Conjectural variation

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In real analysis, a branch of mathematics, Cantor's intersection theorem, named after Georg Cantor, gives conditions under which an infinite intersection of nested, non-empty, sets is non-empty.

Theorem 1: If $(X, d)$ is a non-trivial, complete metric space and ${C_{n}}$ is an infinite sequence of non-empty, closed sets such that $C_{n} \supset C_{n + 1}, \forall n$ and $\lim_{n \to \infty} d i a m (C_{n}) = \sup {d (x, y) : x, y \in C_{n}} \to 0$ . Then, there exists an $x \in X$ such that $⋂_{n = 1}^{\infty} C_{n} = {x}$ .^[1]

Theorem 2: If $X$ is a compact space and ${C_{n}}$ is an infinite sequence of non-empty, closed sets such that $C_{n} \supset C_{n + 1}, \forall n$ , then $⋂_{n = 1}^{\infty} C_{n} \neq \emptyset$ .

Notice the differences and the similarities between the two theorem. In Theorem 2, the $C_{n}$ are only assumed to be closed (and not compact, which is stronger) since given a compact space $X$ and $Y \subset X$ a closed subset, $Y$ is necessarily compact. Also, in Theorem 1 the intersection is exactly 1 point, while in Theorem 2 it could contain many more points. Interestingly, a metric space having the Cantor Intersection property (i.e. the theorem above holds) is necessarily complete (for justification see below). An example of an application of this theorem is the existence of limit points for self-similar contracting fractals.^[2]

Notice that each of the hypotheses above is essential. If the metric space were not complete, then one could construct a nested sequence of non-empty, compact sets converging to a "hole" in the space, i.e. $ℚ$ with the usual metric and the sequence of sets, $C_{n} = [\sqrt{2}, \sqrt{2} + 1 / n]$ . If the sets are not closed, then one can construct sequences of nested sets which have empty intersection, i.e. $ℝ$ with the collection, $C_{n} = (0, \frac{1}{n})$ . The collections $C_{n} = [n, \infty)$ and $C_{n} = [- \frac{1}{n}, 1 + \frac{1}{n}]$ illustrate what may happen when the diameters do not tend to zero: the intersection may be empty, as in the first, or may contain more than a single point, as in the second.

Proof

Theorem 1: Suppose $(X, d)$ is a non-trivial, complete metric space and ${C_{n}}$ is an infinite family of non-empty closed sets in $X$ such that $C_{n} \supset C_{n + 1}, \forall n$ and $\lim_{n \to \infty} d i a m (C_{n}) \to 0$ . Naturally we would like to use the completeness so we will construct a Cauchy sequence. Since each of the $C_{n}$ is closed, there exists a $y_{n}$ in the interior (i.e. positive distance to anything outside $C_{n}$ ) of $C_{n}$ . These $y_{n}$ form a sequence. Since $\lim_{n \to \infty} d i a m (C_{n}) \to 0$ , then given any positive real value, $ϵ > 0$ , there exists a large $N$ such that whenever $n \geq N$ , $d i a m (C_{n}) < ϵ$ . Since, $C_{n} \supset C_{n + 1}, \forall n$ , then given any $n, m \geq N$ , $y_{n}, y_{m} \in C_{n}$ and therefore, $d (y_{n}, y_{m}) < ϵ$ . Thus, the $y_{n}$ form a Cauchy sequence. By the completeness of $X$ there is a point $x \in X$ such that $y_{n} \to x$ . By the closure of each $C_{n}$ and since $x$ is in $C_{n}$ for all $n \geq N$ , $x \in ⋂_{n = 1}^{\infty} C_{n}$ . To see that $x$ is alone in $⋂_{n = 1}^{\infty} C_{n}$ assume otherwise. Take $x^{'} \in ⋂_{n = 1}^{\infty} C_{n}$ and then consider the distance between $x$ and $x^{'}$ this is some value greater than 0 and implies that the $\lim_{n \to \infty} d i a m (C_{n}) \to d (x, x^{'}) > 0$ . Contradiction! Thus the claim follows.

