Question

Find the supremum and infimum of each of the following sets of real numbers

Answer 1

Answer:

$\sup(S) = 3$.

$\displaystyle \inf(S) = \frac{1}{3}$.

Step-by-step explanation:

When factored, $3\,x^{2} - 10\, x + 3$ is equivalent to $(3\, x - 1)\, (x - 3)$.

$3\, x^{2} - 10\, x + 3 < 0$ whenever $\displaystyle x \in \left(\frac{1}{3},\, 3\right)$.

Typically, the supremum and infimum of open intervals are the two endpoints. In this question, $\sup(S) = 3$ whereas $\displaystyle \inf(S) = \frac{1}{3}$.

Below is a proof of the claim that $\sup(S) = 3$. The proof for $\displaystyle \inf(S) = \frac{1}{3}$ is similar.

In simple words, the supremum of a set is the smallest upper bound of that set. (An upper bound of a set is greater than any element of the set.)

It is easy to see that $3$ is an upper bound of $S$:

• For any $x > 3$, $3\,x^{2} - 10\, x + 3 > 0$. Hence, any number that's greater than $3\!$ could not be a member $S$.
• Conversely, $3$ would be greater than all elements of $S\!$ and would thus be an upper bound of this set.

To see that $3$ is the smallest upper bound of $S$, assume by contradiction that there exists some $\epsilon > 0$ for which $(3 - \epsilon)$ (which is smaller than $3\!$) is also an upper bound of $S\!$.

The next step is to show that $(3 - \epsilon)$ could not be a lower bound of $S$.

There are two situations to consider:

• The value of $\epsilon$ might be very large, such that $(3 - \epsilon)$ is smaller than all elements of $S$.
• Otherwise, the value of $\epsilon$ ensures that $(3 - \epsilon) \in S$.

Either way, it would be necessary to find (or construct) an element $z$ of $S$ such that $z > 3 - \epsilon$.

For the first situation, it would be necessary that $\displaystyle 3 - \epsilon \le \frac{1}{3}$, such that $\displaystyle \epsilon \ge \frac{8}{3}$. Let $z := 1$ (or any other number between $(1/3)$ and $3$.)

• Apparently $\displaystyle 1 > \frac{1}{3} \ge (3 - \epsilon)$.

• At the same time, $1 \in S$.

• Hence, $(3 - \epsilon)$ would not be an upper bound of $S$ when $\displaystyle \epsilon \ge \frac{8}{3}$.

With the first situation $\displaystyle \epsilon \ge \frac{8}{3}$ accounted for, the second situation may assume that $\displaystyle 0 < \epsilon < \frac{8}{3}$.

Claim that  $\displaystyle z:= \left(3 - \frac{\epsilon}{2}\right)$ (which is strictly greater than $(3 - \epsilon)$) is also an element of $S$.

• To verify that $z \in S$, set $x := z$ and evaluate the expression: $\begin{aligned} & 3\, z^{2} - 10\, z + 3 \\ =\; & 3\, \left(3 - \frac{\epsilon}{2}\right)^{2} - 10\, \left(3 - \frac{\epsilon}{2}\right) + 3 \\ = \; &3\, \left(9 - 3\, \epsilon - \frac{\epsilon^{2}}{4}\right) - 30 + 5\, \epsilon + 3 \\ =\; & 27 - 9\, \epsilon - \frac{3\, \epsilon^{2}}{4} - 30 + 5\, \epsilon + 3 \\ =\; & \frac{3}{4}\, \left(\epsilon\left(\frac{16}{3} - \epsilon\right)\right)\end{aligned}$.
• This expression is smaller than $0$ whenever $\displaystyle 0 < \epsilon < \frac{16}{3}$.
• The assumption for this situation $\displaystyle 0 < \epsilon < \frac{8}{3}$ ensures that $\displaystyle 0 < \epsilon < \frac{16}{3}\!$ is indeed satisfied.
• Hence, $\displaystyle 3\, z^{2} - 10\, z + 3 < 0$, such that $z \in S$.
• At the same time, $z > (3 - \epsilon)$. Hence, $(3 - \epsilon)$ would not be an upper bound of $S$.

Either way, $(3 - \epsilon)$ would not be an upper bound of $S$. Contradiction.

Hence, $3$ is indeed the smallest upper bound of $S$. By definition, $\sup(S) = 3$.

The proof for $\displaystyle \inf(S) = \frac{1}{3}$ is similar and is omitted because of the character limit.