I have a set
$$B=\{x=(x_1,x_2)\in R^2_+ \mid 2\sqrt{x_1}+x_2\le y\}$$
where $y>0$ is any number.
And I want to show that this set B is closed bounded and convex.
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My Solution is as follows:
First of all, I show this set in figure like that (but I am not sure about correctness, please tell me if it is wrong) Figure of the set is here. Is that figure true?
Secondly I need to show that the set $B$ is closed.
My idea: we know that any set in $R^2_+$ is closed. So, I try to show that the Set $B$ is closed. I guess I need to show that the complement of $B$ in $R^2_+$ is open. I.e. for every $(x_1, x_2)$ not in B $\exists$ r>0 s.t. $B_r(x1, x2) \cap B = \emptyset$.
Thus, let $f(x_1, x_2)=2\sqrt{x_1}+x_2$ $\forall (x1, x2)\in R^2$ is continuous function
So, $$B=\{x=(x_1,x_2)\in R^2_+ \mid f(x_1, x_2)\le y\}=f^{-1}[(0,y)]$$
So, $f:R^2_+\to R $ is continuous function then the set B is closed.
But I am exactly unsure about this proof. Please correct my solution.
Thirdly, I need to show that the set is bounded.
I guess this set is bounded both below and above. Thus, the set B is bounded.
For that, We say that a set is bounded above if there is a number $y$ (an upper bound so that every element in the set is no more than $y,$ and I could say that the set $B$ is bounded below if there is a number $M$ (which is equal to zero in this example) (a lower bound) so that every element in the set is no less than $M=0.$
But is this proof enough to show boundedness? What you think about this?
Finally, the set B is convex
I select two points $G=(x_1’,x_2’)$ and $N=(x_1’’,x_2’’)$ are in the set. Then I have $2\sqrt{x_1‘}+x_2’=2\sqrt{x_1’’}+x_2’’\le y$
So I use the definition
$$a(x_1’,x_2’)+(1-a)(x_1’’,x_2’’)\in B$$
That’s,
$$2[ax_1’+(1-a)x_1’’]^{1/2}+ [ax_2’+(1-a)x_2’’]\le y$$ for all $a\in [0,1]$
But I cannot further prove this is true for all $G$ and $N.$
Please check my answers and if those are wrong, please correct and improve them.
Thanks a lot.
The set is bounded because
0 <= $x_1$ <= $y^2$, 0 <= $x_2$ <= y.
The figure indicates it is not convex.
Consider the line connecting the $x_1$ and $x_2$ intercepts.
When showing the set is closed, the equation for B
ending with [{y}] is wrong. It should be [[0,y]].