Given the prices $p \in \mathbb{R}_{+}^{k}$ and income $y \geq 0$, define the consumer's budget set as the set of feasible consumption bundles: $\beta(p,y) = \{x \in \mathbb{R}_{+}^{k}: <p,x>\leq y \}$. (The set $\mathbb{R}_{+}^{k}$ is the set of non-negative vectors of real numbers.) Show that, if $p_{j} \gt 0$ for all $j = 1, ... , k$, then $\beta(p,y)$ is compact.
I attempted the problem in contra-postive way. Compliment of $\beta(p,y)$ should be open. If you create an open ball radius $\epsilon > 0$ then the ball lies inside the set, which implies the complement is open and further the $\beta(p,y)$ is closed.
As $p_{j} >0 $ and $x \in R^{k}_{+}$ then $<p.x>$ is greater than or equal to $0$ which is lower bound and the $<p.x>\ge y$ and $y>0$. This implies x has upper bound.
Hence, compact.
The budget set is closed and bounded:
closed: if $\{x_n\}\subset\beta(p,y)$ and $x_n\to x$, then because $x_n\cdot p\leq y$ and the weak inequality is preserved under the limit, we have $x\cdot p\leq y$.
bounded: if $P=\min_j p_j>0$, then for any $x\in\beta(p,y)$, $$ y\geq p\cdot x=\sum_j p_j x_j\geq \sum_j P x_j=P(\sum_j x_j)\implies\sum_jx_j\leq\frac{y}{P}. $$ In particular, each $x_j$ is bounded in $[0,y/P]$.