Is it true that any injective function $f: \mathbb{R} \to \mathbb{R}$ is strictly monotone?

Question

Is it true that any injective function $f: \mathbb{R} \to \mathbb{R}$ is strictly monotone?

If yes, how do I prove it? If not, are there any examples of functions that disprove this statement.

I was thinking of $f(x) = {\frac1x}$, which disproves this statement. But I'm not really sure.

Thanks for the help in advance.

You need connectedness of the domain. Your counter is not defined at 0 — Partha, Nov 20 '18 at 04:59
You also need that the function be continuous. Your counterexample is continuous, but the first comment applies. However, if you define $f(0)=0$ in your counterexample, it now works. — Jean-Claude Arbaut, Nov 20 '18 at 05:00
As is, no. Define additionally $f(0)=0$, and it's a correct counterexample. — Jean-Claude Arbaut, Nov 20 '18 at 05:05
How can I define f(0) = 0 since f(x) = 1/x is undefined at x=0. — user10634718, Nov 20 '18 at 05:08
Since $f(x)=1/x$ is undefined at $0$, you can decide to give it any value you wish. Of course this is another function (it's not simply $x\to1/x$ anymore). That is, the new $f$ is defined on $\Bbb R$ to be $f(x)=1/x$ if $x\ne0$, and $f(0)=0$. This one is defined on $\Bbb R$ but discontinuous, whereas your original example was defined on $\Bbb R\backslash{0}$ and continuous. — Jean-Claude Arbaut, Nov 20 '18 at 05:11
As already remarked, you need continuity for this statement. See https://math.stackexchange.com/questions/170147/a-continuous-injective-function-f-mathbbr-to-mathbbr-is-either-strict. — Paul, Nov 20 '18 at 22:59
@Jean-ClaudeArbaut: Your two comments look like an answer; would you mind posting them as one so that we can get this off the unanswered queue? — aleph_two, Dec 31 '18 at 04:56

Jean-Claude Arbaut · Answer 1 · 2021-06-27T09:33:41.580

Your counterexample $f(x)=1/x$ does not work as is, because it's not defined at $0$.

With the function $g:\Bbb R\to\Bbb R$ defined by $g(0)=0$ and $g(x)=1/x$ if $x\ne0$, you have a valid counterexample: it's defined on $\Bbb R$ and injective, but not strictly monotone.

The correct statement would be:

If $f:E\to\Bbb R$ is continuous and injective and $E$ is a connected subset of $\Bbb R$, then $f$ is strictly monotone.

Note that here your $f$ is continuous but it's defined on $(-\infty,0)\cup(0,+\infty)$, which is not connected. The function $g$ defined above is defined on $\Bbb R$, but is not continuous.

Is it true that any injective function $f: \mathbb{R} \to \mathbb{R}$ is strictly monotone?

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