Proving a function is strictly increasing

Question

Suppose the function $g : [0,2] \to \mathbb R$ is continuous and injective on $[0,2]$ and $g(0) < g(2)$, prove that $g$ is strictly increasing on $[0,2]$

I understood that the proof using the Intermediate Value Theorem, claming that $g(0) < g(x) < g(2)$ for all $x \in [0,2]$

However, I decided to use a different approach using quantifiers: Let $I = [0,2]$

Since $g$ is injective, then $(\forall x_1,x_2 \in I)[x_1\neq x_2 \implies g(x_1) \neq g(x_2)]$ is True.

WLOG, I assume that $x_1 < x_2$, then either:

$$(\forall x_1,x_2 \in I)[x_1 < x_2 \implies g(x_1) < g(x_2)]...(1)$$ or

$$(\forall x_1,x_2 \in I)[x_1 < x_2 \implies g(x_1) > g(x_2)]...(2)$$

must be True

I want to prove that $(2)$ is False, hence I will prove that its negated statement $(3)$, is True:

$$(\exists x_1,x_2 \in I)[ \neg (x_1 < x_2 \implies g(x_1) > g(x_2))]...(3)$$

But $(3)$ is clearly True, because consider $x_1=0$ and $x_2=2$, but then $g(0) < g(2)$ as stated in the question.

Therefore $(2)$ is False implies that $(1)$ must be True, and the statement $(1)$ is simply saying that $g$ must be increasing, hence completing the proof.

My lecturer unfortunately said that this proof is not correct and explained it to me a vague manner which I still dont understand.

Appreciate it if anyone can check my proof and see if it is correct. If not, please do give me details specifically on what logical fallacy I have committed at which part of my proof, thank you for your time!

Answer 1

Your attempt fails at the step where you claim:

$\qquad\quad$WLOG, I assume that $x_1 < x_2$, then either: $$(\forall x_1,x_2 \in I)[x_1 < x_2 \implies g(x_1) < g(x_2)]...(1)$$ $\qquad\quad$or $$(\forall x_1,x_2 \in I)[x_1 < x_2 \implies g(x_1) > g(x_2)]...(2)$$ $\qquad\quad$must be True

If you select $x_1,x_2$ with $x_1 < x_2$, you can't then apply a universal quantifier to those variables, since they are now constants.

Your claim could be corrected to

$\qquad\quad$If $x_1 < x_2$, then either: $$g(x_1) < g(x_2)\qquad(1)$$ $\qquad\quad$or $$g(x_1) > g(x_2)\qquad(2)$$

Note: No quantifiers.

Once that error is corrected, you can't get away with showing that option $(2)$ is not possible, simply by noting that $g(0) < g(2)$.

Also note: A clear indication that your proof must be wrong is the fact that you never used the assumption that $g$ is continuous. Without that assumption, the rest of the hypothesis is not enough to force $g$ to be increasing. As an exercise, try to construct a function $g$ which satisfies all conditions of the hypothesis other than continuity, but such that $g$ is not an increasing function on all of $I$.

Answer 2

When you get $$(\forall x_1,x_2 \in I)[x_1\neq x_2 \implies g(x_1) \neq g(x_2)],$$ then next step should be $$(\forall x_1,x_2 \in I)\big[x_1< x_2\implies \big(g(x_1) < g(x_2) \lor g(x_1) > g(x_2)\big)\big]$$

where $\lor$ is within the quantifier, rather than outside it.

I hope this helps $\ddot\smile$

Answer 3

Statements $(1)$ and $(2)$ are not obvious. They do not derive from injectiveness alone; you must use continuity as well. What follows from injectivity without continuity is that

$$ (\forall x_1,x_2 \in I)[x_1 < x_2 \implies g(x_1) < g(x_2) \vee g(x_1) > g(x_2)]...(i) $$

Note it is very different from $(1) \vee (2)$, because the disjunction (the "or") either depends on the chosen $x_1$ or $x_2$ in (i), but not in $(1)\vee(2)$.

The real proof is obtaining that you have (1) or (2). Afterwards, all you need is to notice that $0,2 \in I$ but $g(0) < g(2)$ to discard (2). You did it in a bit of a overcomplicated way, but I think you got the idea.