Proving a set to be closed 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. Can any one please tell me where I am going wrong. Thanks
 A: There are several ways to interpret this question. I will use (mainly) the following interpretation: For each 
(open) complementary interval $I_j$ of the standard Cantor set $C\subset J=[0,1]$ consider the canonical (increasing) affine bijection $f_j: I_j \to I=(0,1)$. Define $C_j:= f_j^{-1}(C)$ (note that this subset is not closed in $J$, however, it becomes closed if we add to it the end-points of the interval $I_j$, which are elements of $C$). Next, define
$$
K:= C\cup \bigcup_{j} C_j.  
$$
Lemma. The set $K$ is closed in $J$. 
Proof. Let us identify the complement $U=J\setminus K$. It is the union of the sets 
$$
U_j= I_j\setminus C_j 
$$
each of which is open in $J$ (as $C_j$ is closed in $I_j$ and $I_j$ is open in $J$). Note that the fact that $C_j$ is not closed in $J$ is irrelevant here.  Therefore, $J\setminus K$ is open and, hence, $K$ is closed in $I$. qed
Note that, by construction, $K$ is also a perfect totally disconnected set, hence, $K$ is homeomorphic to $C$. 
Here is another possible interpretation of the question: On step 1 apply the above construction to the standard Cantor set $C$ and denote the resulting set $K_1\subset J$. Now, apply the same construction to $K_1$ and obtain a new set $K_2$, etc. One, thus, obtains an increasing union of sets $K_m$ each homeomorphic to the Cantor set. Since the maximal length of complementary intervals to $K_m$ tends  to zero as $m\to \infty$, the union
$$
K_\infty= \bigcup_{m} K_m
$$ 
is dense in $[0,1]$. On the other hand, $K_\infty$ has empty interior (Baire's theorem). Hence, $K_\infty$ is not closed.   
