Determine whether the following subsets of $\mathbb{R}$ are bounded.

Question

$A=\{x+\frac{1}{x}:x \in (0,\infty)\}$

$B=\{x^2+xy^2:-2 \leq x \leq 1, -1 \leq y \leq 1\}$

I understand the what it means for a set to be bounded above and below, but how would I go about proving this rigourously?

Answer 1

For B, note that we have $\left|x\right| \leq 2, x^2 \leq 4$ and $y^2 \leq 1$. Thus we get

$\displaystyle\left|x^2 + xy^2\right| \leq \left|x^2 \right| + \left|xy^2\right| \leq 4 + 2 \times 1 = 6$

That means that e.g. $B$ is contained in $(-10, 10)$.

EDIT:

$B$ is bounded below by $-\frac{1}{4}$: assume $x < 0$ (if $x \geq 0$ then $x^2 + xy^2 \geq 0$). Then from $y^2 \leq 1$, multiplying both sides by $x$, we get $xy^2 \geq x$ and thus $x^2 + xy^2 \geq x^2 + x$. Using high school methods we have $x^2 + x \geq -\frac{1}{4}$. (I'm guessing there is a better way to prove this bound but whatever...)

$A$ is bounded below by $2$ because, by AM-GM inequality, we have $x + \frac{1}{x} \geq 2\sqrt\frac{x}{x} = 2$

Answer 2

$A$ obviously isn't bounded as $x$ is not bounded.