{"id":990,"date":"2017-10-06T23:08:05","date_gmt":"2017-10-07T03:08:05","guid":{"rendered":"http:\/\/blog.richmond.edu\/math320\/?p=990"},"modified":"2017-10-08T20:31:52","modified_gmt":"2017-10-09T00:31:52","slug":"the-cantor-set-2","status":"publish","type":"post","link":"https:\/\/blog.richmond.edu\/math320\/2017\/10\/06\/the-cantor-set-2\/","title":{"rendered":"The Cantor Set"},"content":{"rendered":"

By Nick Wan and Elaine Wissuchek<\/p>\n

Introduction to the Cantor Set<\/strong><\/p>\n

<\/p>\n

Cantor set is a special subset of the closed interval [0, 1] invented by a German mathematician Georg Cantor in 1883. In order to construct this set, we need to construct infinitely many subset of \"[0, inductively and take the intersection of all of them. Specifically:<\/p>\n

Let \"I_0. Remove the open middle third \"(1\/3, from \"I_0\".<\/p>\n

Let \"I_1:, precisely \"I_1. Remove middle third open intervals \"(1\/9, and \"(7\/9, from the respective closed intervals whose union is \"I_1\".<\/p>\n

Let \"I_2: say, \"I_2^1\cup.<\/p>\n

Continue this procedure of removal of open intervals from each closed interval \"I^k_j\" where \"1\leq whose union is \"I_j\" where \"j and get \"I_{(j+1)}\" by taking the union of whatever is left from each \"I^k_j\" after the removal of open intervals as before.<\/p>\n

Define the Cantor set \"C:=.<\/p>\n

Another way to view the Cantor set is in terms of ternary expansions:<\/p>\n

Given \"x\in a real number, there is a sequence of integers \"(a_k)^{\infty}_{k=1}, with \"k\in such that the series \"\sum_{k=1}^{\infty} converges to \"x\". In other words, we can write \"x\" in a ternary form: \"x=0.a_1 (base 3). For example, \"\frac{1}{9}=(0.01000...)_3\".<\/p>\n

In the case or our Cantor set, any point \"x\" in \"C\" has the following ternary expansion:<\/p>\n

\"x=\sum_{i=1}^{\infty} with all \"d_k\in\{0,2\}\" from the set \"a_k\".<\/p>\n

For example, the element of the Cantor set \"x=\frac{1}{3}\" corresponds to \"d_k=(0,2,2,2,2,2,2,...)\" and \"x=\frac{2}{3}\" corresponds to \"d_k=(2,0,0,0,0,0,0,...)\".<\/p>\n

Uncountability of the Cantor Set by Diagonalization<\/strong><\/p>\n

We can use Cantor’s diagonalization method to show that the Cantor set is not countable.<\/p>\n

Let \"(v_n)=\{\u00a0(d_k)_n be a sequence so that each \"v_n\" represents an element of the Cantor set. Let all\u00a0\"(v_n). Assume that \"f:\mathbb{N} is a one-to-one correspondence between the natural numbers and the set \"V\" so that the elements of the cantor set are countable.<\/p>\n

Consider\u00a0\"(v_n)^* where \"(v_n)^*=\{\u00a0(d_k)_n^* if \"(d_n)_n and \"(d_k)_n^* if \"(d_n)_n for \"(d_k)_n^* and \"(d_n)_n. Therefore \"(v_n)^*\" will be distinct from any\u00a0\"(v_n) in the \"k\"th position. Thus for \"(v_n)^*, we cannot find an \"n\in\u00a0\mathbb{N}\" for which \"f(n)=(v_n)^*\", so \"f\" is not onto, a contradiction to the one-to-one correspondence of \"f\". Hence by contradicting the definition of countability, that \"\mathbb{N}, \"V\" is uncountable. Since all elements of the cantor set are represented in \"V\", the Cantor set is uncountable.<\/p>\n

The Cantor Set is Closed <\/strong><\/p>\n

Each \"In\" is a union of \"2^n\" closed intervals. As a finite union of closed intervals, it is a closed set in \"[0,. Then \"C\" is a closed subset of \"[0, being the intersection of closed sets \"I_n\" for \"n<\/p>\n

The Cantor Set is perfect<\/strong><\/p>\n

A set is considered perfect if the set is closed and all the points of the set are limit points of the set. For each endpoint in the set \"C\" there will always exist another point in the set within a deleted neighborhood of some radius \"\varepsilon on one side of that point since the remaining intervals at each step are being divided into infinitely small subintervals and since the real numbers are infinitely dense. Likewise, for each nonendpoint in the set there will always exist another point in the set within a deleted neighborhood of some radius \"\varepsilon on both sides of that point. Hence, all deleted neighborhoods of any radius \"\varepsilon around each point of the set \"C\" for which the intersection of that deleted neighborhood and the set are nonempty. Therefore, each point in the set is a limit point of the set, and since the set is closed, the set \"C\" is perfect.<\/p>\n

 <\/p>\n

Dimension of the Cantor Set<\/strong><\/p>\n

Definition: Dimension<\/p>\n