Proof of the Intermediate Value Theorem

The proof relies on the Completness Axiom of the real numbers which guarantees the existence of a least upper bound (supremum) for any non-empty, bounded set of real numbers.

Without loss of generality, assume $f(a)>f(b)$.

Let $m$ : any numer such that $f(b)<m<f(a)$.

Define a set $V$, $V=\{ x \in [a,b] | f(x) > m\}$.

Since, $V$ is non-empty ($f(a)>m \rightarrow x=a$ is in $V$ so, $V \neq \varnothing$)

and $V$ is bounded above (By definition, $V \subset [a,b]$. So, $x\in V \rightarrow x\leq b$. Therefore upper bound of $V=b$),

Completness Axiom guarantees that supremum exists.

Let $c=\operatorname{sup} (V)$.

Since $a\in V$ and $b$ is an upper bound of $V$, $a\leq c\leq b$.

Case I) if $f(c) > m$

Let $\epsilon = f(c) - m$. Then $\epsilon > 0$.

Since $f$ is continuous at $c$, for this $\epsilon$, $\exists\delta > 0$ such that, for any $x\in [a,b]$ with $|x-c|<\delta \rightarrow |f(x)-f(c)|<\epsilon$.

$|f(x)-f(c)|<\epsilon$ can be written as $f(c)-\epsilon < f(x) < f(c)+\epsilon$.

By substituting $\epsilon$ with $f(c)-m$, we can get $m<f(x)$.

So, we can conclude that for all interval $(c-\delta ,c+\delta)$, function $f(x)>m$.

Since $f(b)<m$, $c\neq b$. $c<b$.

With the fact of $c$ being strictly smaller than $b$, we can find a point $x_0$ such that $c<x_0<c+\delta$ and $x_0\in [a,b]$. Then, $f(x_0)>m$.

By the definition of set $V$, $x_0\in V$. However, $x_0>c$.

Since $c$ is supposed to be the supremum of $V$, it contradicts the assumption.

Therefore, $f(c)>m$ is false.

Case II) if $f(c) < m$

Let $\epsilon = m - f(c)$. Then $\epsilon > 0$.

By the continuity of $f$ at $c$, for this $\epsilon$, $\exists\delta>0$ such that, for any $x\in [a,b]$ with $|x-c|<\delta\rightarrow|f(x)-f(c)|<\epsilon$.

$|f(x)-f(c)|<\epsilon$ can be written as $f(c)-\epsilon < f(x) < f(c)+\epsilon$$.

By substituting $\epsilon$ with $f(c)-m$, we can get $f(x)<m$.

This means that for all $x\in (c-\delta , c] \cap [a,b]$, $f(x)<m$.

Therefore no point in this interval can be in set $V$.

This implies that all points $x$ in $V$ must satisfy $x\leq c-\delta.

So, we can think $c-\delta$ as an upper bound of $V$.

But $c-\delta < c$. Since $c$ is supposed to be the least upper bound (supremum), it contradicts the definition of $c$.

Therefore, $f(c)<m$ is false.

Since $f(c) \ngtr m$ and $f(c) \nless m$, by the Law of Trichotomy, $f(c) = m$.