= Solution
A set is <countable set>[countable] if it is finite or bijects with a subset of $\mathbb N$. For any $f:X\to\mathcal P(X)$, the <Cantor diagonal argument> set $D=\{x:x\notin f(x)\}$ differs from every $f(y)$ at $y$. Hence no such map is surjective. Binary sequences are indicator functions of subsets of $\mathbb N$, so $\{0,1\}^{\mathbb N}\cong\mathcal P(\mathbb N)$ is uncountable.
Every member of $L$ corresponds to a unique permutation $(a_0,a_1,\ldots)$ of $\mathbb N$, with $F_n=\{a_0,\ldots,a_{n-1}\}$. Let $p(j)$ be the position of $j$.
For $L_0$, $p(n)<n$ eventually. Choose $N$ after which this holds and then $m$ larger than all $p(0),\ldots,p(N-1)$. The $m$ positions $0,\ldots,m-1$ would contain all $m+1$ values $0,\ldots,m$, a contradiction. Thus \b[$L_0$ is empty], hence countable.
For $L_1$, $p(n)\leq n$ eventually. For every sufficiently large $m$, all values $0,\ldots,m$ occur in positions $0,\ldots,m$, so this initial segment is invariant. Comparing consecutive invariant initial segments gives $p(m)=m$ eventually. These are the finite-support permutations, of which there are countably many. Hence \b[$L_1$ is countable].
Back to article page
