I was hoping someone could tell me how to prove the following problem I was given:
Let $f:[a,b]\to\mathbb{R}$ be a function such that for every $y\in[f(a),f(b)]$ there exists $x\in[a,b]$ such that $y=f(x)$. Then the function is strictly monotone.
Attemp 1:
Suppose $f(a)<f(b)$, and suppose that the function is not strictly increasing.
Then there exists $a',b'\in{(a,b)}$ such that $a'\lt b'$ and $\ f(b')\le{f(a')}$. First, note that $f(b')\not=f(a')$, for otherwise the function is not one-to-one, so suppose $f(b')\lt{f(a')}$.
Since $f$ is Darboux and since $f(a)<f(b')<f(a')$, there exists $a''\in{(a,a')}$ such that $f(a'')=f(b')$, but this is a contradiction since $b'\not\in{(a,a'')}$ and $f$ is injective.
Attempt 2:
Suppose $f(a)<f(b)$, and suppose that the function is not strictly increasing.
Then there exists $a',b'\in{(a,b)}$ such that $a'\lt b'$ and $\ f(b')\le{f(a')}$. First, note that $f(b')\not=f(a')$, for otherwise the function is not one-to-one, so suppose $f(b')\lt{f(a')}$.
First, suppose $f(b')>f(b)$.
Since $f$ is Darboux and since $f(a)<f(b')<f(a')$, there exists $a''\in{(a,a')}$ such that $f(a'')=f(b')$, but this is a contradiction since $b'\not\in{(a,a'')}$ and $f$ is injective.
Now, suppose $f(b')<f(b)$.
Without loss of generality we can assume $f(a')<f(b)$. Similarly, since $f$ is Darboux and since $f(b')<f(a')<f(b)$, there exists $b''\in{(b',b)}$ such that $f(b'')=f(a')$, but this is a contradiction since $a''\not\in{(b',b)}$ and $f$ is injective.
Best Answer
Without loss of generality we can assume $f(a)<f(b)$.
Assume $f$ is not strictly increasing. Then there are $c,d\in[a,b]$ with $c<d$ and $f(c)\ge f(d)$. Since $f$ is injective, we can say that $f(c)>f(d)$.
At this point we don't know whether $f(a)\le f(c)$ or $f(a)>f(c)$.
Suppose $f(a)>f(c)$. Then there is $a'\in[c,b]$ with $f(a)=f(a')$, contradicting injectivity. Similarly we can exclude $f(d)>f(b)$. Hence we have $$ f(a)\le f(c),\quad f(c)>f(d),\quad f(d)\le f(b) $$ and your argument carries over after noticing that we must also have $f(c)\le f(b)$ and $f(a)\le f(d)$ (same reasoning as before).