1-|u|)\). Thus \(\phi\) is one-to-one, and it preserves whether two values differ. The composition \(\phi\circ g\) is continuous because both \(g\) and \(\phi\) are continuous. It is bounded by \(1\), so it has all the stated properties. \(\square\)
Worked Example: Separating a Point from an Interval
Let \(X=\mathbb{R}\) with its standard topology, let \(C=[0,1]\), and take \(x=2\). Define \(f(t)=\min(1,|t-2|)\). Absolute value and the minimum of two continuous real-valued functions are continuous, so \(f\) is continuous. Its values lie in \([0,1]\), so it is bounded.
At the point \(x=2\), \(f(2)=\min(1,|2-2|)=\min(1,0)=0\). For any \(c\in[0,1]\), \(2-c\ge1\), so \(|c-2|=2-c\ge1\) and \(f(c)=\min(1,|c-2|)=1\). Thus \(f\) separates \(2\) from the whole closed interval.
Check: \(0\in C\) gives \(f(0)=\min(1,|0-2|)=\min(1,2)=1\), while \(f(2)=0\). Also, for \(t=3/2\), \(f(3/2)=\min(1,1/2)=1/2\), consistent with the definition and the bound \(0\le f(t)\le1\).
Worked Example: A Discrete Space
Let \(X=\{a,b,c\}\) have the discrete topology, in which every subset is open. Take \(x=a\) and the closed set \(C=\{b,c\}\). Define \(f(a)=0\) and \(f(b)=f(c)=1\). For any subset \(V\subseteq\mathbb{R}\), its preimage \(f^{-1}(V)\) is a subset of \(X\), hence open. This verifies continuity from the definition using preimages of open sets.
The values of \(f\) are \(0\) and \(1\), so \(f\) is bounded. It satisfies \(f(a)=0\) and \(f(z)=1\) for every \(z\in C\). For example, \(f^{-1}(\{0\})=\{a\}\) and \(f^{-1}(\{1\})=\{b,c\}\); both are open in \(X\).
Check: The set \(C\) is closed because its complement \(\{a\}\) is open. The function has the required value \(1\) at both points of \(C\), not merely at one chosen point of that set.
A Metric-Space Construction
In a metric space, one can build a bounded separator directly from distances. If \(F\) is a nonempty closed subset of a metric space \((X,d)\), define \(d(z,F)=\inf\{d(z,w):w\in F\}\). For any \(z,z'\in X\) and \(w\in F\), the triangle inequality gives \(d(z,w)\le d(z,z')+d(z',w)\). Taking the infimum over \(w\) yields \(d(z,F)\le d(z,z')+d(z',F)\). Interchanging \(z\) and \(z'\) gives \(|d(z,F)-d(z',F)|\le d(z,z')\), so \(z\mapsto d(z,F)\) is continuous.
If \(x\notin F\), closedness makes \(X\setminus F\) open. Hence some ball \(B_d(x,r)\), with \(r>0\), lies in \(X\setminus F\). Every \(w\in F\) then satisfies \(d(x,w)\ge r\), so \(d(x,F)\ge r>0\). Consequently the denominator in the following expression never vanishes:
This function is continuous, has values in \([0,1]\), equals \(0\) at \(x\), and equals \(1\) at every point of \(F\). The continuity follows from continuity of the two distance functions and positivity of the denominator; the value bounds follow because both terms in the denominator are nonnegative.
Worked Example: A Closed Half-Plane
Use the Euclidean metric on \(X=\mathbb{R}^2\), let \(x=(0,0)\), and let \(F=\{(u,v):u\ge1\}\). This \(F\) is closed because its complement \(\{(u,v):u<1\}\) is open. The distance from \(z=(u,v)\) to \(x\) is \(\sqrt{u^2+v^2}\), and the distance from \(z\) to \(F\) is \(\max(1-u,0)\): if \(u\ge1\), \(z\in F\); if \(u<1\), the closest point of \(F\) has coordinates \((1,v)\).
