Drexel dragonThe Math ForumDonate to the Math Forum



Search All of the Math Forum:

Views expressed in these public forums are not endorsed by Drexel University or The Math Forum.


Math Forum » Discussions » sci.math.* » sci.math

Topic: Continuous, locally 1-1 function from Reals to Reals is globally 1-1.
Replies: 10   Last Post: Apr 20, 2013 2:57 PM

Advanced Search

Back to Topic List Back to Topic List Jump to Tree View Jump to Tree View   Messages: [ Previous | Next ]
scattered

Posts: 90
Registered: 6/21/12
Re: Continuous, locally 1-1 function from Reals to Reals is globally 1-1.
Posted: Apr 19, 2013 4:31 PM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

On Friday, April 19, 2013 3:03:41 PM UTC-4, hbe...@gmail.com wrote:
> Hi, all:
>
>
>
> I have been able to show that a continuous, locally 1-1 function from
>
>
>
> Reals to Reals is globally 1-1. by locally 1-1 I mean every point x in R
>
>
>
> has a 'hood ( neighborhood) U_x where f|_U_x is 1-1. But I would like to
>
>
>
> know if locally 1-1 enough is alone, without continuity ( I know f being
>
>
>
> locally monotonic is enough).
>
>
>
> For the continuous case, we just show f must be monotonic in every 'hood
>
>
>
> where it is 1-1 . Then we assume there are x,y with f(x)=f(y) . But then
>
>
>
> there is a chain of monotonic 'hoods joining x to y, forcing all these chains
>
>
>
> to be either all increasing or all decreasing ( we encase x,y in [-M,M]
>
>
>
> so that there is a finite cover by locally 1-1 'hoods).
>
>
>
> Is this true if f is just locally 1-1 but not continuous?
>
>
>
> Thanks.


Consider f(x) = x - floor(x)



Point your RSS reader here for a feed of the latest messages in this topic.

[Privacy Policy] [Terms of Use]

© Drexel University 1994-2014. All Rights Reserved.
The Math Forum is a research and educational enterprise of the Drexel University School of Education.