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(#5 in Rudin) Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that
$$
\inf A=-\sup (-A)
$$
Where the $\inf A$ is defined to be the greatest lower bound of $A$.
This question asks to establish a relationship between the infimum (greatest lower bound) of a set of real numbers and the negative of the supremum (least upper bound) of the set of the negatives of those numbers. Here’s the detailed proof:
First, let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ represent the set of all numbers $-x$, where $x$ is in $A$.
To show that $\inf A = -\sup(-A)$, we’ll prove two things:
Proving that $- \sup(-A)$ is a lower bound for $A$:
Suppose $a$ is in $A$. Then $-a$ is in $-A$. Since $\sup(-A)$ is the least upper bound of $-A$, we have $\sup(-A) \leq -a$, or equivalently $- \sup(-A) \geq a$. Since this is true for all $a$ in $A$, $- \sup(-A)$ is indeed a lower bound for $A$.
Proving that for any lower bound $b$ of $A$, we have $b \leq – \sup(-A)$:
Now suppose $b$ is a lower bound for $A$. Then for every $a$ in $A$, we have $b \leq a$, which means $-b \geq -a$. But $-a$ is in $-A$, and since $\sup(-A)$ is an upper bound for $-A$, we have $\sup(-A) \geq -b$ or equivalently, $b \leq – \sup(-A)$.
Since $- \sup(-A)$ is a lower bound for $A$ and every other lower bound $b$ of $A$ is less than or equal to $- \sup(-A)$, it follows that $- \sup(-A)$ is indeed the greatest lower bound for $A$, i.e., $\inf A = -\sup(-A)$. This completes the proof.
(#8 in Rudin) Prove that no order can be defined in the complex field that turns it into an ordered field. Hint: -1 is a square.
First, we need to clarify what an ordered field is. In mathematics, an ordered field is a field together with a total order on its elements that is compatible with the field operations. Specifically, for all $a, b, c$ in the field,
If $a < b$ then $a + c < b + c$.
If $a < b$ and $0 < c$, then $ac < bc$.
Now, consider the complex field. We’re asked to prove that no order can be defined that turns the complex field into an ordered field. We can do this by contradiction, using the fact that $-1$ is a square in the complex field.
Suppose for the sake of contradiction that an order can be defined that turns the complex field into an ordered field. Then consider the number $i$, a member of the complex field. Either $i > 0$ or $i < 0$ (since $i \neq 0$).
If $i > 0$, then $-1 = i^2 > 0$, but this contradicts the property of an ordered field since we know that $-1 < 0$ in any ordered field.
If $i < 0$, then $-1 = i^2 > 0$ again, which contradicts the property of an ordered field, as before.
In either case, we reach a contradiction, so our initial assumption that an order can be defined that turns the complex field into an ordered field must be incorrect. Therefore, no such order exists. This completes the proof.
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