Solutions

(a) True. By the order limit theorem, for any sequence $a_{n}$ that converges to $a$,

$$\forall c \in \mathbb{R}, \forall n \in \mathbb{N}, \quad a_{n} \geq c \implies a \geq c.$$

By applying this theorem on the set $B$, we have

$$\forall b \in B, \forall n \in \mathbb{R}, \quad a_{n} \geq b \implies a \geq b$$

and we can conclude that the limit of $a_{n}$ also is an upper bound for all terms belonging to $B$.

(b) False. We do a simple proof by contradiction. Suppose that the sequence $a_{n}$ has all of its terms in

$$(0, 1)^{c} = (-\infty, 0] \cup [1, \infty)$$

but that its limit $a$ is in $(0, 1)$. By definition, we can then choose $\epsilon = a$ for which

$$\exists N \in \mathbb{N}, \forall n \in \mathbb{N}, \quad n \geq N \implies \abs{a_{n} - a} < a.$$

However, this implies that $a_{n}$ is in $(0, 1)$ for every $n \geq N$ when we assumed that no $a_{n}$ is in $(0, 1)$.

(c) False. We can create sequences that approach an irrational number via only rational ones. The question is: do we have an infinite number of irrational numbers ever closer to it? The answer to that is yes since the rational numbers are dense in $\mathbb{R}$.