I'm not really getting anywhere with the proof for the following:
(a) Let $a$ be a positive infinite decimal expansion. We consider a sequence of a finite decimal expansion
$$a^{(n)} = a_0,a_1a_2\ldots a_n00\ldots$$
Prove that: $\sup\{a^{(n)}: n \geq 1\} = a$
(b) Let $A$ and $B$ be sets consisting of positive infinite decimal expansions, such that for all $a \in A$ and $b \in B$, $a \leq b$.
Show that there exists an infinte decimal expansion $c$, such that $a \leq c \leq b$ for all $a \in A$ and $b \in B$.
I'm not quite sure how to prove this. I guess that in (b) they want to prove completeness.
A) Two things to prove: i) $a = \sum_{k=1}^{\infty}a_k\cdot 10^{-k} \ge \sum_{k=1}^n a_k\cdot 10^k=a^{(n)}$ for any $n$. Thus $a$ is an upper bound of $\{a^{(n)}\}$. And ii) If $d < a$ then there is an $a^(n) > d$. Thus any $d <a$ is not an upper bound of $\{a^{(n)}\}$. So, by definition, $a =\sup\{a^{(n)}\}$.
i) is fairly obvious all $a_j \ge 0$ so $a - a^{(n)} = \sum_{j=n+1}^\infty a_j\cdot 10^{-j} > 0$.
ii) is not so obvious but still easy. If $d < a$ then $a-d >0$ and there is a natural $n$ so that $0 < 10^{-(n-1)} < a-d$. $a - a^{n} = \sum_{j=n+1}^\infty a_j\cdot 10^j \le 10^{-n} < 10^{-(n-1)}$ so $a-10^{-(n-1)} < d < a$. (You may have to prove by induction that $\sum_{j=m+1}^\infty a_j\cdot 10^{-j} \le 10^m$. )
For B).... Let me get back to you on that.