proof any finite sequence of real numbers is bounded For any sequence of real numbers over a finite domain such as,
$ \langle a_n \rangle = a, a+1, \ldots, b $
Can someone show me (in formal terms) how it is I can go about defining a bijection between the range $ \left\{ f(n) : a ≤ n ≤ b \right\} $ and subset of natural numbers so that I can go on to use the fact that any finite set of real numbers is bounded to assert the existence of a maximal and minimal elements of the range.
Or if this is not the way to formally show that maximal and minimal elements of the range exist (obviously they do), can someone show me the way.
Thank you!!
 A: Let $A\subseteq\mathbb{R}$. The real number $\alpha$ is the maximal element of $A$ provided that $\alpha\in A$ and for all $a\in A$ we have $a\leq\alpha$.
For $n\in\mathbb{N}$, define $$\mathbb{N}_n=\{k\in\mathbb{N}:1\leq k\leq n\}$$
By definition, a set $A$ is said to have $n$ elements if there exists a bijection $$f:\mathbb{N}_n\to A$$
We will show that every finite subset of the real numbers has a maximal element. Showing that every finite subset of the real numbers has a minimal element is almost identical.
Define $$S=\{n\in\mathbb{N}:\text{ if }f:\mathbb{N}_n\to A\text{ is a bijection, then $A$ has a maximal element}\} $$ We want to show that $S$ is inductive, that is, if $1\in S$ and if $n\in S$, then $n+1\in S$.
Note: We require $A$ to be non-empty since the empty set cannot have a maximal or minimal element, since there are no elements in the empty set.
Base Case: Let $n=1$ and suppose $f:\mathbb{N}_1\to A$ is a bijection. Then $A$ has $1$ element, call it $f(1)$. Since $f(1)\in A$ and $f(1)\geq a$ for all $a\in A$, then $f(1)$ is the maximal element of $A$. This completes the base case.
Inductive Hypothesis: Let $n=k$. If $f:\mathbb{N}_k\to A$ is bijective, then $A$ has a maximal element.
Inductive Step: Let $n=k+1$. Suppose $f:\mathbb{N}_{k+1}\to A$ is bijective. One can show that the function $\hat{f}:\mathbb{N}_k\to A\setminus\{f(k+1)\}$ defined as $$\hat{f}(n)=f(n)$$is bijective. Then $A\setminus\{f(k+1)\}$ has $k$ elements, so by hypothesis, $A\setminus\{f(k+1)\}$ has a maximal element, call it $\alpha\in A\setminus\{f(k+1)\}$.
One can now distinguish two cases for bijection $f:\mathbb{N}_{k+1}\to A$:
(1) $f(k+1)\leq\alpha$. In this case, the maximal element of $A$ is $\alpha$, since $\alpha\geq a$ for all $a\in A\setminus\{f(k+1)\}$ and $\alpha\geq f(k+1)$.
(2) $f(k+1)>\alpha$. In this case, the maximal element of $A$ is $f(k+1)$ since $\alpha\geq a$ for all $a\in A\setminus\{f(k+1)\}$ and $f(k+1)>\alpha$.
In both cases, $A$ has a maximal element, so $k+1\in S$, which implies $S$ is inductive. This completes the proof.
Also note that this can be generalized to a set $A$ together with a non-strict total order $\leq$ and strict total order $<$, since the Inductive Step required us to compare two elements.
A: As you already said yourself, the statement is obviously true -- and this should usually count as a perfectly fine answer. It is not the point of formal proofs to be joylessly pedantic about things which are indeed clear to us!
One situation where it is useful to ask for a formal proof is this: For some reason you want to develop a formal theory for the mathematics that you are interested in. For this you usually start by expressing some basic facts (axioms) and some modes of reasoning that you are allowed to use (such as induction). Then in order to show that your formal system is strong enough, you can try to formally prove simple true facts (such as boundedness of finite sequences) using only the formal system that you developed. If this is possible, it shows you that you have chosen your axioms and inference rules wisely.
Long story short, to ask for a formal proof of a statement makes only sense if you tell us which are the axioms and inferences that you are allowed to use. If on the other hand you are interested in the truth of a statement, then any convincing argument is allowed, incuding appeal to intuition (which would be appropriate here).
