8

Given a function $f: \mathbb{R} \to \mathbb{R}$, is $x < y \implies f(x) < f(y)$ equivalent to $x < y \iff f(x) < f(y)$.

If $f(x) < f(y)$, then the contradiction occurs only$_\text{(or is it not the "only" case)}$ when $x > y$. If that happens, then $f(x) > f(y)$ and so the two statements are always equivalent.

Am I missing something? Is it always the case or it changes when we change the injectivity, surjectivity, domain or anything else?

dictatemetokcus
  • 585

4 Answers4

13

Yes, they are equivalent. The key here is that "$<$" is a total order on $\mathbb{R}$.

Assume $f$ satisfies the condition $$(*)\quad \mbox{For all $a,b$}, a<b\implies f(a)<f(b)$$ and suppose $f(x)<f(y)$. Clearly this means $\color{red}{\mbox{$x\not=y$, so either $x<y$ or $y<x$}}$. We can't have $y<x$ since then $(*)$ would imply $f(y)<f(x)$, so we must have $x<y$.

This applies to any total order. On the other hand, it breaks down for posets since the implication $x\not=y\implies (x\triangleleft y\mbox{ or }y\triangleleft x)$ no longer needs to hold if $\triangleleft$ is merely assumed to be a partial ordering.

Noah Schweber
  • 245,398
  • 3
    (I was writing a similar answer, with the only difference being that I was going to use the word "trichotomy" when mentioning that exactly one of $x=y$, $x < y$, or $x > y$ is true. I'm writing a comment just to be able to use the word.) – Kyle Miller Jul 24 '21 at 21:09
  • How could I not think of posets! That inclusion of posets in your answer really answers my second question. Thank you. – dictatemetokcus Jul 24 '21 at 21:22
  • @dictatemetokcus Everyone should always be thinking of posets. :P – Noah Schweber Jul 24 '21 at 21:22
4

Yes. Suppose that LHS holds. Let $x,y\in\mathbb{R}$. Suppose that $f(x)<f(y)$. We go to prove that $x<y$ by contradiction. Suppose the contrary that $x \not <y$, then $x\geq y$, i.e., $x=y$ or $x>y$.

If $x=y$, we have $f(x)=f(y)$, contradicting to the given condition. If $x>y$, by the hypothesis (LHS), we have $f(x)> f(y)$, which contradicts to the given condition $f(x)<f(y)$.

Therefore $f(x)<f(y) \Rightarrow x<y$.

Danny Pak-Keung Chan
  • 12,250
2

Suppose $f(a)\gt f(b)$ then we have three possible cases

$a=b$ - well then, by the fact that $f(x)$ is a function we have $f(a)=f(b)$, which is not the presenting case

$a\lt b$ in which case the conditions would give us the contradiction $f(a)\lt f(b)$

Or $a\gt b$, which is the only case which works.

So the reverse implication holds, as required.

Don't be afraid of considering cases when they are not so many in number.

Mark Bennet
  • 100,194
  • I was able to verify that this holds, I understand I should have added all these, but I thought the contradictory part is when $f(x) < f(y) \implies x > y$ and so I added that. The rest are trivially/vacuously true. I was surprised that it holds this way and posted on Stack to find whether this is ALWAYS the case or not. And in the latter, when. But thank you for the clear complete solution! – dictatemetokcus Jul 24 '21 at 21:25
0

If we know that for all $x$ and $y$, $\; x<y\implies f(x)<f(y)$, then we can prove $f(x)<f(y)\implies x<y$ by contraposition. The contrapositive of $f(x)<f(y)\implies x<y$ is $$ x\ge y\implies f(x)\ge f(y) \, . \tag{*}\label{*} $$ If $x=y$, then $\eqref{*}$ clearly holds. If $y<x$, then by assumption $f(y)<f(x)$, and so $f(x)\ge f(y)\blacksquare$

Joe
  • 19,636