Solutions

(a) True. Consider the sequence

$$a_{n} = (0, 1, 0, 1, \dots)$$

with an inifinite number of 1s that diverges.

(b) False. We falsify this statement with a proof by contradiction. Suppose that such a sequence exists; let us denote it by $a_{n}$ and its limit by $a$ where $a \neq 1$. However, if we choose $\epsilon = |1 - a|$, then by definition

$$\exists N \in \mathbb{N}(\forall n \in \mathbb{N}(n \geq N \implies a_{n} \in V_{|1 - a|}(a))).$$

Notice that this is problematic because it implies that no 1s are in $V_{|1 - a|}(a)$ and, consequently, there there is a finite number of 1s (because there could be at most be $N - 1$ 1s). Thus, we encounter a contradiction.

(c) True. Consider the sequence

$$a_{n} = (0, 1, 0, 1, 1, 0, 1, 1, 1, \dots).$$

Clearly, it diverges and we can find $n$ consecutive number of ones for every $n \in \mathbb{N}$.