First of all I don't know why the totally ordered set should be non-empty
If $L$ is totally ordered, and $A \subseteq L$, then $A$ is bounded if and only if there exists $b,c \in L$ such that for all $a \in A, b \le a \le c$.
If $L$ is empty, then there is no $b, c \in L$ to satisfy the definition. It doesn't matter that $A$ is empty too.
It would be nice if someone give me an example of a finite set that has upper/lower bounds but they don't belong to the set
Easy: The set $\{1\}$ has lower bound $0$ and upper bound $2$.
Of course, the supremum and infimum of $\{1\}$ are both $1$, but you didn't say the bounds has to be extreme.
And if you are thinking I am just playing word games, you are right. But that is the real point here: so are you. No one claimed that there are finite sets which do not contain their extrema. Just because the author did not explicitly mention them does not mean they are claiming these do not always exist.
so showing that our subset contains a sup/inf or max/min would not be enough, since we need to consider a more general case :where either the upper/lower bound do belong to the subset or they don't.
No. An upper bound for $A$ is an upper bound for $A$ whether or not it is in $A$. So if $A$ has a maximum, that maximum is an upper bound, and since $A \subseteq L$, that upper bound is in $L$, and therefore $A$ is bounded above in $L$. Similar remarks apply to lower bounds.
You mention suprema and infima, but unlike maxima and minima, suprema and infima do not need to be in the set. In fact that is the difference between maxima and suprema, and between infima and minima. A maximum is a supremum that is contained in the set, and a minimum is an infima that is in the set.
Also, a set can be bounded and not have either one. For example, in the rational numbers $\Bbb Q$, the set $\{x\mid x \in \Bbb Q, x^2 < 2\}$ is bounded, but has no supremum or infimum.
- The base case would be an empty set,an emepty set has both upper/lower bounds, so the theorem is true for $P(0)$
As long as $L$ is not empty, this is true, since any element of $L$ satisfies the conditions to be both an upper and lower bound. If $L = \emptyset$, it is false.
The upper bound of B is either the upper bound of A or a (by assumption upper/lower bound of A exist), if upper bound of B is the upper bound of A then we are done, otherwise it would be a which can be seen that the upper bound of B does exist,so it can be concluded that the preposition is true for every k natural.
You've got the right idea here, but you have not demonstrated it. Effectively, you just said "it's true because it's true".
What you need to do is something like this:
"Let $b$ be an upper bound of $A$. Either $a > b$ or $a \le b$.
If $a > b$, then for all $x \in B$, either $x \ne a$, so $x \in A$ and $x \le b < a$, or $x = a$, so $x \le a$. In either case for all $x \in B, x\le a$, and $a$ is an upper bound.
Otherwise, if $a \le b$, then for all $x \in B$, either $x \in A$, so $x \le b$, or $x = a$, so $x\le b$. In either case, $b$ is an upper bound.
Therefore $B$ has an upper bound."
Best Answer
All polynomials are continuous, which is essentially saying that their graph could be traced without lifting your pencil from the paper. So then, if $\alpha$ is a root of a nonconstant polynomial, $p$,* the polynomial should pass $0$ at $\alpha$ and take a value near $0$ for $x$ near $\alpha$. But we made the assumption that $p$ is nonconstant so $p$ must take small positive/negative values. But then since $p$ is continuous, for a given side of $\alpha$, in the neighbourhood of $\alpha$, it should have the same sign for all values it takes. It should be strictly negative on one side and strictly positive on the other. This is essentially the intermediate value theorem.
By testing values near $\alpha$, we can check their sign and determine whether $p$ is above or below $0$. The root, $\alpha$ must always lie between an $a$ and $b$, such that $p(a)$ and $p(b)$ have opposite signs. When we test a value, $c$, between $a$ and $b$, we can check its sign and discern whether $\alpha$ is in $(a,c)$ or $(c,b)$. This is the method of bisection. We are iteratively improving the bounds by testing values in the interval. I believe there is no general closed formula for the $n$'th bound without a recurrence relation, since it relies on you actively choosing which side of $c$ contains $\alpha$.
* Ignoring roots of even multiplicity; if that is the case, the function would not have a change of sign and the method of bisection would fail.