To prove this, let $y\in A$. Then $-y\in -A$. Since $m = \sup(-A)$ we have $-y\le m$. This implies $y\ge -m$. Thus we have shown that any $y\in A$ is bounded from below by $-m$. This means that $-m$ is a lower bound for $A$.
To prove this, assume that $y \gt -m$ is a larger number. Then $-y \lt m$. By definition, the number $m$ is the least upper bound of $-A$, and therefore $-y$ is not an upper bound for $-A$. It follows that there is an $a\in -A$ with $a\gt -y$. Negate both sides of the inequality and conclude that $-a \gt y$, which implies that $y$ is not a lower bound for $-A$.
Since $F$ consists of real numbers, and since addition and multiplication are defined to be the same as for real numbers, we do not have to check properties that are already true for the real numbers, such as the commutative laws $a+b=b+a$, $ab=ba$, associative laws like $(a+b)+c = a+(b+c)$, and the distributive law $a(b+c) = ab+ac$. We do have to check that the special elements $0$, and $1$ are present in $F$: they are, since \[ 0 = 0+0\sqrt3, \text{ and } 1 = 1 + 0\sqrt3. \] We also have to check that the inverses for addition and multiplication of any number $a+b\sqrt3$ again belong to $F$. For the additive inverse we have \[ -(a+b\sqrt3) = p+q\sqrt3, \text{ with } p= -a, q=-b. \] For the multiplicative inverse we get \[ \frac{1}{a+b\sqrt3} = \frac{a-b\sqrt3}{a^2-3b^2} = p+q\sqrt3 \] with \[ p= \frac{a}{a^2-3b^2}, \qquad q = \frac{-b}{a^2-3b^2}. \] Here it is important to observe that we are not dividing by zero, i.e. that $a^2-3b^2\ne0$. If $a^2-3b^2=0$ were true, then we would have $3=(a/b)^2$, but $\sqrt3$ is not rational, so this cannot be true.
In this solution the following version of the Archimedean property is used: for any $x>0$ there is a positive integer $n\in\mathbb{N}$ such that $10^{-n}\lt x$. That is not exactly the Archimedean property as stated in Rudin’s book. In fact proving that this version also is true is practice problem 7a for the first midterm.
Is $\mathbb Q$ an open subset of $\R$? No. Proof: $0$ is not an interior point of $\mathbb{Q}$ because every interval $(-r, r)$ contains an irrational number. (For example, given $r\gt0$, choose $n\in\mathbb{N}$ so large that $\frac1n\sqrt2 \lt r$.)
Is $\mathbb{Q}$ a closed subset of $\R$? No. For example, $\sqrt2$ is a limit point of $\mathbb Q$, because every interval $\bigl(\sqrt2-r, \sqrt2+r\bigr)$ contains a rational number (Rudin, theorem 1.20b, “$\mathbb Q$ is dense in $\R$”).
Does $\R\setminus\mathbb Q$ have any interior points? No. Let $x\in\R\setminus\mathbb Q$ be given. Then every interval $(x-r, x+r)$ contains a rational number, no matter how you choose $r$. Therefore $x$ is not an interior point of $\R\setminus\mathbb Q$.
Is $\mathbb Z$ an open subset of $\R$? No. We have to disprove that every point in $\mathbb Z$ is an interior point, so we only have to show that one of the points in $\mathbb Z$ is not an interior point. For example, $0$ is not an interior point of $\mathbb Z$ because any interval $(-r, r)$ contains a noninteger.
Is $\mathbb Z$ a closed subset of $\R$? Yes. It is easiest to show that $\R\setminus\mathbb Z$ is open. Let $x\in\R\setminus\mathbb Z$ be given, and let $n$ be the integer for which $n\lt x\lt n+1$. Choose $r=\min(x-n, n+1-x)$. Then the interval $(x-r, x+r)$ is a neighborhood of $x$ which is contained in $\R\setminus\mathbb Z$, so that $x$ is an interior point of $\R\setminus\mathbb Z$.
