Strictly and One to One Function is Continuous

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 [tex]f : \mathbb{R} \to \mathbb{R}[/tex] given by

[tex]f(x) = \left\{ \begin{array}{ll}
1/x & x \neq 0 \\
0 & x = 0
\end{array} \right.[/tex]

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?

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

[tex]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.[/tex]

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
[tex]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.[/tex]

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.

Strictly and One to One Function is Continuous

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