Compactifying the Reals

Consider the set $\mathbb R$ equipped with the usual topology. By the Heine-Borel theorem, $\mathbb R$ is not compact, since though closed, $\mathbb R = \mathbb R \backslash \emptyset$ is obviously not bounded.

However, something strange happens when we switch our perspective just a little bit. Let $S^1 := \{(x, y) : x^2 + y^2 = 1\} \subseteq \mathbb R^2$ denote the unit circle, which is closed and bounded and thus compact in $\mathbb R^2$ . Now consider the rational parameterisation $f : \mathbb R \to S^1$ of $S^1$ defined by

$\displaystyle f(t) = \frac 1{1+t^2} ( 1 - t^2, 2t ).$

Notice that $f$ is continuous and injective, so that $f : \mathbb R \to f(\mathbb R)$ is bijective. In fact, $f(\mathbb R) = S^1 \backslash \{ (-1, 0) \}$ . It is also not hard to show that $f$ is open, and thus is a homeomorphism between $\mathbb R$ and $S^1 \backslash \{ (-1, 0) \}$ . Thus, $\mathbb R \cong S^1 \backslash \{ (-1, 0) \}$ .

What this investigation suggests is that $\mathbb R$ needs one more point in order to be homeomorphic to $S^1$ . Visually, we would like to denote this point with the symbol $\infty$ (but this will, of course, sacrifice some nice properties of $\mathbb R$ —sacrifices we are willing to make from a topological point of view). More poignantly, $\mathbb R$ needs one more point in order to become compact. We will denote this extension by $\mathbb R^* := \mathbb R \cup \{ \infty \}$ . But what kind of topology can we place on $\mathbb R$ so that this is indeed the case?

If we swapped perspectives, we could define the homeomorphism $g = f^{-1} : S^1 \backslash \{ (-1, 0) \} \to \mathbb R$ . What we want to do is append the element $\infty$ to $\mathbb R$ and extend $g$ to a homeomorphism $\tilde g : S^1 \to \mathbb R^*$ , where $\tilde g|_{S^1 \backslash \{ (-1, 0) \}} = g$ and more fundamentally, the right-hand side is equipped with an appropriate topology.

Let’s first suppose $\tilde g$ is a homeomorphism. It seems quite intuitive to define $\tilde g((-1, 0)) = \infty$ , to ensure that $\tilde g|_{S^1 \backslash \{ (-1, 0) \}} = g$ and that $g$ is still bijective. Indeed, we now have $\tilde g$ being bijective. What about continuity? It is clear that open sets of the form $(a, b)$ yield open preimages $\tilde g^{-1}((a, b)) = g^{-1}((a, b))$ . But we need neighborhoods of $\infty$ in our chosen topology too.

Perhaps analysing $S^1$ will be of interest. Basis elements of $S^1$ will resemble $B_d(\mathbf x, r) \cap S^1$ , since we equip $S^1 \subseteq \mathbb R^2$ with the subspace topology. In particular, $U := B_d((-1, 0), r) \cap S^1$ is a neighbourhood of $(-1, 0)$ , and $S^1 \backslash U$ is closed and bounded, and thus compact (by the Heine-Borel theorem). In fact,

$\mathbb R \backslash \tilde g(U) = \tilde g(S^1) \backslash \tilde g(U) = \tilde g(S^1 \backslash U) = g(S^1 \backslash U)$

must be compact since $g$ is a homeomorphism, and since $\tilde g(U) = \mathbb R \backslash (\mathbb R \backslash \tilde g(U))$ , we will extend the topology on $\mathbb R$ to the topology on $\mathbb R \cup \{\infty \}$ using sets of the form

$\mathbb R \backslash C \cup \{\infty\},\quad C \subseteq \mathbb R\quad \text{compact}.$

All that remains is to verify that with this revised topology, that $\tilde g : S^1 \to \mathbb R^*$ is continuous and maps open sets in $S^1$ to open sets in $\mathbb R^*$ to prove the following theorem:

Theorem 1. $\tilde g : S^1 \to \mathbb R^*$ defined by $\tilde g((-1, 0)) = \infty$ and $\tilde g|_{S^1 \backslash \{ (-1, 0) \}} = g$ is a homeomorphism. In particular, $\mathbb R^* = \overline{\mathbb R}$ is compact.

Proof. Let $V \subseteq \mathbb R^*$ be a basic set containing $\infty$ . Then $V = \mathbb R \backslash C \cup \{\infty\}$ for some compact set $C \subseteq \mathbb R$ . Since $g$ is a homeomorphism, $g^{-1}(C) \subseteq S^1 \backslash \{(-1, 0)\}$ is compact and hence closed, and thus

$\begin{aligned} \tilde g^{-1}(V) &= g^{-1}(\mathbb R \backslash C) \cup \{(-1, 0)\} \\ &= (S^1 \backslash \{ (-1, 0) \}) \backslash g^{-1}(C) \cup \{(-1, 0)\} = S^1 \backslash g^{-1}(C) \end{aligned}$

is open, so that $\tilde g$ is continuous, as required. To prove that $\tilde g$ maps open sets to open sets, for any $U = B_d((-1, 0), r) \cap S^1$ , it suffices to check that $\tilde g(U)$ is open. Indeed, $U^- := U \backslash \{(-1, 0)\}$ is open and since $\tilde g$ is bijective,

$\begin{aligned} \tilde g(U) &= \tilde g(U^-) \cup \tilde g(\{(-1, 0)\}) \\ &= g(U^-) \cup \{ \infty \}. \end{aligned}$

By construction, $\mathbb R \backslash g(U^-)$ is closed and can be shown to be bounded, and thus is compact. Hence, $g(U^-) = \mathbb R \backslash C$ for some compact $C \subseteq \mathbb R$ , and

$\tilde g(U) = g(U^-) \cup \{ \infty \} = \mathbb R \backslash C \cup \{ \infty \}$

is open, as required.

Now, in this example, we explicitly took advantage of the Heine-Borel theorem, and therefore will not be able to immediately generalise to other topological spaces. But perhaps, this example will instruct us on the key strategy to do just that, in the next post.

—Joel Kindiak, 1 May 25, 1856H

KindiakMath

Compactifying the Reals

Leave a comment Cancel reply

Compactifying the Reals

Share this:

Leave a comment Cancel reply