How to argue that $\min \left\lVert Ax-b \right\rVert_2$ on $M := \{ x \in \mathbb{R}^m \mid x \ge 0 \}$ has a solution This is related to an old question of mine, I was not sure if I should make a new question, so please tell me if I should edit the old one and I will delete this one.
Consider the following exercise:

Let $A \in \mathbb{R}^{m \times n}$ have rank $n$ and let $b \in \mathbb{R}^m$. Why does the problem $$\min h(x)$$ for  $h(x) := \left\lVert Ax-b \right\rVert_2$ on $M := \{ x \in \mathbb{R}^m \mid x \ge 0 \}$ have a solution?

I understand that $h$ is a continuous function, but $[0,\infty)$ is not a compact interval, so the usual argument "continuous + compact intercal" does not work here. Could you please give me a hint on how to establish the existence of a minimum for $h$ on $M$?
 A: Typo in question: $M$ should be $\mathbb{R}_{+}^{n}$ instead of $\mathbb{R}_{+}^{m}$, also need $m\geq n$ so that $A^{\top}A$ is positive definite.
Pick an arbitrary $x_{0}\in M$, let $M_{r}=\left\{x\in M\middle|h\left(x\right)\in\left[0,h\left(x_{0}\right)\right]\right\}$. It's easy to see that, if $h$ on the restricted set $M_{r}$ has a minimum, it will also be a minimum on $M$. So we only need to prove the existence of minimum on $M_{r}$.
Firstly, it's easy to show $M_{r}$ is closed since $h$ is continuous. What remains is to prove $M_{r}$ is bounded. You can do this by contradiction. Suppose some component in $x$ is unbounded, $\left\|Ax\right\|_{2}$ will go to infinity since $A$ has full column rank. Note that $\left\|Ax-b\right\|_{2}\geq\left|\left\|Ax\right\|_{2}-\left\|b\right\|_{2}\right|$, hence will also go to infinity, contradict with $h\left(x\right)\leq h\left(x_{0}\right)$.
So far you can show $M_{r}$ is closed and bounded hence compact, applying extreme value theorem on the restricted problem completes the proof.
