Solutions

(a) Notice that the sequence $y_{n}$ is decreasing since supposing the opposite would lead to a contradiction. For contradiction, suppose that $y_{n + 1} > y_{n}$. Then,

$$y_{n + 1} > y_{n} \implies \sup \{ a_{k} : k \geq n + 1 \} > \sup \{ a_{k} : k \geq n \}$$

and this would then imply that $y_{n}$ is a smaller upper bound for $\{ a_{k}: k \geq n + 1\}$, directly contradicting the definition of a supremum.

Furthermore, since the sequence $a_{n}$ is bounded below (from simply being bounded), then $y_{n}$ also is. Using $l$ to denote the lower bound of $a_{n}$, that is because

$$\forall n \in \mathbb{N}, \quad l \leq \min_{k \geq n} a_{k} \leq \sup_{k \geq n} a_{n} = y_{n}.$$

Therefore, by applying MCT, we know that $y_{n}$ converges.

(b) We define $\lim \inf y_{n}$ as

$$\lim \inf y_{n} = \inf \{ a_{k} : k \geq n \}.$$

The reason why it always exists for bounded sequences is analog to the reason given in (a). That is, $y_{n}$ is increasing and is bounded above, thereby allowing us to apply MCT.

(c) Consider any bounded sequence $a_{n}$. Then, notice that

$$\forall n \in \mathbb{N}, \inf \{ a_{k} : k \geq n \} \leq \sup \{ a_{k} : k \geq n \} .$$

Additionally, knowing from (a) and (b) that both sequences formed converge (with the inf and sup functions*), we can apply the Order Limit Theorem to obtain

$$\lim \inf \{ a_{k} : k \geq n \} \leq \lim \sup \{ a_{k} : k \geq n \}.$$

A sequence for which is inequality is strict is the alternating sequence

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

Clearly, $\lim \inf a_{n} = 0$ and $\lim \sup a_{n} = 1$.

(d) In the forward direction $(\implies)$, suppose a bounded sequence $a_{n}$ for which

$$\lim \inf a_{n} = \lim \sup a_{n} = l.$$

Then, we can apply the Squeeze Theorem previously proved since

$$\forall n \in \mathbb{N}, x_{n} \leq a_{n} \leq y_{n}.$$

This then means that $a_{n}$ converges to the same limit as $l$.
Now, for the converse $(\impliedby)$, assume that $a_{n}$ converges to $l$. Then, construct the sequences $x_{n}$ and $y_{n}$, defined as

$$x_{n} = \inf \{ a_{k} : k \geq n \}$$

and

$$y_{n} = \sup \{ a_{k} : k \geq n \} .$$