# Expenditure function. Prove that this set is bounded

I need to prove that the following set is bounded in order to derive the expenditure function:

$$e(p,v)=min_x px$$ ST $$\{x \in R^n_+$$ such that $$U(x)\geq v\}$$.

Knowing that $$U(x):R^n \longrightarrow R$$ is a continous function.

I already proved that the set is closed and I think that if the set is closed for being $$U(x)$$ continous, the set must be bounded. However I'm not sure.

The set need not be bounded. To see this, just take $$U$$ to be constant. Then the set will either be empty or equal to $$\mathbb{R}_+^n$$.
It is also possible that no minimum exists. Let $$n=2$$, $$p=(0,1)$$, $$v=1$$, and $$U$$ be given by $$U(x)=U(x_1,x_2)=x_1\cdot x_2$$. Clearly, $$px=0$$ is only possible if $$x_1=0$$, which would lead to a utility of $$0$$. But for every $$\epsilon>0$$, the bundle $$(1/\epsilon,\epsilon)$$ has utility $$1$$ and a price of $$\epsilon$$. So no expenditure minimizing bundle exists. Of course, there is also no expenditure minimzing bundle if there exists no $$x$$ at all such that $$U(x)\geq v$$. This can happen if $$U$$ is bounded above.
However, an expenditure minimizing bundle exists if it holds that prices are strictly positive and there is at least one bundle $$x^*$$ such that $$U(x^*)\geq v$$. To see this, note that minimizing expenditure on the set $$\{x\in\mathbb{R}_+^n\mid U(x)\geq v\}$$ is equal to minimizing expenditure on the set $$\{x\in\mathbb{R}_+^n\mid U(x)\geq v, px\leq px^*\}.$$ The latter set is closed and bounded.