# image of an open set

by zendani
Tags: image
 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.
 PF Patron Sci Advisor Thanks Emeritus P: 15,671 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.
 Sci Advisor P: 1,167 See what f(x)=x2 from ℝ→ℝ does to (-1,1).
P: 15

## image of an open set

thank you micromass and bacle2

now, is there a constant that isn't continous?
PF Patron
Thanks
Emeritus
P: 15,671
 Quote by zendani thank you micromass and bacle2 now, is there a constant that isn't continous?
No. All constant functions are continuous.
 Sci Advisor P: 1,167 "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)?
P: 59
 Quote by Bacle2 "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)?
The preimage of V has to be either X or the empty set.
 Sci Advisor P: 1,167 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?
 P: 59 must be open.
 Sci Advisor P: 1,167 Right. So putting it all together, a constant function is continuous; the inverse image of every open set is open.

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