I have the following problem:
- (continuous function)
Suppose that we consume non-negative amounts of a good $1$ and good $2$, namely, $x_1$ and $x_2$. The utility from consuming $(x_1,x_2)$ is defined by $u(x_1,x_2)=x_1+x_2$. For each $u^*\in \mathbb{R}^1$, let $UC(u^*)=\{(x_1,x_2)\in\mathbb{R}^2\mid u(x_1,x_2)\geq u^*\}$ be the collection of bundles $(x_1,x_2)$ that generate the utility of $u^*$ or higher [...].
I have to prove that for each $u^*$, the set $UC$ is closed but I have no idea how to prove $UC$ is closed in $\mathbb{R}^2$.
Notice that $$UC(u^*)=u^{-1}\big([u^*,\infty)\big).$$ Since $u$ is continuous and $[u^*,\infty)$ is closed (since $\mathscr{C}[u^*,\infty)=(-\infty,u^*)$ is open), we can conclude that $u^{-1}\big([u^*,\infty)\big)$ is closed.
I'm assuming you know that $u$ is continuous if and only if $u^{-1}(F)$ is closed whenever $F$ is closed.
Now, I'd argue that there is a mistake on the problem:
The problem state that $x_1,x_2$ are non-negative. I'd imagine you want to define $u\colon \mathbb{R}^2_+\to \mathbb{R}$. If not, why mention that $x_1$ and $x_2$ are non-negative? In this case, I'd define $UC(u^*)$ as $\{(x_1,x_2)\in \mathbb{R}^2_+\mid u(x_1,x_2)\geq u^*\}$. Then, $UC(u^*)=u^{-1}([u^*,\infty))$ would be closed in $\mathbb{R}^2_+$, but, since $\mathbb{R}^2_+$ is closed in $\mathbb{R}^2$, then $UC(u^*)$ is closed in $\mathbb{R}^2$ (since a closed set in a closed subspace is closed in the original space). The difference is very subtle but it was probably omitted intentionally to not have this somewhat technical element.