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Here is a low tech argument. It seems like not what the book intends, since I don't really need part (a).

Towards a contradiction, suppose that $f(a) > f(b)$. Pick some $r > 0$ so that $f(a) - r > f(b)$. Consider the set $X\subseteq (a,b]$ of points $x\in (a,b)$ so that $f(x) > f(a) - r$. By continuity of $f$ (near $a$), $X$ is non-empty so that $s = \sup X$ is defined. If $s < b$, then $s\in (a,b)$, so $s$ is not a shadow point. This implies there is a $t > s$ such that $f(t) > f(s) > f(b)$. Because $b$ is not a shadow point, for $y > b$, $f(b) \ge f(y)$, so we must have $s < t < b$, so that $t\in X$. The resulting contradiction implies that $s = b$. By continuity of $f$ (near $b$), we get $f(b) \ge f(a) - r > f(b)$. This final contradiction implies that $f(a) = f(b)$.

Here is a low tech argument. It seems like not what the book intends, since I don't really need part (a).

Towards a contradiction, suppose that $f(a) > f(b)$. Pick some $r > 0$ so that $f(a) - r > f(b)$. Consider the set $X\subseteq (a,b]$ of points $x\in (a,b)$ so that $f(x) > f(a) - r$. By continuity of $f$ (near $a$), $X$ is non-empty so that $s = \sup X$ is defined. If $s < b$, then $s\in (a,b)$, so $s$ is a shadow point. This implies there is a $t > s$ such that $f(t) > f(s) > f(b)$. Because $b$ is not a shadow point, for $y > b$, $f(b) \ge f(y)$, so we must have $s < t < b$, so that $t\in X$. The resulting contradiction implies that $s = b$. By continuity of $f$ (near $b$), we get $f(b) \ge f(a) - r > f(b)$. This final contradiction implies that $f(a) = f(b)$.

Here is a low tech argument. It seems like not what the book intends, since I don't really need part (a).

Towards a contradiction, suppose that $f(a) > f(b)$. Pick some $r > 0$ so that $f(a) - r > f(b)$. Consider the set $X\subseteq (a,b]$ of points $x\in (a,b)$ so that $f(x) > f(a) - r$. By continuity of $f$ (near $a$), $X$ is non-empty so that $s = \sup X$ is defined. If $s < b$, then $s\in (a,b)$, so $s$ is not a shadow point. This implies there is a $t > s$ such that $f(t) > f(s) > f(b)$. Because $b$ is a shadow point, for $y > b$, $f(b) \ge f(y)$, so we must have $s < t < b$, so that $t\in X$. The resulting contradiction implies that $s = b$. By continuity of $f$ (near $b$), we get $f(b) \ge f(a) - r > f(b)$. This final contradiction implies that $f(a) = f(b)$.

Here is a low tech argument. It seems like not what the book intends, since I don't really need part (a).

Towards a contradiction, suppose that $f(a) > f(b)$. Pick some $r > 0$ so that $f(a) - r > f(b)$. Consider the set $X\subseteq (a,b]$ of points $x\in (a,b)$ so that $f(x) > f(a) - r$. By continuity of $f$ (near $a$), $X$ is non-empty so that $s = \sup X$ is defined. If $s < b$, then $s\in (a,b)$, so $s$ is not a shadow point. This implies there is a $t > s$ such that $f(t) > f(s) > f(b)$. Because $b$ is not a shadow point, for $y > b$, $f(b) \ge f(y)$, so we must have $s < t < b$, so that $t\in X$. The resulting contradiction implies that $s = b$. By continuity of $f$ (near $b$), we get $f(b) \ge f(a) - r > f(b)$. This final contradiction implies that $f(a) = f(b)$.

Here is a low tech argument. It seems like not what the book intends, since I don't really need part (a).

Towards a contradiction, suppose that $f(a) > f(b)$. Pick some $r > 0$ so that $f(a) - r > f(b)$. Consider the set $X\subseteq (a,b]$ of points $x\in (a,b)$ so that $f(x) > f(a) - r$. By continuity of $f$ (near $a$), $X$ is non-empty so that $s = \sup X$ is defined. If $s < b$, then $s\in (a,b)$, so $s$ is not a shadow point. This implies there is a $t > s$ such that $f(t) > f(s) > f(b)$. Because $b$ is not a shadow point, this means that we have $s < t < b$, so that $t\in X$. The resulting contradiction implies that $s = b$. By continuity of $f$ (near $b$), we get $f(b) \ge f(a) - r > f(b)$. This final contradiction implies that $f(a) = f(b)$.

Here is a low tech argument. It seems like not what the book intends, since I don't really need part (a).

Towards a contradiction, suppose that $f(a) > f(b)$. Pick some $r > 0$ so that $f(a) - r > f(b)$. Consider the set $X\subseteq (a,b]$ of points $x\in (a,b)$ so that $f(x) > f(a) - r$. By continuity of $f$ (near $a$), $X$ is non-empty so that $s = \sup X$ is defined. If $s < b$, then $s\in (a,b)$, so $s$ is not a shadow point. This implies there is a $t > s$ such that $f(t) > f(s) > f(b)$. Because $b$ is a shadow point, for $y > b$, $f(b) \ge f(y)$, so we must have $s < t < b$, so that $t\in X$. The resulting contradiction implies that $s = b$. By continuity of $f$ (near $b$), we get $f(b) \ge f(a) - r > f(b)$. This final contradiction implies that $f(a) = f(b)$.