I'm stuck on the first paragraph in Strang's proof of (part of) the Perron–Frobenius theorem, from his Introduction to Linear Algebra. Why can we assume $t_{\text{max}}$ exists, what guarantees that a maximum of the $t$'s "is attained"?
For a given nonnegative $\mathbf{x}$, I see why there must be a maximum $t$ such that $A \mathbf{x} \geq t \mathbf{x}$. We just start with $t = 0$ and increase it until $t x_i = (Ax)_i$ for some $i$.
But there might be many $\mathbf{x}$ that satisfy $A \mathbf{x} \geq t\mathbf{x}$ for some $t$. And if there are infinitely many, there may be infinitely many maximal $t$'s to choose from, one for each $\mathbf{x}$. In which case I don't see how to guarantee that there's one $t_{\text{max}}$ to rule them all.
Note that the maximum $t$ for each nonnegative $x$ is independent of the scaling of $x$ (by a positive real $r$). I.e., the maximum $t$ such that $A(rx) \ge t(rx)$ is the same as the maximum $t$ for which $Ax \ge tx$, since the two inequalities are equivalent.
Thus we can assume that the $x$s we want to maximize $t$ over all lie on the unit sphere. The nonnegative $x$s on the unit sphere form a closed and thus compact subset of the sphere. It's also I think fairly clear that the maximum $t$ for a given $x$ depends continuously on $x$ (if we add a small vector of length $\epsilon$ to $x$ the maximum $t$ will change by at most the matrix norm of $A$ times $\epsilon$). Then the maximum $t$ depends continuously on a parameter $x$ in a compact set. Thus the supremum of all $t$ across all $x$ is attained for some $x$.