Suppose $f:[a,b]\rightarrow[c,d]$ is one-one and non-decreasing, then $f$ is always increasing.
Is this statement correct? If not, give an example.
The sketch of the proof is as follows: If any flaw in this, please point out.
Now,
Let $x,y \in [a,b]$, We need to prove that $x<y \implies f(x)<f(y)$.
Since $f(x), f(y) \in [c,d]$ either $f(x)=f(y)$ or $f(x)>f(y)$ or $f(x)<f(y)$. Since $f$ is non-decreasing, $f(x)>f(y)$ is not possible. If $f(x)=f(y)$, then $x=y$ as $f$ is one-one, but we assumed that $x<y$, therefore, only possbility is that $f(x)<f(y)$.
The proof is fine. Although it can be cleaner, i.e. shorter. For instance, ()<(), ()=(), or ()>() is always true for real valued functions, regardless of the fact that the image is in [,]. So you can simply say:
Let ,∈[,] with <.
By assumption, ()≤(), otherwise is not non-decreasing. But if ()=(), because is 1-to-1, then =, impossible. Therefore, ()<() whenever <, i.e. is increasing.