The following question was a problem in an Analysis exam:

Let $n \in \mathbb{N}$. Define $A_{n} := \displaystyle \left\{\frac{k}{2^n} \bigg| k \in \mathbb{Z}, 0 \leq k \leq 2^n \right\}$. Let $A_{\infty} = \cup_{n \in \mathbb{N}} A_n$ .

Compute $\overline{A_{\infty}}$ (closure of $A_{\infty}$) and $A_{\infty}^{\circ}$ (the interior of $A_{\infty}$).

I have solved the problem. I got the answer as $\overline{A_{\infty}} = [0,1]$ and $A_{\infty}^{\circ} = \phi$.

But as I wondered about the problem, I observed that $A_{\infty}$ was a strict subset of $\mathbb{Q} \cap [0,1]$ and was still dense in $[0,1]$. This set me thinking; I asked myself if can I get a minimally dense (in $[0,1]$) subset of $A_{\infty}$?

So I defined $P_{\infty} = \cup_{n \in \mathbb{N}} P_n$ where $P_n := \displaystyle \left\{\frac{k}{2^n} \bigg| k \in \mathbb{Z}, k \text{ is prime } , 0 \leq k \leq 2^n \right\}$. And then I defined $D_n := A_n \setminus P_n$ and $D_{\infty}$ appropriately. I could show that $D_{\infty}$ is dense in $[0,1]$ and that it is a strict subset of $A_{\infty}$.

So I have up to two questions: (dense always means dense in $[0,1]$ in the following questions)

0)Does there exist a minimally dense subset of $\mathbb{Q} \cap [0,1]$? If it does, how do we find it?

1) Does there exist a minimally dense subset of $A_{\infty}$? If it does, how do we find it?

2) Is $P_{\infty}$ dense?

Thanks, Isomorphism

  • 7
    $\begingroup$ If $D \subset \mathbb R$ is dense in $\mathbb R$ show that for any $x \in D$ that $D \setminus \{x\}$ is also dense in $\mathbb R$. $\endgroup$ – JSchlather Oct 13 '11 at 3:58
  • 3
    $\begingroup$ What would minimally dense mean? We can always remove any finite set from a set dense in $[0,1]$ and get a set dense in $[0,1]$. If we are careful we can remove more. $\endgroup$ – André Nicolas Oct 13 '11 at 4:02
  • $\begingroup$ Q\{0} is also a strict subset of Q such thah its closure is R $\endgroup$ – Gaston Burrull May 23 '13 at 0:13

We show that the set $P_\infty$ defined by the OP is dense in $[0,1]$.

Let $x\in (0,1)$. We will show that $x$ can be well-approximated by numbers of the form $\frac{p}{2^n}$, where $p$ is prime. Take $n$ very large, and let $\frac{m}{2^n} \in (0,1)$ be a very good approximation to $x$ (absolute value of the error $\lt \epsilon/2$), with $m$ a very large integer. So if $x$ is very close to $0$, we still make sure that the numerator $m$ is large, by taking $2^n$ huge.

A not too difficult result (given the Prime Number Theorem!) about prime gaps is that for any given $\epsilon$, and large enough $k$, there is always a prime between $p_k$ and $p_k(1+\epsilon)$.

Let $q$ be the largest prime which is $\le m$. By the result quoted above, if $m$ is large enough there is a prime $p$ in the interval $q<u<q(1+\epsilon/2)$.

We show that $\frac{p}{2^n}$ is within $\epsilon$ of $x$. It is enough to show that $\frac{p}{2^n}$ is within $\epsilon/2$ of $\frac{m}{2^n}$.

Note that $p>m$, and $p<m(1+\epsilon/2)$. So $$\frac{p}{2^n}-\frac{m}{2^n}<\frac{m(1+\epsilon/2)}{2^n}-\frac{m}{2^n}=\frac{m\epsilon/2}{2^n}<\epsilon/2.$$

(There is a small deliberate gap in the argument, since for $x$ extremely close to $1$, we could have $p>2^n$. Then we use $q/2^n$.)

Comment: There is nothing special about $2^n$ here. The sequence $(2^n)$ can be replaced by any integer sequence $(a_n)$ which is not bounded above. An interesting choice is to let $(a_n)$ be the sequence of primes. The proof is the same.

  • $\begingroup$ I did not the prime gaps result. Thank you. $\endgroup$ – Isomorphism Oct 13 '11 at 8:22

As to questions 0 and 1: No, there is no minimally dense subset of $[0,1]$, and in particular no minimally dense subset of $A_{\infty}$.

Suppose $D$ is any dense subset of $[0,1]$, and let $x\in D$. I claim that $D'=D-\{x\}$ is also dense.

Indeed, let $a\in [0,1]$ and let $\epsilon\gt 0$. We need to show that $(a-\epsilon,a+\epsilon)\cap D'\neq \emptyset$. We know that $(a-\epsilon,a+\epsilon)\cap D\neq\emptyset$. If it contains a point other than $x$, we are done. Otherwise, $(a-\epsilon,a+\epsilon)\cap D = \{x\}$. If $a\neq x$, then taking $\epsilon'=|a-x|/2$ gives a contradiction to $D$ being dense. If $a=x$, then letting $w\neq x$ be any point in $(x-\epsilon,x+\epsilon)\cap [0,1]$ and letting $\epsilon'=|w-x|/2$ gives a point in $[0,1]$ and a neighborhood that does not intersect $D$, a contradiction.

Thus, $D-\{x\}$ is dense as well. Hence, no dense subset of $[0,1]$ is minimal. Inductively, you can remove any finite set from a dense subset and still have a dense subset.

In particular, there is no minimally dense subset of $A_{\infty}$.

  • $\begingroup$ Thank you for your answer and a clear proof. I figured it out a few moments ago. The prime number case turned interesting (as I learnt something new), so I chose the other one as the answer. Thank you for your answer. $\endgroup$ – Isomorphism Oct 13 '11 at 8:21

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