## To construct exotic sets, use limiting arguments

### Quick description

How can one build an uncountable set 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 ? In each case, the easiest approach is to build a sequence of sets and take its union or intersection. If the sets 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 of is dense in if every subinterval of of positive length contains an element of . By contrast, is nowhere dense if there is no interval of positive length such that is dense in . 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 ? 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 will be uncountable if we can find an injection to from the set of all infinite sequences. So perhaps we can build our uncountable set by first building a set , then finding two subsets and of , then finding two subsets and of and two subsets and of , and so on, and defining to be the intersection . If for every sequence we can make sure that the intersection is non-empty, then is uncountable. And this we can do if each is non-empty, closed, and bounded.

How do we ensure that is nowhere dense? For that matter, what sorts of sets should the 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 , inside each of which we have to find two subintervals and . There are countably many stages to this process, so we can ensure that is nowhere dense by means of a just-do-it proof as follows. Enumerate the open intervals with rational end-points as and simply ensure at the th stage that is not a subset of . 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. Incomplete This article is incomplete. More examples to follow.