Simple Proof of Intermediate Value Theorem

Every analysis student knows the Intermediate Value Theorem. I first read the proof in Rudin’s book. While using the general theory of the relation between continuity and connectedness (which is what Rudin uses) to prove the theorem is a fine approach (a great approach, too), there is something truly elegant in the direct proof of the theorem.

Theorem: Let $f$ be a continuous function on the interval $[a,b]$ with $a<b.$ If $f(a)<0<f(b)$ then there exists an $x\in(a,b)$ such that $f(x)=0.$

Before proving the theorem, we have a short lemma.

Lemma: Let $f$ be a continuous function on the interval $[a,b]$ with $a<b.$ If $f(\alpha)>0$ (resp. $f(\alpha)<0$) for any $\alpha\in [a,b],$ then there exists some $\delta>0$ such that $f(x)>0$ (resp. $f(\alpha)<0$) for all $x\in (\alpha-\delta,\alpha+\delta)\cap [a,b].$

Proof of Lemma: We will only prove the lemma for the case $f(\alpha)>0.$ By continuity of $f$ at $x=\alpha$, there exists some $\delta>0$ such that $|f(x)-f(\alpha)|<f(\alpha)/2$ for all $x\in (\alpha-\delta,\alpha+\delta)\cap [a,b].$ So for any $x$ in this region, $f(\alpha)-f(x)<f(\alpha)/2$ meaning $f(x)>f(\alpha)/2>0.$ So we have our result.

Now we can prove the theorem.

Proof of Theorem: Let $E=\{x\in [a,b]:f(x)<0\}.$ Since $E$ is bounded by $b,$ its supremum exists. Let $\alpha=\sup E.$

Firstly, note that $\alpha>a.$ This is because, by our lemma, there exists some $\delta’>0$ such that $f(x)<0$ for all $x\in (a-\delta’,a+\delta’)\cap [a,b].$ So for any $x’\in (a,\min\{a+\delta’,b\}),$ we have $f(x’)<0.$ So by definition of an upper bound, $\alpha\ge x’ > a.$ Secondly, $\alpha\le b$ since it is the least upper bound.

Hence, $\alpha\in (a,b]$ and $f$ is defined on $\alpha.$ We claim that $f(\alpha)=0.$

Assume, for a contradiction, that $f(\alpha)\neq 0.$ If $f(\alpha)<0$ then, by our lemma, there exists some $\delta>0$ such that $f(x)<0$ for all $x\in (\alpha-\delta,\alpha+\delta)\cap [a,b].$ As $f(\alpha)<0<f(b)$ i.e $f(\alpha)\neq f(b)$, we must have $\alpha<b$. So, there exists some $x\in (\alpha,b)$ such that $f(x)<0.$ However, this contradicts the definition of supremum (as an upper bound).

On the other hand, if $f(\alpha)>0$ then, again by our lemma, there exists some $\delta’’>0$ such that $f(x)>0$ for all $x\in (\alpha-\delta’’,\alpha+\delta’’)\cap [a,b].$ For any $x’’\in (\min\{a,\alpha-\delta’’\},\alpha),$ we have $f(x’’)>0.$ Note that this is where the fact that $a<\alpha$ is used.

Now, for any $x\ge x’’$ with $x\in [a,b],$ if $x> \alpha$ then $f(x)\ge 0$ by definition of $\alpha$ as an upper bound and if $x\le \alpha$ then $x\in [x’’,\alpha]\subset (\alpha-\delta’’,\alpha+\delta’’)\cap [a,b]$ so $f(x)>0.$ In other words, for all $x\ge x’’$ we have $f(x)\ge 0.$ Thus, $E$ is bounded above by $x’’.$ This contradicts the definition of supremum (i.e. the least upper bound) as $x’’<\alpha.$

In both cases, we have a contradiction. So, our assumption was wrong. Hence, $f(\alpha)=0$ which proves our theorem.