# $a^*$ 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 $$a^{**}$$.

• $c =\sup\{$ all $x\in [a,b]$ so that $f(x) \le u\}$. That mean no number less than $c$ can be an upper bound of $\{$ all $x\in [a,b]$ so that $f(x) \le u\}$. So $c-\delta$ can not be an upper bound of $\{$ all $x\in [a,b]$ so that $f(x) \le u\}$. So there is a $a^* > c-\delta$ so that $a^* \in \{$ all $x\in [a,b]$ so that $f(x) \le u\}$. And thus $f(a^*)\le u$. Sep 24 at 18:37
• @fleablood But how can you assure that $f(c) < f(a^*) + \epsilon$? Sep 24 at 18:48
• And that is from continuity. If $c-\delta < a^*\le c$ then $|c-a^*| < \delta$ and so $|f(c) - f(a^*)| < \epsilon$ so $-\epsilon < f(a^*) - f(c) < \epsilon$ so $f(c) < f(a^*) + \epsilon < 2\epsilon$. Sep 24 at 18:53
• @fleablood Ah I see, so essentially it follows from the definition of continuity of which we only consider the right handside of the inequality? I guess it did not occur to me that we can only consider half of the inequality. Is that justified in general without having to check if the other side of the inequality is satisfied? Sep 24 at 18:59
• $|x-c|<\delta \iff c-\delta<x<c+\delta\iff x\in (c-\delta,c+\delta)$. By continuity, for every $\epsilon>0$ there is $\delta$ so that $x\in(c-\delta]\subset(c-\delta,c+\delta)\implies |f(x)-f(c)|<\epsilon\implies f(c)< f(x)+\epsilon$. And we don't care about $x\in(c,c+\delta)$ because $x\in(c,c+\delta)\implies f(x)>u$.... and By supremum there exists an $a^*\in(c-\delta, c]$ so that $f(a^*) \le u$. Sep 24 at 21:28

We have $$S=\{x\in[a,b]:f(x)\le u\}$$ and $$c=\sup(S)$$. This means two things:

1. 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 $$c, we have $$f(x)>u$$.
2. Since $$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, 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.
• Regarding 2: The double negatives always confuse me. I find it useful to think "if there were no such element $a^*$ between $x$ and $c$, then $x$ would be a better bound." Sep 24 at 18:41
• @So then does $f(c) < f(a^*) + \epsilon$ then follow from continuity? Since $a^* \in (c-\delta, c] \subset (c-\delta, c+\delta)$ and hence $f(a^*) - \epsilon < f(c) < f(a^*) + \epsilon$ and we only consider the right handside of the inequality? Sep 24 at 18:51

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

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

The $$f(c) < f(a^*) + \epsilon \le u + \epsilon$$.

So for every $$\epsilon > 0$$ we have

$$f(x) \le u + \epsilon$$.

• This has helped me pinpoint my confusion, which is the first part of your answer. Since continuity implies for any $\epsilon > 0$ there exists some $\delta > 0$ such that $|x-c| < \delta$ we have $|f(x) - f(c)| < \epsilon$, how does this imply $x \in (c-\delta, c] \Rightarrow f(c) < f(x) + \epsilon$? How are we sure that points to the left of $c$ will map to points $x$ such that $f(c) < f(x) + \epsilon$? Sep 24 at 18:56
• $|f(x)-f(c)|< \epsilon \implies -\epsilon<f(c)-f(x)<\epsilon\implies f(x)-\epsilon < f(c)<f(x)+\epsilon\implies f(c)<f(x)+\epsilon$. Sep 24 at 21:30
• So By continuity $c-\delta<x\le c\implies c-\delta<x<c+\delta\implies f(x)-\epsilon < f(c)<f(x)+\epsilon\implies f(c)<f(x)+\epsilon$. And by Supremum, there exists an $a^*: c-\delta< a^*\le c$ so that $a^*\in\{x|f(x)\le u\}\implies f(a^*)\le u$. So.... we have $c-\delta <a^*\le c$ so $f(c)< f(a^*)+\epsilon\le u + \epsilon$ for all $\epsilon$. That can only happen if $f(c)\le u$. [Now we need an argument that $f(c)\not< u$ but that's clear by continuity....] Sep 24 at 21:37