Register to reply

Image of an open set

by zendani
Tags: image
Share this thread:
zendani
#1
Mar2-12, 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.
Phys.Org News Partner Science news on Phys.org
Wildfires and other burns play bigger role in climate change, professor finds
SR Labs research to expose BadUSB next week in Vegas
New study advances 'DNA revolution,' tells butterflies' evolutionary history
micromass
#2
Mar2-12, 06:44 AM
Mentor
micromass's Avatar
P: 18,099
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
#3
Mar2-12, 07:28 AM
Sci Advisor
P: 1,169
See what f(x)=x2 from ℝ→ℝ does to (-1,1).

zendani
#4
Mar2-12, 08:28 AM
P: 15
Image of an open set

thank you micromass and bacle2

now, is there a constant that isn't continous?
micromass
#5
Mar2-12, 08:41 AM
Mentor
micromass's Avatar
P: 18,099
Quote Quote by zendani View Post
thank you micromass and bacle2

now, is there a constant that isn't continous?
No. All constant functions are continuous.
Bacle2
#6
Mar2-12, 12:14 PM
Sci Advisor
P: 1,169
"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
#7
Mar2-12, 12:37 PM
P: 59
Quote Quote by Bacle2 View Post
"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
#8
Mar2-12, 02:02 PM
Sci Advisor
P: 1,169
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
#9
Mar2-12, 02:36 PM
P: 59
must be open.
Bacle2
#10
Mar2-12, 07:26 PM
Sci Advisor
P: 1,169
Right. So putting it all together, a constant function is continuous; the inverse image of every open set is open.


Register to reply

Related Discussions
Pre-image open or not? Topology and Analysis 6
Program to open .bin image (C++) Computers 2
The inverse image of an open set is open Calculus & Beyond Homework 6
Why do we specify chart image is open? Special & General Relativity 4
Inverse Image of an Open Set Set Theory, Logic, Probability, Statistics 5