6. A set $A$ of real numbers is said to be dense if every open interval of real numbers contains a point of $A$. For example, Problem 5 shows that the set of rational numbers and the set of irrational numbers are each dense.
(a) Prove that if $f$ is continuous (on $R$) and $f(x) = 0$ for all numbers $x$ in a dense set $A$, then $f(x) = 0$ for all $x$.
Proof:
We will use contradiction to prove the claim. Lets suppose that there exists a number $a$ such that $f(a) \neq 0$. If we can show that $f$ is not continuous at $a$, then, by contradiction, we are done. Lets assume that $f$ is continuous at $a$. Then for all $\epsilon > 0$ there exists a $\delta > 0$ such that for all $x$ in $R$, if $|x - a| < \delta$ then $|f(x) - f(a)| < \epsilon$. Since the statement must be true for all $\epsilon > 0$, if we choose any $\epsilon$ satisfying $0 < \epsilon \le |f(a)|$ (since $f(a) \neq 0, |f(a)| > 0$) then there should exist an appropriate $\delta$. For the purpose of precise demonstration we will choose $\epsilon = \frac{1}{2}|f(a)|$. Now, since $f(x) = 0$ for all $x$ in a dense set $A$ there must exist a number $x_0$ in the open interval $(a - \delta, a + \delta)$ such that $f(x_0) = 0$. Then $|x_0 - a| < \delta$ which implies that $|f(x_0) - f(a)| = |0 - f(a)| = |f(a)| < \epsilon = \frac{1}{2} |f(a)|$. The obvious contradiction that $|f(a)| < \frac{1}{2}|f(a)|$ proves that $f$ cannot be continuous at $a$. But, $f$ is continuous for all real numbers. By contradiction we must admit that there does not exist any number $a$ such that $f(a) \neq 0$. More appropriately, if $f$ is continuous and $f(x) = 0$ for all $x$ in a dense set $A$, then $f(x) = 0$ for all real numbers $x$.
$Q.E.D.$
(b) Prove that if $f$ and $g$ are continuous and $f(x) = g(x)$ for all $x$ in a dense set $A$, then $f(x) = g(x)$ for all $x$.
Proof:
Consider the function $h = f - g$. Since $f$ is continuous and $g$ is continuous, $h$ is certainly continuous (by Theorem 6.1). Then, for all $x$ in $A$, $h(x) = f(x) - g(x) = 0$. Now, since $h(x) = 0$ for all $x$ in a dense set $A$, by part (a), $h(x) = 0$ for all $x$. But this implies that $f(x) = g(x)$ for all $x$, has desired.
$Q.E.D.$
(c) If we assume instead that $f(x) \ge g(x)$ for all $x$ in $A$, show that $f(x) \ge g(x)$ for all $x$.
Proof:
Consider the function $h = f - g$. Since $f$ is continuous and $g$ is continuous, $h$ is certainly continuous (by Theorem 6.1). Then we have, for all $x$ in $A$, $h(x) = f(x) - g(x) \ge 0$. Lets suppose there exists a real number $a$ such that $h(a) < 0$. Since $h$ is continuous for all real numbers, $h$ must be continuous at $a$. Then, by Theorem 6.3, there must exist a $\delta$ such that for all $x$ in $(a - \delta, a + \delta)$, $h(x) < 0$. However, due to the density of the set $A$, there must exist a number $x_0$ in the open interval $(a - \delta, a + \delta)$ such that $h(x_0) \ge 0$. We must admit that there is no open interval such that for all $x$ in $(a - \delta, a + \delta)$, $h(x) < 0$, then $h$ is not continuous at $a$ (by the contrapositive of Theorem 6.3). But this contradicts the fact that $h$ is continuous for all real numbers. Now we must admit that there exists no number $a$ such that $h(a) < 0$. More appropriately, $h(x) \ge 0$ for all real numbers $x$. But this means $f(x) \ge g(x)$ for all $x$, has desired.
$Q.E.D.$
