If I=[0,1] and f: I->R is defined by

f(x)= {x if x is rational, 1-x if x is irrational.

What point of I is f continuous and why?

Also, is f monotone and/or is f one-to-one on I?

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- Apr 3rd 2011, 08:02 AMtn11631continuous? monotone? one to one?
If I=[0,1] and f: I->R is defined by

f(x)= {x if x is rational, 1-x if x is irrational.

What point of I is f continuous and why?

Also, is f monotone and/or is f one-to-one on I? - Apr 3rd 2011, 08:12 AMPlato
- Apr 3rd 2011, 08:32 AMtn11631
- Apr 3rd 2011, 08:48 AMPlato
- Apr 3rd 2011, 09:03 AMtn11631
- Apr 3rd 2011, 09:22 AMPlato
That is correct. It is not monotone.

Consider - Apr 3rd 2011, 09:38 AMtn11631
Yay! Sorry I'm just excited that I managed to pull something out of my brain :) however what about I being one-to-one? I had to look up this definiton becaus e I couldn't remember it but it says that a 1-1 is one that assumes each value in its image exactly once. I wanna say that it is NOT one-to-one but I don't know why and I feel like I'm wrong :/

- Apr 3rd 2011, 09:54 AMPlato
You want to show that if then .

That is easy if both are rational or both irrational.

BUT, what if one is rational and the other irrational? - Apr 3rd 2011, 11:33 AMtn11631
- Apr 3rd 2011, 12:25 PMPlato
- Apr 4th 2011, 07:35 AMtn11631
Oh no now I'm confused. So based on what you said it's making me think yes. However, when i look at the picture of the graph it cannot be that for every point and I thought that was the point of one-to-one. ...also could you possibly explain why for the first part where f is continuous at .5 I thought isaw it but i don't understand it.

- Apr 4th 2011, 07:51 AMPlato
- Apr 4th 2011, 08:26 AMHartlw
continuous at .5

f(.5+d)=.5+d, d rational

f(.5+d)=1-(.5+d) = .5-d, d irrational

l f(.5+d)-.5 l = ldl < e if ldl < e