A property of Lusin sets Here is the translation of a part of the proof of Thm 7 of Sierpinski's Hypothèse du continu that puzzles me :
"Let $Q$ be a Lusin set. [...]  Let $N$ be an uncountable subset of $Q$. Since $Q$ is a Lusin set, there exist, as we know, an interval $I$ such that $N$ is dense in $I$."
I don't see why such an interval actually exist.
 A: By definition every nowhere dense subset of $Q$ is countable. If some $N\subseteq Q$ is not dense in any interval, it is of course nowhere dense, so it must be countable.
A: For clarity, note that the ambient space is $\mathbb{R}$ (otherwise we can not speak of intervals).
"$Q$ is Lusin" means that $Q$ is uncountable and every uncountable subset $N\subset Q$ is non-meager. This means $N$ is not a countable union of nowhere dense sets; in particular, $N$ is not nowhere dense.
We will prove that if $A\subset \mathbb{R}$ is not nowhere dense, then it is dense in some interval $I$. Indeed, consider $\text{cl}(A)$; it has non-empty interior. So there is an open basis set $B\subset \text{cl}(A)$. Since the standard topology on $\mathbb{R}$ is generated by basis sets consisting of open intervals $(a,b)$, $B$ could be taken of this form. But then $A$ is dense in $B$ (since $B\cap \text{cl}(A)=B$), and we may take $I=B$ as our interval. $\square$
Thus "not nowhere dense" means "dense in some interval". Since every uncountable $N\subset Q$ is "not nowhere dense", the result follows.