Theorem 2: Suppose $X$ is a compact topological space and ${C_{n}}$ is an infinite sequence of non-empty, closed sets such that $C_{1} \supseteq \dots C_{k} \supseteq C_{k + 1} \dots,$ . Assume, by contradiction, that $⋂_{n = 1}^{\infty} C_{n} = \emptyset$ . Then we will build an open cover of $X$ by considering the complement of $C_{n}$ in $X$ , i.e. $U_{n} = X ∖ C_{n}, \forall n$ . Each $U_{n}$ is open since the $C_{n}$ are closed. Notice that $⋃_{n = 1}^{\infty} U_{n} = ⋃_{n = 1}^{\infty} (X ∖ C_{n}) = X ∖ ⋂_{n = 1}^{\infty} C_{n}$ , but we assumed that $⋂_{n = 1}^{\infty} C_{n} = \emptyset$ so that means $⋃_{n = 1}^{\infty} U_{n} = X$ . So, there are infinite many $U_{n}$ covering our compact $X$ . That means there exists a large $N$ such that $X \subset ⋃_{n = 1}^{N} U_{n}$ . Notice, however, that $C_{n} \supset C_{n + 1}, \forall n$ implies that $X ∖ C_{n} = U_{n} \subset U_{n + 1} = X ∖ C_{n + 1}, \forall n$ . The only way for the nested and increasing $U_{n}$ to cover $X$ is if there is some index, call it $k$ , such that $X = U_{k}$ . This implies though that $C_{k} = X ∖ U_{k} = \emptyset$ . This is a contradiction since we assumed that the $C_{n}$ were non-empty. Hence, $⋂_{n = 1}^{\infty} C_{n} \neq \emptyset$ .

Notice that in regards to the proof of Theorem 2, we don't need Hausdorffness. At no point in time do we appeal to the nature of points in the space. It is simply a statement about empty or not.

Consider now a metric space $(X, d)$ (not necessarily complete) in which $⋂_{n = 1}^{\infty} C_{n} = x$ whenever ${C_{n}}$ is an infinite sequence of non-empty, closed sets such that $C_{n} \supset C_{n + 1}, \forall n$ and $\lim_{n \to \infty} d i a m (C_{n}) = s u p {d (x, y) : x, y \in X} \to 0$ . Now, let ${x_{k}}$ be a Cauchy sequence in $X$ and take $C_{n} = \overline{{x_{k} : k \geq n}}$ . The bar over the set means that we are taking the closure of the set under it. This guarantees that we are working with closed sets and since they contain the elements of our Cauchy sequence, we know them to be non-empty. In addition, $C_{n} \supset C_{n + 1}$ and since $\forall ϵ > 0, \exists N$ such that when $n, m \geq N, d (x_{n}, x_{m}) < ϵ$ , (note this hold for all indices larger than our large $N$ ) then $d i a m (C_{N}) < ϵ$ . Hence, ${C_{n}}$ satisfies the conditions above and there exists an $x \in X$ such that $⋂_{n = 1}^{\infty} C_{n} = x$ . So, $x$ is in the closure of all of the $C_{n}$ and any open ball around $x$ has non-empty intersection with the $C_{n}$ . Now we will build a sub-sequence of the ${x_{n}}$ , call it ${x_{n_{k}}}$ , where $d (x, x_{n_{k}}) < \frac{1}{k}$ . This implies that ${x_{n_{k}}} \to x$ and since ${x_{k}}$ was Cauchy then it too must converge to $x$ . Since ${x_{k}}$ was an arbitrary Cauchy sequence, $X$ is complete.

References

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Jonathan Lewin. An interactive introduction to mathematical analysis. Cambridge University Press. ISBN 0-521-01718-1. Section 7.8.

↑ "Real Analysis," H.L. Royden, P.M. Fitzpatrick, 4th edition, 2010, page 195
↑ Ergodic Theory and Symbolic Dynamics in Hyperbolic Spaces, T. Bedford, M. Keane and C. Series eds., Oxford Univ. Press 1991, page 225

[1] "Real Analysis," H.L. Royden, P.M. Fitzpatrick, 4th edition, 2010, page 195

[2] Ergodic Theory and Symbolic Dynamics in Hyperbolic Spaces, T. Bedford, M. Keane and C. Series eds., Oxford Univ. Press 1991, page 225

[1]

[2]

Conjectural variation

Proof

References

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