# Budget Set- closed and boundedness

I am fairly new to economics, and we were introduced to budget sets, The professor mentioned that the budget set $$B(p,w) = \{x \in R^{l}_{+}: px \leq w\}$$ is non empty and closed - I could prove the latter (the closed part) by taking sequences, but how do I claim that it is non empty?

Also, if the price of one commodity goes to zero, what happens to boundedness of $$B(p,w)$$? - I know zero prices means "free goods", and person can purchase infinite amounts of it. but how to prove it rigorously?

The budget set is always defined given a price vector $$p=(p_i)_{i\leq l}$$ (it seems like $$l$$ is the number of goods in your problem) and an income $$w$$. We usually implicitly assume that prices are strictly positive and the income is finite. Otherwise, as you correctly point out, we end up with infinite consumption being optimal and we don't even have an economic problem (loosely defining economics as the science of incentives regarding the allocation SCARCE goods).
A set is bounded if there exists an upper and a lower bound. We have the lower bound of $$x_i\geq 0$$ for all goods $$i$$ by construction of the problem. We can only have an upper bound on any quantity $$x_i$$ if $$p_i>0$$ and a finite $$w$$. Suppose some finite $$\overline x= (\overline x_i)_{i \leq l}$$ were an upper bound and $$p_i=0$$ for some $$i$$. Then you can see that this $$\overline x_i$$ cannot be the upper bound, because $$\overline x_i+\epsilon$$ is also in the budget set for any $$\epsilon>0$$.
EDIT: Sorry, I ignored the question on non-emptyness. Take any finite income and positive prices and note that $$x$$ is a continuous choice. Take the maximum price $$\overline p= \max_i (p_i)_{i\leq l}$$ and then consider a quantity vector $$\widehat x= (\widehat x_i)_{i\leq l}$$ with $$\widehat x_i= \frac{w}{l \overline p}$$ then $$\sum \widehat x_i p_i \leq \sum \widehat x_i \overline p = l \frac{w}{l \overline p} \underline p=w$$. So quantity vector $$\widehat x$$ is in the budget set such that it cannot be empty.