# A continuous, injective function $f: \mathbb{R} \to \mathbb{R}$ is either strictly increasing or strictly decreasing.

I would like to prove the statement in the title.

Proof: We prove that if $$f$$ is not strictly decreasing, then it must be strictly increasing. So suppose $$x < y$$.

And that's pretty much how far I got. Help will be appreciated.

Prove the contrapositive instead: if $$f$$ is not strictly increasing and not strictly decreasing, then it is not one-to-one.

For example, say there are points $$a\lt b\lt c$$ such that $$f(a)\lt f(b)$$ and $$f(b)\gt f(c)$$. Either $$f(a)=f(c)$$ (in which case $$f$$ is not one-to-one), or $$f(a)\lt f(c)$$, or $$f(c)\lt f(a)$$.

If $$f(a)\lt f(c)\lt f(b)$$, then by the Intermediate Value Theorem there exists $$d\in (a,b)$$ such that $$f(d)=f(c)$$; hence $$f$$ is not one-to-one.

Now, there are other possibilities (I made certain assumptions along the way, and you should check what the alternatives are if they are not met).

• Thanks, I knew you went for the kill with the IVT; just wasn't sure how to set it up. Jul 13, 2012 at 3:19
• Since it is strictly increasing and the domain is unbounded hence the function should be unbounded also? Jul 17, 2019 at 18:40
• @Upstart: Certainly not. $\arctan(x)$ is continuous, strictly increasing, with unbounded domain, but its values always lie between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$. Jul 17, 2019 at 19:27

Consider $g\colon \{(x,y)\mid x<y\}\mapsto\mathbb R$, defined by $g(x,y):=f(x)-f(y)$. Clearly $g$ is continuos. Since the domain of $g$ is connected and $g$ has no zeroes, the image of $g$ is an interval not containing $0$.

• How to show that $g(x, y)$ is continuous? Also whats the result you are using for later statement " Every connected set map to connected set if continuos? " Sep 24, 2021 at 20:36
• This is a very nice and slick proof of that statement! Thanks! Apr 13 at 20:50

Suppose $f : \mathbb{R} \rightarrow \mathbb{R}$ is continuous and not strictly increasing. Then there exists two points such that $f(a) = f(b)$, or there exists three points $a < b < c$ such that $f(a) < f(b)$ and $f(b) < f(c)$. The first case contradicts injectivity. Suppose the second, without loss of generality, suppose that $f(b) - f(a) \leq f(c) - f(b)$. Then $f(b) \leq f(b) - (f(b) - f(a)) = f(a) \leq f(c)$. By the intermediate value theorem, there exists $d$ such that $b < d < c$ such that $f(d)= f(a)$. This contradicts injectivity.

Since $$f$$ is one-to-one, for $$a we have $$f(a)\neq f(b)$$. We first consider the case when $$f(a). I claim that in this case $$f$$ is strictly increasing.

First, note that for any $$x\in(a,b), f(a). If not, then since $$f$$ is 1-1, must have $$f(x). But then by the IVT there is some $$c\in(x,b)$$ so that $$f(c)=f(a)$$, contradicting $$f$$ being 1-1.

Now suppose for contradiction that $$f$$ is not strictly increasing. So there is some $$x,y\in I$$, $$x with $$f(y). By the previous paragraph, we also have $$f(a). So by the IVT, there is some $$c\in (a,x)$$ with $$f(c)=f(y)$$, contradicting $$f$$ being 1-1.

Thus $$f$$ is strictly increasing if $$f(a). If $$f(b), a similar argument (with all inequalities reversed) shows $$f$$ is strictly decreasing.

• The third paragraph is a non-sense. Oct 9, 2019 at 10:42