Perfect Subsets of the Real Line

Let $P \subseteq \Real$ be perfect. We construct an injection from the set $\Cant$ of all infinite binary sequences into $P$. An infinite binary sequence $\xi = \xi_0 \xi_1 \xi_2 \dots$ can be identified with a real number $\in [0,1]$ via the mapping

Note that this mapping is onto. It follows that the cardinality of $P$ is at least as large as the cardinality of $[0,1]$. The Schröder-Bernstein Theorem (for a proof see e.g. Jech (2003)) implies that $|P| = |\R|$.

To construct the desired injection, choose $x \in P$ and let $\eps_0 = 1 = 2^0$. Since $P$ is perfect, $P \cap U_{\eps_0}(x)$ is infinite. Let $x_0 \neq x_1$ be two points in $P \cap U_{\eps_0}(x)$, distinct from $x$. Let $\eps_1$ be such that $\eps_1 \leq 1/2$, $U_{\eps_1}(x_0), U_{\eps_1}(x_1) \subseteq U_{\eps_0}(x)$, and $\overline{U_{\eps_1}(x_0)} \cap \overline{U_{\eps_1}(x_1)} = \emptyset$, where $\overline{U}$ denotes the closure of $U$.

We can iterate this procedure recursively with smaller and smaller diameters, using the fact that $P$ is perfect. This gives rise to a so-called Cantor scheme, a family of open balls $(U_\sigma)$ satisfying certain nesting conditions. Here the index $\sigma$ is a finite binary sequence, also called a string. A Cantor scheme is defined by the following properties.

1. $\diam(U_\sigma) \leq 2^{-|\sigma|}$, where $|\sigma|$ denotes the length of $\sigma$.
2. If $\tau$ is a proper extension of $\sigma$, then $\Cl{U_\tau} \subseteq U_\sigma$.
3. If $\tau$ and $\sigma$ are incompatible (i.e. neither extends the other), then
   $$U_\tau \cap U_\sigma = \emptyset.$$
4. The center of each $U_\sigma$, call it $x_\sigma$, is in $P$.

Let $\xi$ be an infinite binary sequence. Given $n \geq 0$, we denote by $\xi\Rest{n}$ the string formed by the first $n$ bits of $\xi$, i.e.

The finite initial segments give rise to a sequence $x_{\xi\Rest{n}}$ of centers. By properties (1.) and (2.), this is a Cauchy sequence. By (4.), the sequence lies in $P$. Since $P$ is closed, the limit $x_\xi$ is in $P$. By (3.), the mapping $\xi \mapsto x_\xi$ is well-defined and injective.

Let $C \subseteq \Real$ be uncountable and closed. We say $z \in \Real$ is a condensation point of $C$ if

Let $D$ be the set of all condensation points of $C$. Note that $D \subseteq C$, since every condensation point is clearly a limit point and $C$ is closed.

Furthermore, we claim that $D$ is perfect. It is straightforward to verify that $D$ is closed. To see that every point of $D$ is a limit point, suppose $z \in D$ and $\eps > 0$. Then, by assumption, $U_\eps(z) \cap C$ is uncountable. We would like to conclude that $U_\eps(z) \cap D$ is uncountable, too, since this would mean in particular that $U_\eps(z) \cap D$ is infinite. The conclusion holds if $C \setminus D$ is countable.

To show that $C\setminus D$ is countable, assume that $y \in C \setminus D$. Then, for some $\delta > 0$, $U_\delta(y) \cap C$ is countable. We can find an interval $I(y) \subseteq U_\delta(y)$ that contains $y$ and has rational endpoints. There are at most countably many intervals with rational endpoints and hence for each $y \in C \setminus D$ there are at most countably many choices for $I(y)$. Thus, we have

The right hand side is a countable union of countable sets, hence countable.

Finally, the fact that $C \setminus D$ is countable also implies that $D$ is non-empty. We have therefore verified that $D$ is a perfect subset of $C$.