Show that the following property of the real numbers is generic: Let $b \in \mathbb{N}$ and $a=a_{1}a_{2}...a_{k}$ be a finite sequence of natural numbers such that $a_{i} \in \{0, 1, ..., b-1\}$
Show that a generic element $x \in \mathbb{R}$ repeats $a$ infinite times in the development of $x$ in base $b$.
I have been looking for a countable infinity of closed sets with empty interiors that compose the set of elements with the compliment of this property but am being thrown off in my head with the base $b$ and I'm not sure why.
Any ideas or clues on which sets to go for? Should I be using a different version of Baire's theorem?
*I thought of maybe looking at the stuff in between the $a$ repeats but I keep getting my mind stuck...
 A: In the comments to the question you define "generic" to mean that the numbers without this property form a meagre set.
It is enough to show the stronger property that for a generic real, the digit sequence $a_1\ldots a_k$ occurs infinitely many times at positions in the base-$b$ expansion that are multiples of $k$. Therefore, if we move to base $b^k$ instead, what we really need to show is that the set $B$ of numbers that contain only finitely many $a$s in their $b$-ary expansion is meagre.
Lemma. Let $B_0$ be the set of numbers that have no $a$s in their base-$b$ expansion to the right of the fraction point. Then $B_0$ is nowhere dense.
Proof. Let $x=n+[0.x_1x_2x_3\ldots]_b$ be an arbitrary positive real number. We can find intervals abitrarily close to $x$ in which every number contains an $a$, namely $$( n+[0.x_1x_2\ldots x_ma]_b, n+[0.x_1x_2\ldots x_ma]_b + b^{-(m+1)})$$ for as high an $m$ as we want. These intervals are outside $\overline{B_0}$, so $x$ is not in the interior of $\overline{B_0}$. The case for negative $x$ is similar with some strategic minuses thrown in.
Lemma. For every $q\in \mathbb N$, the set $B_q = \{xb^{-q}\mid  x\in B_0\}$ is nowhere dense. Proof. Immediate by the previous lemma.
Now, every number that contains finitely many $a$s is in one of the $B_q$s (Namely, let $q$ be one more than the index of the last $a$ in the number's expansion). Therefore $B=\bigcup_q B_q$. As a countable union of nowhere dense sets, it is by definition meagre.

($B_0$ and therefore $B_q$ and $B$ also have Lebesgue measure zero).
