I am having a little trouble understanding one of the steps in this proof.

From Stephen Abbott’s *Analysis*:

Using AoC to prove the IVT:

TO simplify matters, consider $f$ as a continuous function which satisfies $f(a)<0<f(b)$

and show that $f(c) =0 $ for some $c in (a,b)$. First let

Clearly we can see $K$ satisfies the axiom of completeness, and thus we can let $c = sup K$

There are three cases to consider: $f(c) > 0, f(c)<0 space text{and} space f(c) = 0$

**Part where I have difficulty understanding the proof**

. The author states:

“Since $c$ is the least upper bound of $K$, we can rule out the first two cases.

“

I don’t understand how we can conclude this immediately?

Thinking about it, if $f(c)>0$, then $c notin K$ so this would contradict $c$ being a least upper bound.(Right?)

However, how can we rule out $f(c)<0$? Here $c in K$, which is a valid least upper bound?

## Answer

As Wojuvu said, it does require an extra step of proof.

First of all, the claim

Thinking about it, if $f(c)>0$, then $cnotin K$ so this would contradict c being a least upper bound.(Right?)

is not quite true. The least upper bound of a set $A$ does not need to be an element of the set. (Take the set $[0,1)$ and its least upper bound $1$, for example.) We need to do something a bit more delicate.

To rule out the two cases $f(c)>0$ and $f(c)<0$ in the proof you will need to use the continuity of $f$. (Note that the theorem requires a continuous mapping, as can be seen with some examples of step-functions, and we have not yet used continuity.) For both cases the idea is similar: if, for example you would have $f(c)>0$, then you should be able to find an $varepsilon > 0$ such that for all points $z in (c – varepsilon,c) $ you have $f(z) > 0$. And **this** would be a contradiction with $c$ being the least upper bound.