Minimally dense subsets of $[0,1]$? [duplicate]

I'm curious about the notion of a minimally dense subset of a given space (I've been using $[0,1]$, but if others are interesting, I'm interested in that too!). Here are two questions.

• Does there exist a set $A$ that is dense in $[0,1]$, but for all infinite $B\subset A$, $A\setminus B$ is not dense in $[0,1]$? I suspect that the answer to this question is no, but I haven't been able to prove it. I'm also curious about what modifications to this statement can create a notion of "minimally dense" subsets.

• For all $\alpha\in \mathbb{R}\setminus\mathbb{Q}$, there exist infinitely many rational numbers $\frac{m}{n}$ with $|\alpha-\frac{m}{n}|<\frac{1}{n^2}$. In his analysis book, Browder remarks that this theorem is a way that one can quantify density. Is there a way to use statements like this to come up with a "minimally dense" subset of [0,1]?

Let me know if there are other ways of talking about this notion, or whether you think there is something wrong with the idea of a minimally dense set in general. Thanks!

marked as duplicate by Shaun, Asaf Karagila♦ general-topology StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Aug 13 '18 at 16:17

No, if $A$ is dense in $[0,1]$ then there exists an infinite set $B\subset A$ such that $A\setminus B$ is dense in $[0,1]$.
For example, say $(a_n)\subset A$ is a sequence with $a_{n+1}<a_n$ and $a_n\to 0$. Let $B=\{a_{2k+1}\}$. Then $A\setminus B$ is dense:
Since $a_{2k}\to0$, $A\setminus B$ contains a sequence that converges to $0$. Suppose that $x\in[0,1]$ and $x\ne0$. Choose a sequence $(b_n)\subset A$ with $b_n\to x$. Since $a_n\to0$ and $b_n\to x\ne0$ it follows that $B\setminus A$ contains all but finitely many of the $b_n$; hence $A\setminus B$ contains a sequence converging to $x$.