I am trying to understand the fact that $e(p, v(p,y)) = y$. There is a proof in the text Advanced Microeconomic Theory (Jehle and Reny) that states the following:
Because $u(·)$ is strictly increasing on $R_n^+$, it attains a minimum at $x = 0$, but does not attain a maximum. Moreover, because $u(·)$ is continuous, the set $U$ of attainable utility numbers must be an interval. Consequently, $U = [u(0), u^b)]$ for $u^b > u(0)$, and where $u^b$ may be either finite or $+∞$.
To prove, fix $(p, y) ∈ R^{++}_n × R^+$. We know $e(p, v(p, y)) ≤ y$. We would like to show in fact that equality must hold. So suppose not, that is, suppose $e(p, u)<y$, where $u = v(p, y)$. Note that by definition of $v(·), u ∈ U$, so that $u<u^b$. By the continuity of $e(·)$ from, we may therefore choose $ε > 0$ small enough so that $u + ε<u^b$, and $e(p, u + ε)<y$.
I don't understand the point below:
Note that by definition of $v(·)$, $u ∈ U$, so that $u<u^b$.
$v(·)$ denotes the maximum utility that can be achieved for given prices and wealth, and I do not see why we cannot have $u = u^b$ such that $u ≤ u^b$ since $u^b$ may be finite. What am I missing here?