Prove that if $f$ is strictly increasing at each point of (a,b), then $f$ is strictly increasing on (a,b).

Question

If $c \in (a,b)$ we can say that $f$ is strictly increasing at c if there exists $\delta \gt 0$ such that if $c-\delta \lt x \lt c$, then $f(x) \lt f(c)$ and if $c \lt x \lt c + \delta$, then $f(c) \lt f(x)$.

We say that $f$ is strictly increasing on (a,b) if whenever $x,y \in (a,b)$ with $x \lt y$, we have $f(x) \lt f(y)$

Hint: Argue by contradiction. Then there exists $c,d \in (a,b)$ with $c \lt d$ and $f(c) \ge f(d)$. Now consider $$\text{lub} \{x|a\lt x \lt d \; \text{and} \; f(x) \ge f(d)$$ I know that we have an IF,THEN type of proof so I can assume it is continuous at each point, but I am unsure as how to begin proving it using delta (or by contradiction using least upper bound hint). Any help with how to begin this problem would be appreciated.

You can see the second proof given in this answer: https://math.stackexchange.com/a/960585/72031 — Paramanand Singh, Jan 18 '18 at 03:27
For your hint let the lub described in hint be the number $s$. Then $c\leq s\leq d$. Use the fact that $f$ is strictly increasing at $s$ and then derive a contradiction. — Paramanand Singh, Jan 18 '18 at 03:30
For deriving a contradiction show that $s=d$. Also note that this has nothing to do with continuity as you are suspecting. — Paramanand Singh, Jan 18 '18 at 03:39

Paramanand Singh · Accepted Answer · 2018-01-18T04:09:05.207

I am expanding my comment into an answer. The thing you call lub is also called supremum by many authors and I will use this term. Let $$A=\{x\mid a<x<d, f(x) \geq f(d) \}, \, s=\sup A$$ and by assumption the set described above is non-empty (it contains $c$) and is bounded above by $d$ so by completeness property the supremum $s$ exists and $c\leq s\leq d$.

We next show that $s=d$. Suppose on the contrary that $s<d$. Then $f$ is increasing at $s$ and therefore there is a $\delta_s>0$ such that if $s-\delta_s<x<s$ then $f(x) <f(s) $ and if $s<x<s+\delta_s$ then $f(s) <f(x) $. And clearly we can choose $s+\delta_s<d$. Now by definition of supremum there is an element $p\in A$ such that $s-\delta_s<p$ and hence $f(p) \geq f(d)$ (by definition of $A$) and from what we have described above we have $f(s) \geq f(p) $. Thus $f(s) \geq f(d) $ and thus for all points $x$ with $s<x<s+\delta_s$ we have $f(x) >f(s) \geq f(d) $ so that these points $x$ belong to $A$. This is a contradiction as these points are greater than $s=\sup A$. Thus $s=d$ and then $s\notin A$.

Next we get another contradiction because we have a point $p\in A$ with $s-\delta_s<p<s$ such that $f(p) \geq f(d) $ and $f(p) <f(s) =f(d) $. And this completes our proof.

You should also try to prove this nice result using Heine Borel theorem which is especially designed to obtain global properties from local ones.

Prove that if $f$ is strictly increasing at each point of (a,b), then $f$ is strictly increasing on (a,b).

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