
#1
Mar212, 06:17 AM

P: 15

Hi
the image of an open set by a function continues IS an open set??? i think that if we have a constant function , the image of an open set does not have to be open. 



#2
Mar212, 06:44 AM

Mentor
P: 16,580

Indeed, the image of an open set does not have to be open in general. Your example of a constant function is a good one.
A function that does satsify that the image of an open set is open, is called an open function. 



#3
Mar212, 07:28 AM

Sci Advisor
P: 1,168

See what f(x)=x^{2} from ℝ→ℝ does to (1,1).




#4
Mar212, 08:28 AM

P: 15

image of an open set
thank you micromass and bacle2
now, is there a constant that isn't continous? 



#6
Mar212, 12:14 PM

Sci Advisor
P: 1,168

"now, Is there any constant that is not continuous?"
Try this: f:X>Y , f(x)=c , constant. Use the inverse open set definition: V open in Y; then you have two main options: i)V contains c. ii)V does not contain c. What can you say about f^{1}(V)? 



#7
Mar212, 12:37 PM

P: 59





#8
Mar212, 02:02 PM

Sci Advisor
P: 1,168

Right; don't mean to drag it along too much, but:
What does the continuity version of open sets say? The inverse image of an open set.... What follows, then? 



#9
Mar212, 02:36 PM

P: 59

must be open.




#10
Mar212, 07:26 PM

Sci Advisor
P: 1,168

Right. So putting it all together, a constant function is continuous; the inverse image of every open set is open.



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