# Proof of Intermediate Value theorem which I dont understand

Intermediate Value Theorem. Let $f : [a, b] → R$ be continuous and suppose that $u$ lies between $f(a)$ and $f(b)$. Then there is a point $c$ between $a$ and $b$ where $f(c) = u$.

Proof part 1:

*Assume that $f(a) < u < f(b)$ and let $A$ be the set $\{x \in [a, b] : f(x) ≤ u\}.$ This set is non-empty since it contains $a$ and is bounded above by $b.$ Let $s$ be its least upper bound. The aim is to show that $f(s) = u$. *

Why are we trying to show $f(s)=u$? I understand that the value $u$ needs to have some value $c$ assigned to it between $a$ and $b$ but why does the supremum $s$ matter here. I have looked at the whole proof by the way but to start with I don't understand this. Also what does $f(x)≤u$ mean here I don't understand the use of it.

• Have you tried to draw it? Drawing some potential functions and the 2 sets you talked about for every functions. It would be of great help. – Max Ft Mar 23 '18 at 19:13

## 4 Answers

We have to show that such $c$ exists. The proof says that $s_1:=\sup\{x \in [a, b] : f(x) \leq u\}$, which is explicit real number contained in $(a,b)$, is a candidate for $c$. Another possible candidate could be $s_2:=\inf\{x \in [a, b] : u\leq f(x)\}$. Now that we have a candidate we have "only" to verify that it satisfies the required property, i.e. $f(s_1)=u$ or $f(s_2)=u$ (note that $c$ may be not unique).

• Oh, I get it now, Thanks! Also, do you mind answering another question related to the part 2 of this proof I dont want to make another topic just for this: Suppose $f(s) < u$. Note that that $s$ cant equal $b$ because we know $f(b) > u$. If we put $e = u − f(s)$ then for some $δ > 0$ $|f(x) − f(s)| < e$ as long as $|x − s| < δ$. I dont understand why they let $e= u − f(s)$ and how the next inequality is derived from this equality – NoteBook Mar 23 '18 at 19:26
• In part 2) we have if $f(s)<u$ then $u-f(s)>0$. Call this positive number $\epsilon$ and use the definition of continuity of $f$ at $s$ : there is some $\delta>0$ such that if $|x-s|<\delta$ then $|f(x) − f(s)| < \epsilon$. – Robert Z Mar 23 '18 at 19:32

The informal thinking behind this proof is roughly the following:

Think of the graph of the function $y=f(x)$. Now draw a line at height $y=u$. Try to visualize the set $A:=\{x \in [a,b]:f(x) \leq u\}$. This consists of all $x$ for which the graph is below or at the level of this line.

The idea now is to consider what happens at the boundary (just in a naive sense, not necessarily topologically) of this set $A$. There we switch from being in the set $A$ to not being in the set $A$. Or in other words from being below the line to being above the line. Since we have a continuous function this means, we need to be on the line, so there $f(x) = u$.

Now how to find a point on this boundary? This is simple, we take the rightmost point of $A$, which is the least upper bound $s:= \sup A$. Then to the left of $s$, we are in $A$ and to the right we are not in $A$. So we are on the boundary, as desired.

From this idea it is not hard to write down the details of the formal proof.

• Thanks I understand it this way too when I visualize it – NoteBook Mar 23 '18 at 19:34

They should have said "Let $c$ be its least upper bound" and then worked on showing that $f(c) = u.$

And they could just as easily have let $c$ be the greatest lower bound of $\{ x\in[a,b] : f(x) \ge u \}.$

I like to think of such problem graphically. Picture a simple continuous function in your head or draw it.

Let's think about this one. $\sup\{x ∈ [a, b] : f(x) ≤ u\}$

Well, $f(b)>u$ so f starts above u at the point b. Then imagine you are following its path 'backwards' starting from point b. The biggest x such that $f(x) ≤ u$ actually correspond to the first time f hit the value u following its path 'backwards'.

Likewise $\inf\{x ∈ [a, b] : f(x) ≤ u\}$ will give you the first point where your function will hit u if you follow its path forward starting from point a.