Why Bounded Functions Extend
The universal property of the Stone–Čech compactification extends maps whose target is compact Hausdorff. The real line is Hausdorff but not compact, so that property does not directly apply to an arbitrary map into \(\mathbb{R}\). Boundedness supplies the missing condition: the range of a bounded real-valued function lies in a compact interval. Applying the universal property to that interval gives the extension.
Let \(X\) be a completely regular Hausdorff space, and write \(j:X\to\beta X\) for its Stone–Čech compactification. Thus \(\beta X\) is compact Hausdorff, \(j\) is an embedding, and \(j(X)\) is dense in \(\beta X\). We use the universal property in the following form: if \(K\) is compact Hausdorff and \(g:X\to K\) is continuous, there is a unique continuous map \(G:\beta X\to K\) such that \(G\circ j=g\).
Proof. If \(X\) is empty, both \(\beta X\) and the domain of \(f\) are empty, and there is exactly one map from \(\beta X\) to \(\mathbb{R}\). Now suppose \(X\) is nonempty. Boundedness means that some \(M\geq 0\) satisfies \(|f(x)|\leq M\) for every \(x\in X\). Consequently \(f\) takes values in the compact Hausdorff interval \([-M,M]\), and is continuous as a map to that interval. The universal property gives a unique continuous \(F:\beta X\to[-M,M]\) with \(F\circ j=f\). Composing \(F\) with the inclusion \([-M,M]\subseteq\mathbb{R}\) gives the required extension \(\overline{f}\).
For uniqueness as a map into \(\mathbb{R}\), suppose \(G,H:\beta X\to\mathbb{R}\) are continuous and agree on \(j(X)\). The set \(E=\{p\in\beta X:G(p)=H(p)\}\) is closed: if \(G(p)\ne H(p)\), disjoint open intervals around their values have inverse images whose intersection is a neighborhood of \(p\) on which the maps still differ. Since \(E\) contains the dense set \(j(X)\), it equals \(\beta X\). Thus \(G=H\). \(\square\)
The proof also covers the degenerate bound \(M=0\): then \(f\) is identically zero, its target interval is the compact singleton \(\{0\}\), and its extension is identically zero.
The Exact Range of the Extension
The extension does more than exist: its image is exactly the closure in \(\mathbb{R}\) of the original range. This describes which real values occur on the compactification, without claiming that the values at particular new points can usually be calculated directly.
Proof. The set \(\overline{f}(\beta X)\) is compact because it is the continuous image of the compact space \(\beta X\). A compact subset of a Hausdorff space is closed, so this image is closed in \(\mathbb{R}\). To recall why compact subsets are closed here, let \(C\) be compact in a Hausdorff space and let \(y\notin C\). For each \(c\in C\), choose disjoint open neighborhoods \(U_c\) of \(c\) and \(V_c\) of \(y\). Finitely many of the \(U_c\) cover \(C\). The intersection of the corresponding finitely many \(V_c\) is an open neighborhood of \(y\) disjoint from \(C\). Hence the complement of \(C\) is open. Since \(\overline{f}(\beta X)\) is closed and contains \(f(X)=\overline{f}(j(X))\), it contains \(\overline{f(X)}\).
For the reverse inclusion, let \(y=\overline{f}(p)\) for some \(p\in\beta X\). Every open neighborhood \(O\) of \(y\) has an inverse image \(\overline{f}^{-1}(O)\) that is open and contains \(p\). Density of \(j(X)\) implies that this inverse image meets \(j(X)\). For a point \(j(x)\) in that intersection, \(f(x)=\overline{f}(j(x))\in O\). Thus every open neighborhood of \(y\) meets \(f(X)\), which means \(y\in\overline{f(X)}\). Both inclusions prove the equality. \(\square\)
Proof. By the proposition, the range of \(\overline{f}\) is the closure of the range of \(f\). Taking the closure of a bounded subset of \(\mathbb{R}\) does not change its supremum of absolute values: the closed range is contained in \([-S,S]\), where \(S=\sup_{x\in X}|f(x)|\), and it contains \(f(X)\). Therefore its supremum of absolute values is exactly \(S\). \(\square\)
Worked Extensions
Worked Example: A Function on the Real Line with Both Endpoint Values
Take \(X=\mathbb{R}\) with its standard topology and \(f(x)=x/(1+|x|)\). The real line is completely regular Hausdorff, and \(f\) is continuous because the denominator is positive and absolute value is continuous. We determine the range before applying the theorem.
If \(x\geq0\), then \(f(x)=x/(1+x)\), so \(0\leq f(x)<1\). If \(x<0\), then \(f(x)=x/(1-x)\), so \(-1<f(x)<0\). Conversely, for any \(y\in(-1,1)\), set \(x=y/(1-|y|)\). The denominator is positive, and \(x\) has the same sign as \(y\). Hence \(|x|=|y|/(1-|y|)\), and substitution gives \(f(x)=\bigl(y/(1-|y|)\bigr)\big/\bigl(1+|y|/(1-|y|)\bigr)=y\). Thus \(f(\mathbb{R})=(-1,1)\).
