The One-Point Compactification Of Rn Is The Sphere

Adding One Point to Euclidean Space

Euclidean space has no point at infinity, but its one-point compactification adds a single point whose neighborhoods contain all points outside some compact set. Stereographic projection gives a concrete way to see what this construction looks like: the added point becomes the north pole of a sphere.

Fix an integer \(n\geq 1\). Give \(\mathbb{R}^n\) its standard Euclidean topology and metric. Write \(\alpha\mathbb{R}^n=\mathbb{R}^n\cup\{\infty\}\) for its one-point compactification. We use the topology in which the open sets not containing \(\infty\) are the usual open subsets of \(\mathbb{R}^n\), and a set containing \(\infty\) is open exactly when it has the form \(\{\infty\}\cup(\mathbb{R}^n\setminus K)\) for some compact \(K\subseteq\mathbb{R}^n\). In particular, an open neighborhood of \(\infty\) contains every point outside a compact set.

Let \(\mathbb{R}^{n+1}=\mathbb{R}^n\times\mathbb{R}\), with its Euclidean topology, and write its points as \((u,t)\). The unit sphere and its north pole are

Here \(S^n\) has the subspace topology inherited from \(\mathbb{R}^{n+1}\). We use the Heine–Borel theorem: a subset of Euclidean space is compact if and only if it is closed and bounded.

The Formula and Its Inverse

For a finite point \(x\), put \(r^2=\|x\|^2\). The squared norm of the first coordinate of \(\Phi(x)\) is \(4r^2/(r^2+1)^2\), while the square of its last coordinate is \((r^2-1)^2/(r^2+1)^2\). Their sum is \[ \frac{4r^2+(r^2-1)^2}{(r^2+1)^2} =\frac{r^4+2r^2+1}{(r^2+1)^2}=1. \] Thus \(\Phi(x)\in S^n\). Its first coordinate is nonzero whenever \(x\ne 0\); when \(x=0\), its last coordinate is \(-1\). Hence no finite \(x\) maps to \(N\).

For any \((u,t)\in S^n\setminus\{N\}\), we have \(t\ne 1\), so the following expression is defined:

To check this inverse, the sphere equation gives \(\|u\|^2=1-t^2\), and therefore \[ \left\|\frac{u}{1-t}\right\|^2 =\frac{1-t^2}{(1-t)^2} =\frac{1+t}{1-t}. \] For \(x=u/(1-t)\), it follows that \(\|x\|^2+1=2/(1-t)\). Substitution into the formula for \(\Phi\) gives first coordinate \(u\) and last coordinate \(t\). This proves that every point other than \(N\) is hit, and that the displayed formula is a two-sided inverse away from \(N\).

Continuity, Including at Infinity

At each finite point, every coordinate of \(\Phi(x)\) is a quotient of continuous functions, and the denominator \(\|x\|^2+1\) is positive. Thus \(\Phi\) is continuous there. At \(\infty\), we must check the topology’s compact-complement neighborhoods rather than rely only on the formula at finite points.

Proof. Set \(r^2=\|x\|^2\). The first \(n\) coordinates of \(\Phi(x)-N\) have squared norm \(4r^2/(r^2+1)^2\). The last coordinate of this difference is \((r^2-1)/(r^2+1)-1=-2/(r^2+1)\), whose square is \(4/(r^2+1)^2\). Adding gives \[ \|\Phi(x)-N\|^2 =\frac{4r^2+4}{(r^2+1)^2} =\frac{4}{r^2+1}. \] \(\square\)

Proof of the theorem. We have already shown that \(\Phi\) is a bijection. It remains to prove that it and its inverse are continuous. Continuity of \(\Phi\) at finite points follows from the coordinate formulas. For continuity at \(\infty\), let \(O\) be an open neighborhood of \(N\) in \(S^n\). Since \(S^n\) has the subspace topology, there is a \(\delta>0\) such that every point of \(S^n\) at distance less than \(\delta\) from \(N\) belongs to \(O\). If \(\|x\|>2/\delta\), the lemma gives \[ \|\Phi(x)-N\|=\frac{2}{\sqrt{\|x\|^2+1}} <\frac{2}{\|x\|}<\delta. \] The closed ball \(K=\{x:\|x\|\leq 2/\delta\}\) is compact by Heine–Borel. Therefore \(\{\infty\}\cup(\mathbb{R}^n\setminus K)\) is an open neighborhood of \(\infty\), and \(\Phi\) maps it into \(O\). This proves continuity at \(\infty\).

