Need help with proof of intermediate value theorem 
However i need hint on how to disprove that $f(c)$ cannot be less than zero.
Thanks
 A: If $f(c) > 0$, then since $f$ is continuous, there is some positive $\epsilon$ such that $$\lvert x-c \rvert < \epsilon \implies \lvert f(x)-f(c) \rvert < f(c)$$
Since $f(c) > 0$, $\lvert f(x)-f(c) \rvert < f(c)$ means that $0 < f(x) < 2f(c)$, so $f(x) > 0$.
Thus, look at $c-\frac{\epsilon}2$.


*

*We have all $c-\frac{\epsilon} 2 \leq x \leq c$ is not in $K$ because $\lvert x-c \rvert < \epsilon \implies f(x) > 0$.

*We also have all $x > c$ is not in $K$ since $c$ is an upper bound of $K$.


Therefore, $x \geq c-\frac{\epsilon}{2} \implies x \notin K$, so $c-\frac{\epsilon}{2}$ is a upper bound of $K$, contradicting that $c$ is the least upper bound.

If $f(c) < 0$, then since $f$ is continuous, there is some positive $\epsilon$ such that $$\lvert x-c \rvert < \epsilon \implies \lvert f(x)-f(c) \rvert < -f(c)$$ Since $f(c) < 0$, $\lvert f(x)-f(c) \rvert < -f(c)$ means that $0 > f(x) > 2f(c)$, so $f(x) < 0$.
Thus, look at $c+\frac{\epsilon}{2}$. Since $\lvert c+\frac{\epsilon}{2}-c \rvert < \epsilon$, we have $f(c+\frac{\epsilon}{2}) < 0$ and thus $c+\frac{\epsilon}{2} \in K$, which contradicts that $c$ is an upper bound of $K$.

In my opinion, this is a pretty significant part of the proof, so I'm not sure why it's left out, but I guess that it was meant to be an exercise by the reader.