The theorem gives a unique extension \(\overline{f}:\beta\mathbb{R}\to\mathbb{R}\). The sequences \(x_n=n\) and \(x_n=-n\), for \(n\in\mathbb{N}\), give \(f(n)=n/(1+n)\to1\) and \(f(-n)=-n/(1+n)\to-1\). Therefore \(\overline{f(\mathbb{R})}=[-1,1]\), and the exact-range proposition yields \(\overline{f}(\beta\mathbb{R})=[-1,1]\). In particular, the extension attains both endpoint values, although neither is attained by \(f\) on \(\mathbb{R}\).
Check: the range calculation gives \(|f(x)|<1\) for every real \(x\), while the two displayed sequences approach the two endpoints. Its range closure is therefore precisely \([-1,1]\), as required by the proposition.
Worked Example: Extending the Sine Function
Let \(X=\mathbb{R}\) and \(f(x)=\sin x\). This function is continuous and satisfies \(-1\leq\sin x\leq1\), so it has a unique continuous extension \(\overline{f}:\beta\mathbb{R}\to\mathbb{R}\).
For every \(n\in\mathbb{Z}\), \(\sin(\pi/2+2\pi n)=1\) and \(\sin(3\pi/2+2\pi n)=-1\). Together with the bound \(-1\leq\sin x\leq1\), these values show that \(f(\mathbb{R})=[-1,1]\). Its range is already closed. The exact-range proposition now gives \(\overline{f}(\beta\mathbb{R})=[-1,1]\).
Check: the extension’s image cannot contain values outside \([-1,1]\), because its image equals the closure of \(f(\mathbb{R})\); it contains every value in \([-1,1]\), because the original function already takes all those values.
Worked Example: A Bounded Function on a Noncompact Interval
Give \(X=(0,1)\) the subspace topology inherited from \(\mathbb{R}\), and set \(f(x)=x(1-x)\). The interval is completely regular Hausdorff, and \(f\) is continuous. The identity \(f(x)=\frac14-(x-\frac12)^2\) shows that \(0<f(x)\leq\frac14\). The value \(1/4\) occurs at \(x=1/2\). Every \(y\) with \(0<y\leq1/4\) occurs: the equation \(x(1-x)=y\) has roots \(x=(1\pm\sqrt{1-4y})/2\), which lie in \((0,1)\). The value \(0\) does not occur. Thus \(f(X)=(0,1/4]\), and its closure is \([0,1/4]\).
There is a unique continuous extension \(\overline{f}:\beta X\to\mathbb{R}\), and the exact-range proposition gives \(\overline{f}(\beta X)=[0,1/4]\). In particular, some point of \(\beta X\) is sent to \(0\). Such a point is not in \(j(X)\), since \(f(x)>0\) for every \(x\in(0,1)\). The remainder \(\beta X\setminus j(X)\) is nonempty: otherwise \(j(X)=\beta X\), so \(X\) would be compact. But the open cover \(\{(1/n,1):n\geq2\}\) of \((0,1)\) has no finite subcover; a finite subcollection has a largest index \(N\) and misses every \(x\in(0,1/N]\).
Check: the range contains \(1/4\) and every value strictly between \(0\) and \(1/4\), but not \(0\). Its closure is therefore \([0,1/4]\), consistent with the extension’s image.
Algebra and a Common Pitfall
Extensions respect basic pointwise operations. For bounded continuous \(f,g:X\to\mathbb{R}\), let \(\overline{f}\) and \(\overline{g}\) be their extensions. Then \(\overline{f+g}=\overline{f}+\overline{g}\) and \(\overline{fg}=\overline{f}\,\overline{g}\). Indeed, \(f+g\) and \(fg\) are bounded and continuous, so their extensions exist. The functions \(\overline{f}+\overline{g}\) and \(\overline{f}\,\overline{g}\) are continuous on \(\beta X\), and on \(j(X)\) they restrict respectively to \(f+g\) and \(fg\). Uniqueness of the extension makes each pair equal. For any real scalar \(c\), the same argument gives \(\overline{cf}=c\overline{f}\).
Common pitfall. Boundedness cannot be dropped when the desired target is \(\mathbb{R}\). For \(X=(0,1)\), the continuous function \(f(x)=1/x\) is unbounded: at \(x=1/n\), for each integer \(n\geq2\), its value is \(n\). If a continuous extension \(F:\beta X\to\mathbb{R}\) existed, the compactness of \(\beta X\) would make \(F(\beta X)\) compact and hence bounded in \(\mathbb{R}\). But \(F(j(x))=1/x\) for all \(x\in(0,1)\), contradicting the unbounded values \(n\). The universal property does not directly apply to this map because its range is not contained in any compact interval.
Check Your Understanding
For each question, identify the role played by boundedness, compactness, or density.
- Why does a bounded continuous \(f:X\to\mathbb{R}\) define a continuous map into some compact interval \([-M,M]\)?
- If \(f(X)=[2,5)\), what is the image of its Stone–Čech extension?
- Explain why two continuous real-valued maps on \(\beta X\) that agree on \(j(X)\) must agree everywhere.
- For \(f(x)=x(1-x)\) on \((0,1)\), which value in the extension’s image is not attained on \(j(X)\)?
- Why can the unbounded function \(f(x)=1/x\) on \((0,1)\) have no continuous real-valued extension to \(\beta(0,1)\)?