IVT Proof

For N=0

1. Assume $f$ is cont on $[a,b]$ and $f(a)<0<f(b)$
2. We need to show that there is a set of $(a,a+\delta )$ such that, $\forall x\in (a,a+\delta ),f(x)<0$ or in other words, that $S=\{x\in (a+\delta )|f(x)<0\}$ is non-empty
   1. Let $\delta >0$ be arbitrary
   2. since $f$ is continuous on $[a,b]$, then $f$ is continuous at $a$
   3. Thus, we can say that * : $\forall ϵ>0,\exists \delta >0$ s.t $|x-a|<\delta ⟹|f(x)-f(a)|<ϵ$
   4. Invoke * with $-f(a)$ as $-f(a)>0$
      1. Thus, $\exists \delta >0$ s.t $|x-c|<\delta ⟹|f(x)-f(a)|<-f(a)$
      2. This means $\exists \delta >0$ s.t $|x-a|<\delta \Rightarrow f(x)<0$
      3. So on, $\exists \delta ,(a-\delta )\cup (a+\delta )$, $f(x)<0$
      4. So, on $\exists \delta ,(a+\delta ),f(x)<0$
      5. This means $S=\{x\in (a+\delta )|f(x)<0\}$ set is non-empty
3. Since $S$ is non-empty, by completenes axiom, $S$ has a supremum $s$
4. Note that $s\in (a,b)$ because $s$:
   1. f must be continuous at $s$
   2. and $a<0$ which would mean if $s=a$, then $s\in S$
   3. and $b>0$ which would mean $s$ is not the least upper bound
5. Then, $s=0$ as:
   1. $\forall x\in S,s\ge x$ which means $x\ge 0$
   2. $\forall x$ that is an upper bound on s, meaning it is on interval $(a+\delta ,b)$, $s<x$ so this means $x\le 0$
   3. Thus, $s=0$
6. as this supremum that = 0 always exists, $\exists x\in (a,b)$ s.t $f(x)=0$

For Arbitrary N

1. Suppose $f$ is continuous on $[a,b]$
2. Let $N$ be arbitrary number that follows $f(a)<N<f(b)$
3. Let $g(x)$ = $f(x)-N$
4. Note that $g(x)$ is continuous on $[a,b]$
5. Also note that $g(a)=f(a)-N$ and as $f(a)<N⟹f(a)-N<0$ thus $g(a)<0$
6. Also note that $g(b)=f(b)-N$ and as $f(b)>N⟹f(b)-N>0$ thus $g(b)>0$
7. Then, by IVT on $n=0$, $\exists x\in (a,b)$ s.t $g(x)=0$
   1. Then, $\exists x\in (a,b)$ s,t $0=f(x)-N$
   2. $\exists x\in (a,b)$ $N=f(x)$
8. Therefore $\forall N\in (f(a),f(b)),\exists x\in (a,b)$ s.t $f(x)=N$ $\square$