Open Set Image: Is Constant Function's Image Open?

[zendani]

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.

 

[micromass]

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.

 

[Bacle2]

See what f(x)=x² from ℝ→ℝ does to (-1,1).

 

[zendani]

thank you micromass and bacle2

now, is there a constant that isn't continous?

 

[micromass]

thank you micromass and bacle2

now, is there a constant that isn't continous?

No. All constant functions are continuous.

 

[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)?

 

[inknit]

"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.

 

[Bacle2]

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?

 

[inknit]

must be open.

 

[Bacle2]

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