Proove that the set A= \{ $x^{2}+x\cdot y^{2}$ : $-2 \leq x \leq 1, -1\leq y \leq 1$ } is bounded above and below Let A be the set  A= { $x^{2}+x\cdot y^{2}$ : $-2 \leq x \leq 1, -1\leq y \leq 1$ } 
Determine wether the set A, subset of the reals is bounded above, below both or neither. Rigorously justify your answer.
Notes : I'm not allowed to use limits or calculus . So far I've found A is bounded above and below simply by minimising and maximising the values of x and y, however I don't think this is rigorous enough. Is there any more formal proof showing that the set is bounded above and below ? Any help is much appreciated.
 A: You only need a lower and upper bound, not the greatest and least respectively.For that, $$|x^2+xy^2|\leq |x|^2+|x||y|^2\leq4+2=6$$, and you are done.
A: A pedestrian approach:
1) Lower bound of $A = x^2 + xy^2$ , where $-2 \leq x \leq 1$ and $-1\leq y \leq 1$.
Let $ \alpha: = y^2$, then $0\leq \alpha \leq 1$.
Consider:
A($\alpha$ )= $x^2$ + $\alpha$ x =
= $( x + \alpha /2)^2$ - $\alpha^2$  / 4.
This is a family of parabolas, $\alpha$ the parameter, with 
minimum = - $\alpha ^2$ / 4, $\\$
at ($-\alpha$ /2, -$\alpha ^2 $ / 4).
Choose $\alpha$ = 1 (y= +1 or y = -1) to find the curve with the smallest minimum of the family of  $\alpha$ curves.
Finally:
$$A(\alpha) = (x+\alpha /2)^2 - \alpha ^2 / 4   \geq  \\  
 -\alpha ^2 / 4 \geq - 1/4 .$$
2) Upper bound for A in the given x,y interval,
 a bit of patchwork.
Consider the intervals  - 2 $\leq$ x $\leq$ 0,  and  -1 $\leq$ y $ \leq$ 1 ,  I.e. $ \alpha$ $\geq$ 0.
$A(\alpha) = x^2 + \alpha x \leq A(\alpha=0) = x^2$ , recall that $\alpha \geq 0$, so the left hand side is smaller than $ x^2$ for negative x.
A($\alpha$) $\leq x^2 $, for - 2 $\leq$ x $\leq 0 $,
Setting x = - 2  gives A($\alpha$) $\leq$ A($\alpha$ = 0) = 4   in the intervals considered.
Almost done.
Now consider the remaining interval 0 $\leq$ x $\leq$ 1, $\alpha$ as before.
$$A(\alpha) = x^2 + \alpha x \leq A(\alpha = 1) = \\ =  x^2 + 1x ,$$ where all the x terms are positive (or zero) and $\alpha = 1$.
The maximum for A($\alpha=1$) = 2 at x=1.
For the complete interval -2 $\leq$ x $\leq$ 1, with $\alpha$  (y) in the given range:
Maximum = max( 2,4) = 4 .
A: Your $x\in[-2,1]$ and $y\in[-1,1]$ is a compact set. $x^2+xy^2$ is a continuous function so that both minimizing and maximizing $x^2+xy^2$ over the set has a solution. Set lower and upper bound equal to $x^2+xy^2$ evaluated at the minimum and maximum respectively.
