2.3. Perfect Sets
Theorem 2.2.17 guarantees that an infinite subset of a compact set has at least one limit point. However, this is a somewhere statement, not an everywhere statement: it does not say that every point of the set participates in the accumulation. An infinite compact set may still contain many isolated points. Perfectness imposes a much stronger local picture: a perfect set has accumulation everywhere within the set.
Definition 2.3.1. Let \(M\) be a metric space, and let \(A\subseteq M\). A point \(x\in A\) is said to be isolated in \(A\) if
\[B_r(x)\cap A=\{x\}\]for some \(r>0\). That is, \(x\) is an isolated point of \(A\) if and only if \(x\) is not a limit point of \(A\). The set \(A\) is said to be perfect if it is closed and has no isolated points. Closedness gives \(A'\subseteq A\), while the absence of isolated points gives \(A\subseteq A'\). Hence
\[A\text{ is perfect}\iff A=A'.\]Remark 2.3.2. As openness and closedness, perfectness may depend on the ambient space. For example, \(\mathbb Q\) has no isolated points and is closed in itself, so it is perfect. In \(\mathbb R\), however, \(\mathbb Q\) is not perfect in \(\mathbb R\). Indeed, since \(\mathbb Q\) is a proper subset of \(\mathbb R\), we may choose \(x\in \mathbb R\setminus \mathbb Q\). Then, \(x\) is a limit point of \(\mathbb Q\), since Theorem 1.1.10 guarantees that every interval \((x-r,x+r)\) has a rational. Thus \(\mathbb Q\) is not closed in \(\mathbb R\). Therefore, despite having no isolated points, \(\mathbb Q\) is not perfect in \(\mathbb R\).
This example of \(\mathbb Q\) also shows that having no isolated points is not, by itself, enough to force a set to be uncountable. The missing ingredient here is closedness. The next theorem shows that, in Euclidean space, closedness turns this local crowding into a strong global conclusion.
Theorem 2.3.3. Every nonempty perfect subset of \(\mathbb R^k\) is uncountable.
Proof. Let \(P\) be a nonempty perfect subset of \(\mathbb R^k\). Since \(P\) has limit points, \(P\) is infinite by Corollary 2.1.9. Suppose, for contradiction, \(P\) is countable, and denote the points of \(P\) as \(\mathbf p_1,\mathbf p_2,\mathbf p_3,\dots.\) We now construct a sequence \((V_n)\) of open balls. Define \(V_1\) to be any open ball around \(p_1\). Suppose \(V_n\) has been constructed so that \(V_n\) has a point of \(P\). Then, we may choose a point \(\mathbf y\in V_n \cap P\) other than \(p_n\). Indeed, even if \(\mathbf p_n\in V_n\cap P\), since \(V_n\) is open, there is a ball \(B_r(\mathbf p_n)\subseteq V_n\). Since \(\mathbf p_n\) is a limit point of \(P\), \(B^{\ast}_r(\mathbf p_n)\) has a point of \(P\), so such a \(\mathbf y\) exists. Let \(B_s(\mathbf y)\subseteq V_n\), and let
\[0 < t < \min \lbrace s,\|\mathbf p_n-\mathbf y\|\rbrace.\]Then we define \(V_{n+1} = B_t(\mathbf y)\). Since \(\mathbf y\in V_{n+1}\cap P\) so that \(V_{n+1}\) satisfies our induction hypothesis, the construction can proceed. Put \(K_n=\overline {V_n} \cap P\). Then, each \(K_n\) is nonempty and compact. Moreover, since \(\overline{V_{n+1}}\subseteq V_n\subseteq\overline{V_n}\), we have \(K_{n+1}\subseteq K_n\). Thus Corollary 2.2.12 gives a point \(\mathbf x\in \bigcap_{n=1}^{\infty} K_n\). This point has survived every stage of the construction. But since \(\mathbf x\in P\), the assumed enumeration gives \(\mathbf x=\mathbf p_m\) for some \(m\). This contradicts the fact that the \((m+1)\)-st ball was chosen precisely so that \(\mathbf p_m\notin K_{m+1}\).\(\square\)
The theorem gives a strong conclusion from perfectness, but familiar perfect sets such as closed intervals in \(\mathbb R\) do not make the result feel particularly surprising. The Cantor set is a more revealing example.
Definition 2.3.4. (Cantor set). Let \(C_0\) be the interval \([0,1]\). Removing the open interval \(\left(\frac 1 3, \frac 2 3\right)\) from \(C_0\), let \(C_1\) be the union of the intervals
\[\left[0,\frac 1 3\right], \left[\frac 2 3, 1\right].\]Removing the open middle thirds of the intervals, let \(C_2\) be the union of the intervals
\[\left[0, \frac{1}{9}\right], \left[\frac{2}{9}, \frac{1}{3}\right], \left[\frac{2}{3}, \frac{7}{9}\right],\left[\frac{8}{9}, 1\right].\]Continuing in this way, we obtain a sequence of sets \(C_n\). Then, the set \(C\) defined by
\[C = \bigcap_{n=1}^{\infty} C_n\]is called the Cantor set.
At every finite stage it becomes increasingly fragmented, and the surviving intervals shrink toward zero length. Nevertheless, the limiting set has no isolated points and it therefore remains uncountable. We now show that no point of \(C\) is isolated. Since each \(C_n\) is compact and \(C_{n+1}\subseteq C_n\), Corollary 2.2.12 shows that \(C\) is nonempty. Moreover, \(C\) is closed as an intersection of closed sets. Fix \(x\in C\) and \(r>0\). At the \(n\)-th stage, \(x\) lies in one of the component intervals of \(C_n\), whose length is \(3^{-n}\). Both endpoints of this interval remain in all later stages and therefore belong to \(C\). Choose an endpoint \(y_n\neq x\). Then
\[0<|x-y_n|\leq 3^{-n}.\]For sufficiently large \(n\), we have \(3^{-n}<r\), and hence
\[y_n\in B_r^*(x)\cap C.\]Thus every point of \(C\) is a limit point of \(C\). Since \(C\) is also closed, it is perfect. Theorem 2.3.3 now shows that \(C\) is uncountable.
References
- Rudin, W. (1976). Principles of mathematical analysis (3rd ed.). McGraw-Hill Education.