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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 :)

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  • $\begingroup$ The production function $f$ here is convex. It is the set $Y$ that is not convex. $\endgroup$ Jan 1, 2023 at 10:41

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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 $x\geq 0$).

It is not convex, as you can take two points such that the segment joining them doesn't belong entirely to it.

$V(y)= \{x: x \geq \sqrt y \}$ 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 $y$.

A half-line is a convex set: if you take any two points of it, the segment joining them belongs to the half-line.

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