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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?

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Even if $u^b$ is finite, it can never be achieved. This is what is meant by "does not attain a maximum". Rather, $u(x)$ approaches $u^b$ from below as $x \to \infty$. This is because $u$ is strictly increasing. If we had $u(x_*) = u^b$ for some $x_*$, then we would have $u(x_* + 1) > u^b$ and $u^b$ could not be a bound. This is why $U$ is written as the half-open interval $[u(0), u^b)$.

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