Last Thursday, I was introducing to my students the notions of limit superior and limit inferior. In particular, I was emphasizing that a sequence of reals may not converge, but and always exist in the extended real number system. After giving some fairly basic examples of such sequences, one student asked me what the is of the sequence defined by
I was a bit taken aback by the question because, aside from , I did not (and do not) know off the top of my head what the values are. Nevertheless, I thought it was a great question–I love getting asked questions that require me to think a bit before answering–and asked the student if I could get back to him.
Well, it turns out that another one of my students also thought it was a great question and was eager to find out the answer himself. He did some investigating on Math.StackExchange, which led to the answer that
More interestingly, the set is dense in . Proving this result is the goal of today’s post.
Lemma 1. If is irrational, then the set
is dense in .
Proof. Since every real number can be written as the difference
where and , we may assume that . Fix , and let be given. Replacing by a smaller , if necessary, we may assume that . Choose sufficiently large so that . Partition the interval into intervals of length given by
By the pigeonhole principle, there exist two distinct natural numbers such that
for some and natural numbers . Since is irrational, . Without loss of generality, we may assume so that
Observe that , since is irrational. Since the disjoint half-open intervals
contain , for satisfying , we see that there exists such that
Setting and , we conclude that .
Lemma 2. is irrational.
The following proof is due to British mathematician Mary Cartwright.
Proof. For each , define
We can integrate by parts twice to obtain a recurrence relation for , with . Indeed,
For each , set , so that the above recurrence relation can be restated as
By induction, it follows that for each ,
where and are polynomials with integer coefficients and degree . Suppose that is rational so that , where . Since , we see that the RHS of the equation
is an integer. Since the integrand in the definition of is nonnegative and bounded by , we see that for all . As , the expression
So for sufficiently large , there exists an integer , which is a contradiction.
Now that we know that is irrational, and therefore is irrational, we can complete the proof that countable set is dense in . Fix . Since is a continuous function, by the intermediate value theorem, there exists such that . By lemma 2, there exist integers such that , for given. Since is continuous, for given, there exists such
which implies that
where we use the periodicity of .