I would like to prove the statement in the title.
Proof: We prove that if $f$ is not strictly decreasing, then it must be strictly increasing. So suppose $x < y$.
And that's pretty much how far I got. Help will be appreciated.
I would like to prove the statement in the title.
Proof: We prove that if $f$ is not strictly decreasing, then it must be strictly increasing. So suppose $x < y$.
And that's pretty much how far I got. Help will be appreciated.
Prove the contrapositive instead: if $f$ is not strictly increasing and not strictly decreasing, then it is not one-to-one.
For example, say there are points $a\lt b\lt c$ such that $f(a)\lt f(b)$ and $f(b)\gt f(c)$. Either $f(a)=f(c)$ (in which case $f$ is not one-to-one), or $f(a)\lt f(c)$, or $f(c)\lt f(a)$.
If $f(a)\lt f(c)\lt f(b)$, then by the Intermediate Value Theorem there exists $d\in (a,b)$ such that $f(d)=f(c)$; hence $f$ is not one-to-one.
Now, there are other possibilities (I made certain assumptions along the way, and you should check what the alternatives are if they are not met).
Consider $g\colon \{(x,y)\mid x
Suppose $f : \mathbb{R} \rightarrow \mathbb{R}$ is continuous and not strictly increasing. Then there exists two points such that $f(a) = f(b)$, or there exists three points $a < b < c$ such that $f(a) < f(b)$ and $f(b) < f(c)$. The first case contradicts injectivity. Suppose the second, without loss of generality, suppose that $f(b) - f(a) \leq f(c) - f(b)$. Then $f(b) \leq f(b) - (f(b) - f(a)) = f(a) \leq f(c)$. By the intermediate value theorem, there exists $d$ such that $b < d < c$ such that $f(d)= f(a)$. This contradicts injectivity.
Since $f$ is one-to-one, for $a we have $f(a)\neq f(b)$. We first consider the case when $f(a)
First, note that for any $x\in(a,b), f(a)
Now suppose for contradiction that $f$ is not strictly increasing. So there is some $x,y\in I$, $x
Thus $f$ is strictly increasing if $f(a)