I came across the following problems during my course of study in real analysis:
If $(A_n)$ is an increasing sequence of sets with union $A$ (i.e. $A_n \uparrow A$), then show that $A_n \to A$. State and prove the analogous proposition for decreasing sequences.
So $A_1 \subseteq A_2 \subseteq \cdots \subseteq A_n \subseteq \cdots$ and $A_1 \cup A_2 \cup \cdots \cup A_n \cup \cdots = A$. We want to show that $\text{lim inf} \ A_n = \text{lim sup} \ A_n = A$. Now if $x \in A_n$ for any $n$ then $x \in A$ by definition of union. Hence the result follows?
The analogous result would be: $A_n \downarrow A$ and $\bigcap A_n = A$, then $A_n \to A$? I suppose we could also use the intersection/union definition of limit supremums and limit infimums.
If $(a_n)$ and $(b_n)$ are null sequences, show that $(a_{n}b_{n})$ is a null sequence.
So for every $\epsilon >0$, $|a_n| \leq \epsilon$ for all $n > N_1$ and $|b_n| \leq \epsilon$ for all $n> N_2$. So choose $N_3 = \max \{N_1, N_2 \}$. Thus for every $\epsilon >0$, $|a_n b_{n}| \le \epsilon^{2} < \epsilon$ for all $n > N_3$?
Do these ideas sound correct?