Whether a set is closed or not Denote by $C_{[0,1]}$ the ternary Cantor set on $[0,1]$. Now consider $[0,1] \setminus C_{[0,1]}$. It contains open intervals. Now define Cantor sets on all these open intervals by simply translating and dilating the standard Cantor set. Denote them as $C_{[a_i,b_i]}$. Now is the set $F=C_{[0,1]} \cup \bigcup C_{[a_i,b_i]}$ closed?
My try: I wanted to show that its complement is open. I argued that the complement of $F$ consists of open intervals. So $F^c$ consists of unions of open intervals. Hence $F$ is closed. But my professor says it is not so. Is it because the open intervals are getting smaller every time? I do not understand where i am going wrong. Can any one please tell me where I am going wrong. Thanks

Reworked solution after clearing up a misunderstanding of the question
Take a point $x \notin F$. Then especially, it is not in $C_{[0, 1]}$, so it must be contained in some open interval $(a, b)$ in the complement of $C_{[0, 1]}$ (assume for simplicity that $(a, b)$ is maximal, so that $a, b \in C_{[0, 1]}$).
It is clear that $F\cap [a, b] = C_{[a, b]}$ is a (translated and uniformly shrinked) Cantor set. Since $x \notin F$ and therefore not in $C_{[a, b]}$, we have an interval $(c, d)\subseteq (a, b)$ containing $x$ that does not intersect $F\cap[a, b]$. Hence $(c, d)$ does not intersect $F$, and since $x$ was arbitrary, this shows that $F$ is closed.
