Prove by induction on $n$ that any set of $n$ reals is bounded
Working:
I approached the problem by splitting it into three cases and proving each case, it seems a bit tedious to me how I did it, so I was wondering if there's a quicker method. I also want to confirm whether my arguments are valid for each case. Here goes:
Base case: Consider when $n = 1$, i.e., the set $A = \{a\}$ where $a \in \mathbb{R}$. Then for $\epsilon \in \mathbb{R}^+ \cup \{0\}$, $a+\epsilon$ is an upper bound for $A$ and $a-\epsilon$ is a lower bound for $A$, hence $A$ is bounded.
Inductive hypothesis: Assume that any set of $k$ reals is bounded.
Proof: By the inductive hypothesis, let $B = \{a_1, a_2, \cdots, a_k\}$ where $a_i \in \mathbb{R}$ for $i = 1, 2, \cdots, k$. Then there exists some $u \in \mathbb{R}$ such that $a \le u$ for all $a \in B$ and similarly, there exists some $l \in \mathbb{R}$ such that $l \le a$ for all $a \in B$.
Now assume we add one more element, $a_{k+1}$, to $B$ and form the set $C = B \cup \{a_{k+1}\}$, now consider the following three cases:
1) If $\min(B) \le a_{k+1} \le \max(b)$ then $C$ is still bounded above by $u$ and below by $l$, hence $C$ is bounded.
2) If $a_{k+1} > \max(B)$, then $a_{k+1} \le u + |a_{k+1} - \max(B)|$, then $u + |a_{k+1} - \max(B)|$ is an upper bound for $C$, while $l$ is still a lower bound for $C$, hence $C$ is bounded.
3) If $a_{k+1} < \min(B)$, then $l - |\min(B) - a_{k+1}| \le a_{k+1}$, then $|\min(B) - a_{k+1}|$ is a lower bound for $C$ while $u$ is still an upper bound for $C$, hence $C$ is bounded.
Thus, we have shown that any set of $k+1$ reals is bounded, hence we are done.
Are my arguments valid? Are there any quicker ways of proving this inductively? Thanks.
Your answer is nice. A quick way to prove the inductive step:
Let $C= B\cup \{a_{k+1}\}$ and since $B$ is bounded then there's $M>0$ such that $$|b|\le M,\quad \forall b\in B$$ and let $M'=M+|a_{k+1}|$ then it's easy to see that $$|c|\le M',\quad \forall c\in C$$ hence $C$ is bounded.