Question

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

S = {3x 2 − 10x + 3 < 0}​

Answer 1

`\sup(S) = 3`.

`\displaystyle \inf(S) = (1)/(3)`.

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((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) = (1)/(3)`.

Below is a proof of the claim that
`\sup(S) = 3`. The proof for
`\displaystyle \inf(S) = (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 (1)/(3)`, such that
`\displaystyle \epsilon \ge (8)/(3)`. Let
`z := 1` (or any other number between
`(1/3)` and
`3`.)

• Apparently
  `\displaystyle 1 > (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 (8)/(3)`.

With the first situation
`\displaystyle \epsilon \ge (8)/(3)` accounted for, the second situation may assume that
`\displaystyle 0 < \epsilon < (8)/(3)`.

Claim that
`\displaystyle z:= \left(3 - (\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 - (\epsilon)/(2)\right)^(2) - 10\, \left(3 - (\epsilon)/(2)\right) + 3 \\ = \; &3\, \left(9 - 3\, \epsilon - (\epsilon^(2))/(4)\right) - 30 + 5\, \epsilon + 3 \\ =\; & 27 - 9\, \epsilon - (3\, \epsilon^(2))/(4) - 30 + 5\, \epsilon + 3 \\ =\; & (3)/(4)\, \left(\epsilon\left((16)/(3) - \epsilon\right)\right)\end{aligned}`.
• This expression is smaller than
  `0` whenever
  `\displaystyle 0 < \epsilon < (16)/(3)`.
• The assumption for this situation
  `\displaystyle 0 < \epsilon < (8)/(3)` ensures that
  `\displaystyle 0 < \epsilon < (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) = (1)/(3)` is similar and is omitted because of the character limit.