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Image of an open set

  1. Mar 2, 2012 #1
    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: Mar 2, 2012
  2. jcsd
  3. Mar 2, 2012 #2

    micromass

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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.
     
  4. Mar 2, 2012 #3

    Bacle2

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    See what f(x)=x2 from ℝ→ℝ does to (-1,1).
     
  5. Mar 2, 2012 #4
    thank you micromass and bacle2

    now, is there a constant that isn't continous?
     
  6. Mar 2, 2012 #5

    micromass

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    No. All constant functions are continuous.
     
  7. Mar 2, 2012 #6

    Bacle2

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    "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: Mar 2, 2012
  8. Mar 2, 2012 #7
    The preimage of V has to be either X or the empty set.
     
  9. Mar 2, 2012 #8

    Bacle2

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    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?
     
  10. Mar 2, 2012 #9
    must be open.
     
  11. Mar 2, 2012 #10

    Bacle2

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    Right. So putting it all together, a constant function is continuous; the inverse image of every open set is open.
     
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