$a^*$ and $a^{**}$ in wikipedia's proof of IVT I am following the proof of the IVT on Wikipedia, and have one point of confusion. In the proof they say

By the properties of the supremum, there exists some $a^* \in (c-\delta, c]$ that is contained in $S$, and so
$$f(c) < f(a^*) + \epsilon \leq u + \epsilon$$
Picking $a^{**} \in (c, c+\delta)$, we know that $a^{**} \not \in S$ because $c$ is the supremum of $S$. This means that
$$f(c) > f(a^{**}) - \epsilon > u -\epsilon$$

What bothers me is how this proof is splitting the half intervals. How are we sure that for example $a^*$ exists? If this comes from continuity, how do we not know that such an $a^*$ only  exists in the other half of the interval? Likewise for $a^{**}$.
 A: We have $S=\{x\in[a,b]:f(x)\le u\}$ and $c=\sup(S)$. This means two things:

*

*Since $c$ is an upper bound for $S$, any number larger than $c$ is not in $S$. That is, for any $x$ in the domain with $c<x$, we have $f(x)>u$.

*Since $c$ is the least upper bound for $S$, any number smaller than $c$ is not an upper bound for $S$. In other words, if $x<c$, then since $x$ is not an upper bound there must be some element $a^*\in S$ which is larger than $x$. But $c$ is still an upper bound for $S$, so $x<a^*<c$.

A: 
Continuity means that for any $\epsilon > 0$ there exists a $\delta>0$ so that for all $|x-c|< \delta$ we have $|f(x)-f(c)|< \epsilon$.

So if $c-\delta < x \le c$ then $|x-c| < \delta$ and $|f(x)-f(c)| < \epsilon$ so $f(c) < f(x) + \epsilon$.

And $c =\sup \{y| f(y) \le u\}$ means that $c$ is an upper bound of  $\sup \{y| f(y) \le u\}$ and that any number less than $c$ will not be an upper bound of $\sup \{y| f(y) \le u\}$.

$c-\delta < c$ so $c-\delta$ is not an upper bound of $\sup \{y| f(y) \le u\}$.  And $c$ is an upper bound of $\sup \{y| f(y) \le u\}$.  That means there must exist a $a^*$ so that $c-\delta < a^* \le c$ where $a^* \in \sup \{y| f(y) \le u\}$.
So $a^* \in (c-\delta, c]$.  And we have $a^* \in \sup \{y| f(y) \le u\}$ so $f(a^*) \le u$.
So that means $f(a^*) + \epsilon \le u + \epsilon$.
And so we have

*

*$c-\delta < a^* \le c$

*For all $x\in(c-\delta, c]$ that $f(c) < f(x) +\epsilon$ so $f(c) < f(a^*) +\epsilon$.

*$f(a^*) +\epsilon \le u+ \epsilon$
The $f(c) < f(a^*) + \epsilon \le u + \epsilon$.
So for every $\epsilon > 0$ we have
$f(x) \le u + \epsilon$.
