Theorem: Let $f$ be continuous on $[a,\,b]$ and assume $f(a)<f(b)$. Then for every $k$ such that $f(a)<k<f(b)$, there exists a $c\in[a,\,b]$ such that $f(c)=k$.
proof: $f$ continuous at $a\implies$ for $\varepsilon=k-f(a)>0$, $\exists\delta>0$ s.t. $$|f(x)-f(a)|<\varepsilon=k-f(a)\quad\forall x\colon |x-a|<\delta.$$ Consider the set $H=\{x\in[a,\,b]\colon f(x)<k\}\not=\emptyset \implies c=\sup{(H)}$. Show $f(c)=k$, suppose $f(c)<k\iff k-f(c)>0$. We know $f$ is continuous at $c$ so $\forall\varepsilon>k-f(c)>0$ $\exists\delta>0:|f(x)-f(c)|<\varepsilon=k-f(c)$ when $|x-c|<\delta$. $\implies f(x)-f(c)<k-f(c)$. Say $x=c+\delta/2\implies f(x)<k\implies c+\delta/2\in H$ which contradicts the fact $c=\sup{(H)}$, since $\delta>0$. Same proof works if $f(c)>k$, thus $f(c)=k$.
This is a proof for the intermediate value theorem given by my lecturer, I was wondering if someone could explain a few things:
- What is the set $H$, what does it define?
- Why does contradicting the fact that $c=\sup{(H)}$ prove that $f(c)\not<k$?
- How would I continue to actually finish this proof, ie. show $f(c)\not>k$?