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.


4 Answers 4


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).

  • $\begingroup$ Thanks, I knew you went for the kill with the IVT; just wasn't sure how to set it up. $\endgroup$ Jul 13, 2012 at 3:19
  • $\begingroup$ Since it is strictly increasing and the domain is unbounded hence the function should be unbounded also? $\endgroup$
    – Upstart
    Jul 17, 2019 at 18:40
  • $\begingroup$ @Upstart: Certainly not. $\arctan(x)$ is continuous, strictly increasing, with unbounded domain, but its values always lie between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$. $\endgroup$ 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 $0$.

  • $\begingroup$ 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? " $\endgroup$
    – M Desmond
    Sep 24, 2021 at 20:36
  • $\begingroup$ This is a very nice and slick proof of that statement! Thanks! $\endgroup$ 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 $a<b$ we have $f(a)\neq f(b)$. We first consider the case when $f(a)<f(b)$. I claim that in this case $f$ is strictly increasing.

First, note that for any $x\in(a,b), f(a)<f(x)$. If not, then since $f$ is 1-1, must have $f(x)<f(a)$. 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 $x,y\in I$, $x<y$ with $f(y)<f(x)$. By the previous paragraph, we also have $f(a)<f(x)$. 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 $f(a)<f(b)$. If $f(b)<f(a)$, a similar argument (with all inequalities reversed) shows $f$ is strictly decreasing.

  • $\begingroup$ The third paragraph is a non-sense. $\endgroup$
    – user26857
    Oct 9, 2019 at 10:42

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