I am now in engineering mathematics class and it goes over some basic set theory. Since I haven't had any experience with set theory, three statements leave me confused. Thanks for your help!
Statement:
1) General comment (not theorem or axiom): Set > Group > Ring > Field
2) Imagine set like set of Natural number is not aligned like on axis like we are all taught early but the elements jumble around without order.
3) Definition: A subset A of an ordered field F is bounded from below iff there exists p in F such that (for all x in A) (p <= x)
Question:
1) Why as convention "bounded set" is used instead of "bounded field"?
2) Could a set being "not ordered" and "bounded" at the same time?
3) Regarding optimization view point, is polytope view of a bounded convex set totally correct? Is polytope view just to help imagining? I.e. Polytope view implied ordered-ness on a set by implying spatial location of set element.
A set is simply a collection of objects. Sometimes, such as with natural or real numbers, a set can have an ordering on it. Other sets have no such ordering. One example is the set of complex numbers. Or 3-dimensional space. You can organize these sets somewhat, but not completely (at least without have an ordering with weird and undesirable properties). There are all sorts of sets that do not normally come with an order.
A group is a set and a binary operation on that set which obeys certain properties. Thus any Group is a Set. But a set does not have to be a group. Consider the set of red-headed men. What binary relation exists on this set that you can use to combine two red-headed men to produce another red-headed man? (And remember that there also has to be another red-headed man that when combined with the last one will produce the first!) So Groups are Sets, but Sets are not always Groups.
A Ring is a group with a second binary operation that distributes over the first. Thus any ring is a Group, but a group need not have a second binary relation stuck to it. A Group does not have to be a Ring.
A Field is a Ring where the second binary operation is almost a group operation in its own right. So every Field is a Ring, but a Ring does not have to be a field. For example, if I look at the set $Q =\{0, 1, 2, 3\}$, can make it a Group by giving it the operation $\oplus$, where I define $a \oplus b$ by first taking $a + b$, and if the result is greater than $3$, I also subtract $4$. I can further make $Q$ a Ring if I include the operation $\otimes$ defined by similarly reducing $ab$ by $4$ or $8$ if necessary to get the result back in $Q$: $$\begin{matrix}\begin{array}{c|ccc} \oplus & 0 & 1 & 2 & 3\\ \hline 0 & 0 & 1 & 2 & 3\\ 1 & 1 & 2 & 3 & 0\\ 2 & 2 & 3 & 0 & 1\\ 3 & 3 & 0 & 1 & 2\end{array} \qquad&\begin{array}{c|ccc} \otimes & 0 & 1 & 2 & 3\\ \hline 0 & 0 & 0 & 0 & 0\\ 1 & 0 & 1 & 2 & 3\\ 2 & 0 & 2 & 0 & 2\\ 3 & 0 & 3 & 2 & 1\end{array}\end{matrix}$$
To see that this ring is not a Field, note that in a field every non-zero element must have a multiplicative inverse, but looking at the $\otimes$ table, you can see that $2$ does not have one.
Thus every Field is a Ring. Every Ring is a Group. Every Group is a Set. But Sets do not need to be Groups, Groups do not need to be Rings, and Rings do not need to be Fields.
Why "Bounded set" not "Bounded Field"? Because it is the set that has a bound on it, not the field! If I take a set $A$ of real numbers (say let $A = \{x \in \Bbb R | 1 \le x^2 \le 4\}$, for example), and then note that there is a number $n$ such that $n \le x$ for all $x \in A$, (in this case, $n = -4$ is one example), does this fact tell me something useful about the set $A$, or something about the real numbers $\Bbb R$? It tells me something about $A$. It says that $A$ only goes so far down, but no farther. For the real numbers as a whole, it really doesn't say much. $\Bbb R$ is not bounded by $-4$. $\Bbb R$ has infinitely many numbers in it that are less than $-4$. So it is the set $A$ that is bounded, not the field $F$.
Can a set be bounded without being ordered? Yes, and no. By the particular definition you have given, the answer is no, as that definition makes explicit mention of $A$ being the subset of an ordered field, and the condition "bounded" is defined in terms of that ordering. But we can, and do, define "bounded" more generally. In particular, subsets of $\Bbb R^n$ are called "bounded" if there is some real number $R$ such that every point of the set is within a distance $R$ of the origin. Note that $\Bbb R^n$ is not ordered for $n > 1$.
For your final question, I don't know what you mean by the "polytope view".