Uncountable Proper Dense Subsets of Real Line

Question

We know that the set of irrational numbers is an uncountable proper dense subset of real numbers. Also the set of transcendental numbers is a proper uncountable subset of irrationals which is dense in real line.

My question is what is next? And how far we can continue these uncountable proper dense chain of subsets?

Do we have an infinite chain of uncountable proper subsets which are dense in real line?

Answer 1

If $S$ is a dense subset of $\Bbb R$ then there is a countable $T \subset S$ where $T$ is dense in $\Bbb R.$ Proof: Let $J$ be the set of non-empty bounded open real intervals with rational endpoints. For $j\in J$ let $f(j)\in j\cap S$. Let $T=\{f(j):j\in J\}.$

If $S$ is an uncountable dense subset of $\Bbb R,$ let $T$ be a countable subset of $S$ such that $T$ is dense in $\Bbb R.$ Let $U=S\backslash T.$

Now $U$ is uncountable. It is a consequence of the Axiom of Choice, (which was already used in the first paragraph to get the function $f$), that $U$ has an uncountable subset $V$ which can be well-ordered by a relation $<_w$ such that for all $v\in V$ the set $\{v'\in V:v'<_wv\}$ is countable.

So for $v\in V$ let $S(v)=S\backslash \{v'\in V:v'<_wv\}.$ Every $S(v)$ is dense in $\Bbb R$ because $S(v) \supset T.$ And if $v_1<_wv_2$ then $S(v_1)\supsetneqq S(v_2).$

BTW. If $v_0$ is the $<_w$-least member of $V$ then $S(v_0)=S.$

Answer 2

If by chain, you mean a collection of subsets totally ordered by inclusion, then $\{(a,\infty):a\in\Bbb R\}$ is an easy example, and is order-isomorphic to $\Bbb R$.