For continuity of the inverse at a point other than \(N\), use the formula \(u/(1-t)\): it is continuous where \(1-t\ne 0\), and the topology on \(\alpha\mathbb{R}^n\) restricted to \(\mathbb{R}^n\) is the Euclidean topology. It remains to check continuity at \(N\). Let \(W\) be an open neighborhood of \(\infty\) in \(\alpha\mathbb{R}^n\). By definition, \(W=\{\infty\}\cup(\mathbb{R}^n\setminus K)\) for a compact \(K\subseteq\mathbb{R}^n\). If \(K\) is empty, \(W=\alpha\mathbb{R}^n\) and there is nothing to check. Otherwise \(K\) is bounded: the open balls \(B(0,m)\), for \(m\in\mathbb{N}\), cover \(K\); compactness gives a finite subcover, so some \(M\geq 0\) satisfies \(\|x\|\leq M\) for every \(x\in K\).

For \(z=(u,t)\in S^n\setminus\{N\}\), write \(x=u/(1-t)\) and \(d=\|z-N\|\). The sphere equation gives \(d^2=2(1-t)\), and the inverse formula gives \[ \|x\|^2=\frac{1+t}{1-t}=\frac{4}{d^2}-1. \] Consequently, if \(d<2/\sqrt{M^2+1}\), then \(\|x\|^2>M^2\), so \(x\notin K\). The open sphere neighborhood of \(N\) defined by this distance bound maps under \(\Phi^{-1}\) into \(W\), and \(N\) itself maps to \(\infty\in W\). This proves continuity of the inverse at \(N\). Thus \(\Phi\) is a homeomorphism. \(\square\)

What Spherical Neighborhoods Mean at Infinity

This calculation explains the topology at the added point: smaller spherical caps around \(N\) correspond to exteriors of larger compact balls in \(\mathbb{R}^n\). In particular, the formula does not create a separate point for each direction of escape. Every direction is represented by points on the sphere approaching the same north pole.

Common pitfall. On the line, one might incorrectly treat \(+\infty\) and \(-\infty\) as two distinct added points. A related computational error is to replace \(x^2\) by \(x|x|\) when \(x\) is negative. At \(x=-2\), that incorrect replacement would put \(-4\) in place of \(x^2=4\), giving the wrong last coordinate \((-4-1)/(-4+1)=5/3\). The actual formula gives \(((-2)^2-1)/((-2)^2+1)=3/5\). More generally, the lemma gives distance \(2/\sqrt{x^2+1}\) from the image to \(N\), which tends to \(0\) along both tails. The one-point compactification therefore identifies both ways of tending to infinity with the same point.

Proof. The inverse formula shows that every point of \(S^n\setminus\{N\}\) is the image of a finite point, while the formula assigns \(\infty\) to \(N\). Thus the image of \(\mathbb{R}^n\) is exactly \(S^n\setminus\{N\}\). Every point of that set belongs to the image, and every open neighborhood of \(N\) contains images of finite points: for example, take \(x=me_1\), where \(e_1=(1,0,\ldots,0)\), and let \(m\in\mathbb{N}\) grow. The lemma gives distance \(2/\sqrt{m^2+1}\), which tends to \(0\). Hence the image is dense. The sphere is closed and bounded in Euclidean space, so Heine–Borel makes it compact; it is Hausdorff as a subspace of the Euclidean space \(\mathbb{R}^{n+1}\). Homeomorphisms preserve these properties. \(\square\)