Solutions

(a) $x_{n} = (0, 1, 0, 1, \dots)$ is zero-heavy. If you choose $M = 1$, then

$$\forall N \in \mathbb{N}, \exists n \in \mathbb{N}, N \leq n \leq N + 1 \implies (x_{n} = 0 \wedge x_{n + 1} = 1) \vee (x_{n} = 1 \wedge x_{n + 1} = 0).$$

(b) Yes. For any $a_{i}$, if we choose $N = i$, we then have $\exists a_{i} = 0: i \leq i \leq i + M,$ implying that there is an infinite number of 0s.

(c) No. Consider the sequence $x_{n} = (0, 1, 0, 1, 1, 0, 1, 1, 1, \dots)$

(d) A sequence is not zero-heavy if for all $M \in \mathbb{N}$, there exists $N \in \mathbb{N}$ which for all $n$ satisfying $N \leq n \leq N + M$, $x_{n} \neq 0$.