I attach the Cantor-like set construction here:
Let $\alpha \in (0, 1)$. Define the "middle-$\alpha$" of a closed interval by $$ M_\alpha([a, b]) = (-\alpha, \alpha) \frac{b - a}{2} + \frac{b + a}{2}. $$ If $S = \bigcup_{k = 1} ^m I_k$ is a disjoint union of closed intervals, then $M_{\alpha}(S) = \bigcup_{k = 1} ^m M_{\alpha}(I_k)$.
Given a sequence $(\alpha_j)_{j = 1} ^\infty$, construct cantor set as such: Let $S_0 = [0, 1]$ and for $j = 1, 2, \cdots$, let $S_j = S_{j - 1} - M_{\alpha_j}(S_{j - 1})$.
Let $S = \bigcap_{j = 1} ^\infty S_j$ (the $(\alpha_j)$-Cantor set). Then $S$ is closed and nowhere dense in $[0, 1]$ and $m(S) = \lim_{j \to \infty} m(S_j) = \prod_{j = 1} ^\infty (1 - \alpha_j) > 0$ iff $\sum \alpha_j < \infty$.
My question is:
At step 3, how does one conclude that the Cantor-like set $S$ is nowhere dense? An argument for $\frac{1}{3}$-Cantor set is that if there was an open interval in $S$, then the measure of $S$ would have been non-zero, absurd. However, this argument wouldn't work here by the immediate comment after saying $S$ is nowhere dense in the construction. Intuitively, any open intervals should not be in this set as we are "cutting" (through intersecting) the set infinitely many times, and eventually this cutting would get to any fixed open interval. However, how does one make this argument rigorous?
Hint:
One can show that a set is nowhere dense if and only if its complement contains a dense open (see here, say).
Can you show that the complement of a (fat) cantor set contains a dense open set?
I hope this helps ^_^