Connection between Monotone and One-to-one Functions

[hadron23]

Hello,

I was curious about the following point. I know that if a function is monotone, then it is one to one (meaning for x1 != x2, then f(x1) != f(x2) ).

But what about the converse? I can't seem to think of a counter-example.

 

[Moo Of Doom]

How about $$f : \mathbb{R} \to \mathbb{R}$$ given by

$$f(x) = \left\{ \begin{array}{ll}<br /> 1/x & x \neq 0 \\<br /> 0 & x = 0<br /> \end{array} \right.$$

It's injective, but not monotone (because it's not continuous ;) ).

 

[hadron23]

Thanks!

So, is that to say, that any injective, continuous function is strictly monotone?

 

[JG89]

Correct.

 

[Bhatia]

Thanks!

So, is that to say, that any injective, continuous function is strictly monotone?

If f is bijective and monotone with both its range and domain being a closed and bounded interval...then what can we say about the continuity of f ?

 

[micromass]

If f is bijective and monotone with both its range and domain being a closed and bounded interval...then what can we say about the continuity of f ?

This thread is 2 years old. I guess the OP already found it by now.

 

[Bhatia]

This thread is 2 years old. I guess the OP already found it by now.

Hello, Micromass

Thanks for your reply.

If f is injective and continuous then it is strictly monotone...that is clear

My question is : If f is montone then is f continuous ?

 

[micromass]

No, there are a lot of counterexamples, for example

$$f:\mathbb{R}\rightarrow\mathbb{R}:x\rightarrow \left\{\begin{array}{c} x~\text{if}~x\leq 0\\ x+1~\text{if}~x>0\\ \end{array}\right.$$

This is injective and strictly increasing, but not continuous. If f is required to be surjective however, then it will be continuous.

Hint: next time you may get a faster reply if you just start a new topic about it!

 

[Bhatia]

No, there are a lot of counterexamples, for example

$$f:\mathbb{R}\rightarrow\mathbb{R}:x\rightarrow \left\{\begin{array}{c} x~\text{if}~x\leq 0\\ x+1~\text{if}~x>0\\ \end{array}\right.$$

This is injective and strictly increasing, but not continuous. If f is required to be surjective however, then it will be continuous.

Hint: next time you may get a faster reply if you just start a new topic about it!

Thanks for your help with an example. I appreciate it.

Thanks for yesterday's reply too.

Regards,
Bhatia

 

[disregardthat]

A function which is injective and continuous need not be monotonic. E.g. f : R-{0} --> R defined by f(x) = 1/x is continuous and injective, but not monotonic.

If however f : A --> B is injective and continuous, A and B are totally ordered topological spaces and A is connected, then f will be monotonic. The key here is that R itself in the standard topology is connected, and so continuous injective functions f : R --> R will be monotonic.

 

[Bhatia]

A function which is injective and continuous need not be monotonic. E.g. f : R-{0} --> R defined by f(x) = 1/x is continuous and injective, but not monotonic.

If however f : A --> B is injective and continuous, A and B are totally ordered topological spaces and A is connected, then f will be monotonic. The key here is that R itself in the standard topology is connected, and so continuous injective functions f : R --> R will be monotonic.

Thanks for the insight. I get what you mean now.