Solutions

(a) $(-1)^{n}$ is frequently in the set $\{1\}$ because

$$\forall N \in \mathbb{N}, \quad a_{N + 1} \in \{1\} \vee a_{N} \in \{ 1\}.$$

(b) A sequence that is eventually in a set implies that it is frequently in it, also. To prove this assertion formally, suppose $a_{n}$ is eventually in the set $A \subseteq \mathbb{R}$. By definition,

$$\exists N_{1} \in \mathbb{N}: \forall n \in \mathbb{N}, \quad n \geq N_1 \implies a_{n} \in A.$$

Therefore,

$$\forall N \in \mathbb{N}: N \leq N_1, \quad n = N_{1} \implies a_{n} \in A$$

and

$$\forall N \in \mathbb{N}: N > N_1, \quad n = N \implies a_{n} \in A.$$

(c) $a_{n}$ converges to $a$ iif. for every $\epsilon > 0$, $a_{n}$ is eventually in $V_{\epsilon}(a)$.

(d) $x_{n}$ is frequently in $(1.9, 2.1)$ but not eventually necessarily in it. To prove that $x_{n}$ is frequently in that interval, notice that

$$\forall N \in \mathbb{N}, \exists n \in \mathbb{N}: n \geq N \implies a_{n} \in (1.9, 2.1)$$

because if the above assertion was false, then it would imply that there is a finite number of 2s, which is a contradiction. On the other hand, to prove that $x_{n}$ does not eventually in the interval, consider

$$x_{n} = (0, 2, 0, 2, 0, 2, \dots).$$

Clearly, this contrived sequence is not eventually in $(1.9, 2.1)$.