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I have only taken an excerpt from the book from Spivak 3rd edition page 220 in his "Inverse Function" chapter.

At the end of the 3rd paragraph, he says that

Then $f$ takes on some value $f(x) > y$ (because $\alpha$ is the least upper bound)

What does the sup have to do with anything? And why $f(x) > y$? Why not $f(x) < y$? I am guessing (forgive my stupidity here) his purpose in the third paragraph is to tell us that the domain of $f^{-1}$ must also take the form $(a,b)$ or $(a, \infty)$ (assuming $f$ is increasing)

EDIT Let's say $f$ has the domain $(a, b)$ and codomain $(f(c), \infty)$. Then does that mean $f^{-1}$ has a domain $(f(c), \infty)$ and codomain $(a,b)$?

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If $y < \alpha$, then $y$ is not an upper bound since $\alpha$ is the least upper bound. What does it mean for $y$ to not be an upper bound of $A$? Does that answer your question? – Ink Jun 22 '13 at 3:09

Because $y < \alpha$ and $\alpha$ is the least upper bound of $A$, this means that $y$ is not an upper bound of $A$. So some value in $A$ must properly be bigger (but $\le \alpha$ of course). So (by the definition of $A$) there is some $c \le x < b$ with $f(x) > y$.

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The author is trying to establish that $y$ is a value that $f$ takes for some argument. Directly finding such an argument is not possible, so the Intermediate Value Theorem will be called upon to provide one. As one already has the value $f(c)<y$, what is needed for this is a value $f(x)>y$; this question of utility is why one is not searching for an argument $x$ with $f(x)<y$ (although they certainly exist). Next, to be able to find an argument $x$ with $f(x)>y$ at all, it is important that such $x$ exists, in other words that $y$ is not an upper bound for the values of $f$. This is why $y$ was taken to be strictly less than the least upper bound$~\alpha$; without this hypothesis the argument could not work. But now this hypothesis gives the existence of $x$ with $f(x)>y$; having$f(c)<y<f(x)$, it suffices to apply the IVT and to conclude that $y$ is a value of $f$.

This show values $y$ with $f(c)<y<\alpha$ occur as values of $f$; since $\alpha$ is an upper bound for the values of $f$, it is also clear that values $z$ with $\alpha<z$ do not occur as values of $f$. This leaves only the question whether $\alpha$ itself occurs as value of$~f$, and the las sentence of the paragraph shows it does not.

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