Thus a separating function is \(f(u,v)=\sqrt{u^2+v^2}/\bigl(\sqrt{u^2+v^2}+\max(1-u,0)\bigr)\). At \((0,0)\), the numerator is \(0\), while the second term in the denominator is \(1\), so \(f(0,0)=0\). If \((u,v)\in F\), then \(\max(1-u,0)=0\); also \((u,v)\ne(0,0)\), so the numerator is positive and \(f(u,v)=1\).
Check: At \((0,1)\), the numerator is \(\sqrt{0^2+1^2}=1\), and the second denominator term is \(\max(1-0,0)=1\). Therefore \(f(0,1)=1/(1+1)=1/2\), which lies in \([0,1]\). At \((2,0)\in F\), the numerator is \(2\) and the second term is \(\max(1-2,0)=0\), so \(f(2,0)=2/(2+0)=1\).
From Separating Functions to Stone–Čech
A compactification of \(X\) is a compact Hausdorff space \(K\) together with an embedding \(e:X\to K\) whose image is dense in \(K\). A continuous real-valued function on a compact space is bounded: its image is compact in \(\mathbb{R}\), and compact subsets of \(\mathbb{R}\) are bounded. Therefore any continuous function on \(K\), when restricted along \(e\), gives a bounded continuous real-valued function on \(X\).
The Stone–Čech compactification of a completely regular Hausdorff space is the compactification designed so that every bounded continuous real-valued function on \(X\) extends continuously over it. The separation theorem explains why this is a rich supply of coordinates: these functions already distinguish every pair of points of \(X\), and distinguish each point from any closed set that omits it. Boundedness makes their ranges compatible with compact target intervals.
Such extensions, when they exist, are unique. Indeed, suppose \(g,h:K\to\mathbb{R}\) are continuous and agree on \(e(X)\). The function \(g-h\) is continuous, so its zero set is closed because \(\{0\}\) is closed in \(\mathbb{R}\). This zero set contains the dense set \(e(X)\), and a closed set containing a dense subset must be all of \(K\). Hence \(g=h\) everywhere. This proves uniqueness, but it does not construct the compactification or establish the extension property; those are separate parts of the Stone–Čech theory.
A Common Pitfall
A common error is to use \(g(t)=t\) on \(\mathbb{R}\) as an example of a bounded continuous function that separates points. It is continuous and distinguishes any two different real numbers, but it is not bounded: for every proposed bound \(M\), choosing \(t=M+1\) gives \(|g(t)|=M+1>M\). The bounded function \(\phi(t)=t/(1+|t|)\) corrects the problem. For example, \(\phi(2)=2/(1+2)=2/3\), while \(\phi(3)=3/(1+3)=3/4\); the values differ, and for every \(t\), \(|\phi(t)|=|t|/(1+|t|)<1\).
The opposite mistake is to read “functions distinguish points” as saying that one function must be injective on the entire space. The separation theorem makes a pairwise assertion: for each distinct pair, there is a suitable function, which may vary with the pair. It also separates points from closed sets, not automatically arbitrary pairs of disjoint closed sets. The exact separation condition matters whenever function values are used to describe a topology or a compactification.
Check Your Understanding
For each question, identify the separation condition being used and check the function values or hypotheses carefully.
- State the point-and-closed-set condition in the definition of complete regularity. What values must the separating function take?
- Why does the Hausdorff hypothesis imply that \(\{y\}\) is closed, and how does this give a bounded function separating distinct points \(x\) and \(y\)?
- For \(g:X\to\mathbb{R}\) continuous, explain why \(g/(1+|g|)\) is bounded and why it preserves whether two values of \(g\) are different.
- In the Euclidean metric, use the distance formula to separate \(x=(0,0)\) from the closed set \(F=\{(u,v):u\ge2\}\).
- Why must two continuous extensions to a compactification that agree on the embedded dense copy of \(X\) agree everywhere?