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Image of an open set

by zendani
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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.
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micromass
#2
Mar2-12, 06:44 AM
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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
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P: 1,172
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
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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
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P: 1,172
"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: 1,172
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: 1,172
Right. So putting it all together, a constant function is continuous; the inverse image of every open set is open.


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