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

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SUMMARY

The discussion centers on the properties of constant functions in relation to open sets in topology. It is established that the image of an open set under a constant function is not necessarily open, as demonstrated through examples. Constant functions are confirmed to be continuous, meaning their inverse images of open sets are also open. The participants, micromass and bacle2, clarify the definitions and implications of these concepts effectively.

PREREQUISITES
  • Understanding of open sets in topology
  • Familiarity with the concept of continuous functions
  • Knowledge of inverse images in function mapping
  • Basic grasp of mathematical notation and functions
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  • Study the definition and properties of open functions in topology
  • Explore the implications of the inverse image theorem
  • Learn about the characteristics of continuous functions in various contexts
  • Investigate examples of functions that are not open and their properties
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Mathematicians, students of topology, and anyone interested in the properties of functions and their mappings in mathematical analysis.

zendani
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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:
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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.
 
See what f(x)=x2 from ℝ→ℝ does to (-1,1).
 
thank you micromass and bacle2

now, is there a constant that isn't continous?
 
zendani said:
thank you micromass and bacle2

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

No. All constant functions are continuous.
 
"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:
Bacle2 said:
"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.
 
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?
 
must be open.
 
  • #10
Right. So putting it all together, a constant function is continuous; the inverse image of every open set is open.
 

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