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Proposition: $f:[a,b] \to \mathbb{R}$ is one to one only if $f$ is strictly monotone.

Proof: Suppose, the statement is false. Then, for some $c, d \in (a,b)$, $f(d)=\sup f=M, f(c)= \inf f=m $ [Without loss of generality. Suitable adjustments are made when either of $c, d$ coincides with on of the boundary points]. $[c,d] \subset [a,b]$. We need to show that for at least one $\xi \in [c,d]$, $f(\xi)=f(\eta), \eta \in [a,b]\cap[c,d]^c=I$.

Again, suppose, there does not exist such a point. Hence, $\forall \eta \in I$, $f(\eta)\notin [m,M]\implies f(\eta) < m $ or, $f(\eta)>M$, a contradiction .

Hence, both $c, d$ must coincide with $a$ and $b$ [in some order].

Is this correct?

Subhasis Biswas
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    Your reference to "suitable adjustments" is not at all sufficient. For example, your proof gives no information for the function $f(x)=x^3-x$, $-2 \le x \le +2$, where $f(-2) = \text{inf}(f)=-8$ and $f(2)= \text{sup} f = 8$. I think that $\text{inf}(f)$ and $\text{sup}(f)$ are not particularly relevant to this question. – Lee Mosher Apr 04 '19 at 23:20
  • I completely missed that part. I omitted the case where local maxima and minima exists. – Subhasis Biswas Apr 04 '19 at 23:27
  • It can be done in a much simpler way. Have a look at https://math.stackexchange.com/questions/1129412/how-can-i-prove-that-a-continuous-injective-function-is-increasing-decreasing – skylark Apr 05 '19 at 04:47

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