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    Middle-thirds Cantor set is compact, totally disconnected, perfect, and null; Borel–Cantelli lemma

    E-L3F4O

    Area
    Real Analysis
    Topics
    Borel-Cantelli, Cantor Set, Measure Theory
    Status
    draft
    • Show that that Cantor middle-thirds set is compact, totally disconnected, and perfect, with outer measure zero.

    • Prove the Borel-Cantelli lemma.

    Solution

    Let \(C_0 = [0,1]\), let \(C_n\) be obtained from \(C_{n-1}\) by deleting the open middle third of each of its intervals, and let \(C = \bigcap_{n \ge 0} C_n\). By induction, \(C_n\) is the disjoint union of \(2^n\) closed intervals of length \(3^{-n}\), the stage-\(n\) intervals.

    1. \(C\) is compact.

    Each \(C_n\) is a finite union of closed intervals, hence closed, so \(C\) is a closed subset of the compact set \([0,1]\).

    2. \(m^*(C) = 0\).

    \(C \subseteq C_n\) and \(m^*(C_n) = 2^n 3^{-n}\), so monotonicity gives \(m^*(C) \le (2/3)^n\) for every \(n\).

    3. \(C\) is totally disconnected.

    2. So every connected subset of \(C\) has at most one point.

    4. \(C\) is perfect: every point of \(C\) is a limit point of \(C\).

    4.1. Every endpoint of a stage-\(k\) interval lies in \(C\).

    Deleting an open middle third from a closed interval keeps both endpoints, and each endpoint of a stage-\(k\) interval is an endpoint of a stage-\(j\) interval for every \(j \geq k\). So it lies in every \(C_j\).

    4.2. For \(x \in C\) and \(n \geq 0\) there is \(y \in C\) with \(0 < \abs{x - y} \leq 3^{-n}\).

    Let \(I\) be the stage-\(n\) interval containing \(x\). \(I\) has two endpoints at distance \(3^{-n}\) from each other, so at least one of them differs from \(x\); it lies in \(C\) by step 1 and in \(I\), so its distance from \(x\) is at most \(3^{-n}\).

    4.3. Q.E.D.

    Step 2 gives points of \(C \setminus \theset{x}\) within \(3^{-n}\) of \(x\) for every \(n\). Since \(C\) is closed by step 1, \(C\) is perfect.

    5. Borel–Cantelli lemma: if \((E_n)\) is a sequence of measurable sets in a measure space \((X, \mu)\) with \(\sum_n \mu(E_n) < \infty\), then \(\mu(\limsup_n E_n) = 0\).

    \(\limsup_n E_n = \bigcap_N \bigcup_{n \ge N} E_n\) is measurable. For every \(N\), \(\limsup_n E_n \subseteq \bigcup_{n \ge N} E_n\), so monotonicity and countable subadditivity give \(\mu(\limsup_n E_n) \le \sum_{n \ge N} \mu(E_n)\). The right-hand side is the tail of a convergent series and tends to \(0\) as \(N \to \infty\).

    6. Q.E.D.

    Steps <1>1–<1>4 prove compactness, outer measure zero, total disconnectedness, and perfectness of \(C\); step 5 proves the Borel–Cantelli lemma.