Is this proof that every strictly increasing $f\colon\mathbb{R\to R}$ is injective flawed?

Question

The problem is as follows:

A function $f:\mathbb{R}\to\mathbb{R}$ is called strictly increasing if $$\forall x,y\in\mathbf{R},\ x<y\implies f(x)<f(y).$$ Show that any strictly increasing function is injective.

The provided solution is as follows:

$\ \ $ $\text{S}\scriptstyle{\text{OLUTION}}:$ Suppose that $x_1,x_2\in\mathbb{R}$ are such that $f(x_1)=f(x_2)$. Then it is not true that $x_1<x_2$ (for then $f(x_1)<f(x_2)$) and so $x_1\geq x_2$. Similarly $x_2\geq x_1$, and so $x_1=x_2$. Thus $f$ is injective.

I've been pondering over this proof, and it seems to me that it might be logically incorrect.

In the definition of a strictly increasing function, we have that $x < y \implies f(x) < f(y)$. So we have that $A \implies B$, where $A = x < y$ and $B = f(x) < f(y)$. So it is a one-way implication.

Then, in the solution proof, it begins by assuming that $f(x_1) = f(x_2)$. But it then claims that this implies that it is not true that $x_1 < x_2$, since we would otherwise have had $f(x_1) < f(x_2)$. However, as I just stated, the definition for a strictly increasing function is provided as $x < y \implies f(x) < f(y)$ -- a one-way implication from $x < y$ to $f(x) < f(y)$. But isn't the solution proof attempting to use $f(x) < f(y) \implies x < y$ instead, since it is using the fact that we have $f(x_1) = f(x_2)$ to claim that it is then not true that $x_1 < x_2$?

So, in order for this solution proof to be correct, wouldn't the definition provided for a strictly increasing function have to also include the reverse one-way implication, so that we would have $x < y \iff f(x) < f(y)$?

I would greatly appreciate it if people could please take the time to clarify this. If I'm misunderstanding anything, I would appreciate an explanation so that I may understand where my misunderstanding lies.

Answer 1

I wouldn't say the proof is very well written, but it is logically correct. The definition of an increasing function is $$\hbox{if}\quad x_1<x_2\quad\hbox{then}\quad f(x_1)<f(x_2)\ .$$ As you point out, the converse of an implication is not logically equivalent to the original implication. So we cannot assume without proof that the converse is true.

However, this proof uses not the converse but the contrapositive, $$\hbox{if}\quad f(x_1)\not<f(x_2)\quad\hbox{then}\quad x_1\not<x_2\ ,$$ which is equivalent to the definition and is therefore automatically true. Thus:

• assume $f(x_1)=f(x_2)$;
• then $f(x_1)\not<f(x_2)$;
• so $x_1\not<x_2$;
• so $x_1\ge x_2$

and so on. Alternatively, you could view it as a proof by contradiction: let $f(x_1)=f(x_2)$; suppose $x_1<x_2$; since $f$ is increasing, we have $f(x_1)<f(x_2)$; but this is not true; so $x_1\ge x_2$.

Comment. In this case you can use the definition of an increasing function to show that the converse of the definition actually is true (try it). However as far as I can see it doesn't help you with the current problem.

Answer 2

$P\implies Q$ implies (and classically is equivalent to) $\neg Q \implies \neg P$. The latter formula is called the contrapositive of the former. So from $x < y \implies f(x)<f(y)$ we get $\neg(f(x)<f(y))\implies\neg(x<y)$. That is, $f(x)\geq f(y)\implies x\geq y$.

In this case, we have the assumption $f(x_1)=f(x_2)$. This implies $\neg(f(x_1)<f(x_2))$, i.e. $f(x_1)\geq f(x_2)$. We can thus use the contrapositive to infer $\neg(x_1<x_2)$ or $x_1\geq x_2$. Now swap $x_1$ and $x_2$ and use the same argument and we get $x_2\geq x_1$. Thus $x_2=x_1$.