the proof of the supremum of the continuity function

Question

I totally have no idea about doing this question.Could someone help me?I'd appreciate it!

Assume $f(x)$ is continuous on $[a, b]$ for some real numbers $a < b$ and that $f(a) < y < f(b)$.

Define the set $S := \{z \in [a, b] | \forall x \in [a, z], f(x) < y\}$.

Let $c = \sup S$. Prove $c < b$.

Answer 1

We only have to show that $c\neq b$. For this, we prove that $f(c) < f(b)$. First of all, let $\varepsilon := f(b)-y$. As $f$ is continuous at $c$, there exists $\delta > 0$ such that $|x-c|<\delta$ implies $|f(x)-f(c)|<\varepsilon$. As $c = \sup S$, there exists $z\in S$ such that $|z-c| < \delta$. Hence, $|f(z)-f(c)| < \varepsilon$ and therefore $$ f(c) < f(z) + \varepsilon < y + \varepsilon = f(b). $$

Answer 2

Directly from the definition of continuity. There is $\delta>0$ such that for $x\in (b-\delta,b)$ $f(x)>y$. Hence $\sup(S)\leq b-\delta<b$.

In fact you only need continuity at $b$.