image of an open set

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

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

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.