# Convexity of production sets and input requirement sets

The following question is from Microeconomic Analysis by Hal R Varian.

True or false? If V(y) is a convex set, then the associated production set Y must be convex.

The solution available says;
False. There are many counterexamples. Consider the technology generated by a production function f(x) = x^2. The production set is Y = {(y, −x) : y ≤ x^2} which is certainly not convex, but the input requirement set is V (y) = {x : x ≥ √y} which is a convex set.

I am unable to understand how f(x) = x^2 is not convex, and V(y) is convex. Could someone please explain? Please give the intuition too :)

• The production function $f$ here is convex. It is the set $Y$ that is not convex. Jan 1 at 10:41

To see why $$Y$$ is not convex and $$V(y)$$ is convex, remember the definition of convex set, and then consider what are the two sets $$Y$$ and $$V(y)$$.
A set (of $$R^n$$ or of a more general vector space) is convex if the segment that joins any two points of the set is entirely contained in the set.
The set $$Y=\{(y, -x):y\leq x^2 \}$$ is a portion of plane: it is the part of plane between the graph of the function $$y=x^2$$ and the $$x$$ axis, if $$y\geq 0$$ (only the negative part, if $$x\geq 0$$).
$$V(y)= \{x: x \geq \sqrt y \}$$ is, instead, an interval, a portion of line, more precisely, it is the half-line $$[\sqrt y, +\infty)$$: the set of all $$x$$ equal or greater than $$\sqrt y$$, for a given $$y$$.