The following is an exercise from Calculus by Spivak.
Consider the set $S=\lbrace x\in\mathbb{R} : x^2+x-1<0\rbrace$. What is $\sup(S)$ and $\inf(S)$?
My Attempt for $\sup(S)$: Let $f(x)=x^2+x-1$. We show that $f(x)$ is continuous at all points $a\in\mathbb{R}$. Let $\epsilon>0$ be arbitrary. If $|x-a|<1$ then $-1<x-a<1$ and $a-1<x<1+a$ so $x+a+1<2a+1$. Choose $\delta_0=\min\lbrace 1,\frac{\epsilon}{|2a+1|}\rbrace$. Then $|x-a|<\delta_0$ implies \begin{align*} |f(x)-f(a)| & =|x^2+x-a^2-a| \\ & =|x^2-a^2+x-a| \\ & =|(x-a)(x+a)+(x-a)| \\ &=|(x-a)(x+a+1)| \\ & <\frac{\epsilon}{|2a+1|}\cdot|x+a+1| \\ & <\epsilon \end{align*} Therefore, $\lim_{x\to a}f(x)=f(a)$ for all $a\in\mathbb{R}$. Let $a_0=\frac{-1+\sqrt{5}}{2}$. Because $\lim_{x\to a_0}f(x)=f(a_0)$ we have $\lim_{x\to a_0^-}f(x)=f(a_0)$. Then for any $\epsilon>0$ there exists $\delta>0$ such that $a_0-x<\delta$ implies $|f(x)-f(a_0)|=|f(x)|<\epsilon$. If $x\in\mathbb{R}$ satisfies $a_0-x<\delta$ then $a_0-\delta<x$.
Two problems:
I'm stuck after here, and I don't know if I'm proceeding with the correct methods at all.
Even if this is not the right path for the proof, is my proof of the continuity of $f(x)$ valid?
Just solve $x^2+x-1 = 0$, and call the roots $a, b$, say $a < b$. Then $S = (a,b)$, and $\sup(S) = b, \inf(S) = a$