[Math] Minimal dense subset of $\mathbb{Q} \cap [0,1]$

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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

Best Answer

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 $A_{\infty}$.

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