Is the set $A =\{x: x^2 < 3x\}$ bounded, find $\sup A$ and $\inf A$ if they exist?

Question

Is the set $A = \{x: x^2 < 3x\}$ bounded? Find $\sup A$ and $\inf A$ if they exist.

My answer is no, it is unbounded and so it has no $\sup$ and $\inf$. Am I correct?

Answer 1

$$ x^2 - 3x = x(x-3) < 0 $$

So there are $3$ options.

$x<0$ then both $x$ and $x-3$ are negative so the inequality is not satisfied.

$0<x<3$ then the $x$ factor is positive but $x-3$ is negative so the inequality is satisfied.

$x>3$ then both factors are positive so the inequality is not satisfied.

$A=(0,3)$. It is bounded, has $\inf=0$ and $\sup=3$.

Answer 2

Note that$$x^2 - 3x < 0$$

$$x(x-3) < 0$$

Hence $$0<x<3$$

Note that $0$ is a lower bound. Try to find out the infimum and supremum.

Answer 3

If $x \ge 3$ then $x^2 = x \cdot x \ge 3x$ and $x \not \in A$ so $3$ is an upper bound.

And if $0 < x < 3$ then $x^2 = x\cdot x < 3x$ so $x \in A$ so $(0,3) \subset A$ and no $a < 3$ is an upper bound of $A$. So $3 = \sup A$ by definition.

If $x < 0$ then $3x < 0$ and $x^2 > 0$ and $x^2 \ge 3x$ so $x \not \in A$. So $A$ is bounded below by any negative number. Forthermore $0^2 = 3*0$ so $0 \not \in A$. So $A$ is bounded below by $0$.

If $a > 0$ then $\frac a2 < a$ and $\frac a2 \in A$ so no $a > 0$ is an lower bound so $\inf A = 0$ by definition.