To construct exotic sets, use limiting arguments

Quick description

How can one build an uncountable set $S$ that is nowhere dense, or a large supply of dense open sets, or a bounded path of infinite length, or a set of Hausdorff dimension $1\frac 12$ ? In each case, the easiest approach is to build a sequence of sets $S_1,S_2,\dots$ and take its union or intersection. If the sets $S_n$ approximate the property you want, then the union or intersection may have it exactly.

Prerequisites

Basic set theory and real analysis, countability

Example 1

A subset $X$ of $\R$ is dense in $[a,b]$ if every subinterval of $[a,b]$ of positive length contains an element of $X$ . By contrast, $X$ is nowhere dense if there is no interval $[a,b]$ of positive length such that $X$ is dense in $[a,b]$ . It might seem that nowhere dense sets would have to be pretty small, but they can in fact be uncountable.

How does one build an uncountable subset of $\R$ ? The obvious way is to make sure that it contains some interval of positive length, but that option is not available if we want our subset to be nowhere dense. Another approach is to bear in mind that a set $X$ will be uncountable if we can find an injection to $X$ from the set of all infinite $01$ sequences. So perhaps we can build our uncountable set $X$ by first building a set $W$ , then finding two subsets $W_0$ and $W_1$ of $W$ , then finding two subsets $W_{00}$ and $W_{01}$ of $W_0$ and two subsets $W_{10}$ and $W_{11}$ of $W_1$ , and so on, and defining $X$ to be the intersection $\bigcap_{k=1}^\infty \bigcup_{\epsilon_1,\dots,\epsilon_k} W_{\epsilon_1\epsilon_2\dots\epsilon_k}$ . If for every $01$ sequence $\epsilon_1,\epsilon_2,\dots$ we can make sure that the intersection $\bigcap_{k=1}^\infty W_{\epsilon_1\epsilon_2\dots\epsilon_k}$ is non-empty, then $X$ is uncountable. And this we can do if each $W_{\epsilon_1\epsilon_2\dots\epsilon_k}$ is non-empty, closed, and bounded.

How do we ensure that $X$ is nowhere dense? For that matter, what sorts of sets should the $W_{\epsilon_1\epsilon_2\dots\epsilon_k}$ be? Since all we know so far is that it would be good if they were non-empty, closed and bounded, we may as well begin by trying the simplest such sets, namely closed intervals. Then at each stage of our construction we shall have closed intervals of the form $W_\epsilon$ , inside each of which we have to find two subintervals $W_{\epsilon 0}$ and $W_{\epsilon 1}$ . There are countably many stages to this process, so we can ensure that $X$ is nowhere dense by means of a just-do-it proof as follows. Enumerate the open intervals with rational end-points as $(r_1,s_1),(r_2,s_2),\dots$ and simply ensure at the $k$ th stage that $(r_k,s_k)$ is not a subset of $\bigcup_{\epsilon_1,\dots,\epsilon_k} W_{\epsilon_1\epsilon_2\dots\epsilon_k}$ . This is easy to do.

General discussion

The usual solution to this problem would be to observe that the Cantor set is an example. And indeed, what we have done is exactly this, except that instead of removing the middle third at each stage we have removed intervals of unspecified lengths.

This article is incomplete. More examples to follow.

Comments

Nowhere dense sets

Wed, 03/06/2009 - 21:39 — Anonymous (not verified)

In fact, nowhere dense sets can not only be uncountable, they can can positive measure. See the wikipedia page on the Smith-Volterra-Cantor set. This set is a good source for counterexamples.

Post new comment

(Note: commenting is not possible on this snapshot.)