Topology - Nowhere Dense Sets, Perfect Sets in the Real Line R

- Let G be an open subset of R. Recall that G is a countable union of disjoint open intervals. We call (a, b) a component interval of G if (a, b) is contained in G and a not in G, b not in G.
Let A be a nonempty closed subset of R. Then G = R - A is open in R. Show that if A is nowhere dense and perfect, then for any two component intervals (a_i, b_i) and (a_j, b_j) of G with b_i ]ess than a_j, there is a component interval (a_l, b_l) of G such that b_j less than a_l, b_l less than a_j.
- Let A be a nonempty perfect subset of R. Then G = R - A is open in R. If for any two component intervals (a_i, b_i) and (a_j, b_j) of G with b_i ]ess than a_j, there is a component interval (a_l, b_l) of G such that b_j less than a_l, b_l less than a_j,. Does the set A have to be nowhere dense?
- Note that a decimal x in [0, 1] can be written as 0.r_1r_2...r_n.... Let
A = {x = 0.r_1r_2...r_n... in [0, 1] : r_i is not 6 for all positive integers i}.
Show that A is perfect.