I know that budget set is defined as $B(p,b) = \{x \in R_+^n: p.x \leq b \}$.
If $p \geq 0$ and $b \geq 0$, this budget set is called Walrasian budget set or the competitive budget set.
I also know that if $p>>0$, then Walrasian budget set is compact.
But why do we need the assumption that $p >> 0$? Why is not it compact for some $p_i = 0$?
If all $p_i > 0$, then each $x\in B(p, b)$ has $x_i \leq b_i / p_i$ for each $i$. The set $B(p, b)\subset \mathbb{R}^n$ is then closed and bounded, and thus compact. If some $p_i = 0$, then the projection $\pi_i:B(p, b) \to \mathbb{R}$ on to the $i$th coordinate is the noncompact set $\mathbb{R}^+$, so $B(p, b)$ itself is noncompact.