Image of an open set

1. Mar 2, 2012

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.

Last edited: Mar 2, 2012
2. Mar 2, 2012

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.

3. Mar 2, 2012

Bacle2

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

4. Mar 2, 2012

zendani

thank you micromass and bacle2

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

5. Mar 2, 2012

micromass

No. All constant functions are continuous.

6. Mar 2, 2012

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)?

Last edited: Mar 2, 2012
7. Mar 2, 2012

inknit

The preimage of V has to be either X or the empty set.

8. Mar 2, 2012

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?

9. Mar 2, 2012

inknit

must be open.

10. Mar 2, 2012

Bacle2

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