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

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Discussion Overview

The discussion centers on the properties of functions, particularly focusing on whether the image of an open set under a constant function is open. Participants explore concepts related to continuity and the behavior of constant functions in the context of open sets.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • Some participants propose that the image of an open set under a constant function does not have to be open.
  • Others agree that the image of an open set does not have to be open in general, citing constant functions as examples.
  • One participant mentions that a function that does satisfy the condition of mapping open sets to open sets is called an open function.
  • There is a question raised about the existence of a constant function that is not continuous, with a later reply asserting that all constant functions are continuous.
  • Participants discuss the inverse image definition of open sets, considering cases where the open set contains or does not contain the constant value.
  • One participant suggests that the preimage of an open set under a constant function must be either the entire domain or the empty set.
  • Another participant emphasizes that the continuity version of open sets implies that the inverse image of every open set must be open.

Areas of Agreement / Disagreement

Participants generally agree that the image of an open set under a constant function is not open, and that all constant functions are continuous. However, there is ongoing exploration regarding the implications of these properties and the definitions involved.

Contextual Notes

The discussion includes assumptions about the definitions of open sets and continuity, and the implications of these definitions on the behavior of constant functions. Some mathematical steps and definitions remain unresolved or are not fully explored.

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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