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Continuous Function- Open Sets

  1. Apr 16, 2014 #1
    1. The problem statement, all variables and given/known data
    I'm trying to do a problem, and in order to do it I need to find a function f:RR which is continuous on all of R, where A[itex]\subseteq[/itex]R is open but f(A) is not. Can anyone give an example of a function that satisfies these properties? I think once I have an example I'll understand things a lot better. Thanks.
     
  2. jcsd
  3. Apr 16, 2014 #2

    Dick

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    Try a function with a maximum or a minimum and think about it. A lot of functions work.
     
  4. Apr 16, 2014 #3
    Oh ok would f(A)=sin(A) work?
     
  5. Apr 16, 2014 #4

    Dick

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    It would. Now you have to give an example of an open set A such that f(A) isn't open.
     
  6. Apr 16, 2014 #5
    If I take A to be the interval (-[itex]\pi[/itex],[itex]\pi[/itex]) which is open, then sin(A) would be on the interval [-1,1], which is closed. Am I understanding this correctly?
     
  7. Apr 16, 2014 #6

    Dick

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    Almost, but sin((-pi,pi))=(-1,1) which is an open interval. The endpoints aren't included. Try sin((0,pi)).
     
  8. Apr 16, 2014 #7

    jbunniii

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    That's not true. ##\sin(\pi/2) = 1##, ##\sin(-\pi/2) = -1##

    Any open interval containing ##[-\pi/2,\pi/2]## will work.
     
  9. Apr 16, 2014 #8

    Zondrina

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    I believe something like:

    ##f: x → a^2 - (x-a)^2## should fit those requirements.
     
  10. Apr 16, 2014 #9
    I don't get why sin((-[itex]\pi[/itex],[itex]\pi[/itex])=(-1,1). I thought because sin([itex]\pi[/itex]/2)=1 and since [itex]\pi[/itex]/2[itex]\in[/itex] (-[itex]\pi[/itex],[itex]\pi[/itex]) and also because sin(-[itex]\pi[/itex]/2)=-1 and since -[itex]\pi[/itex]/2[itex]\in[/itex] (-[itex]\pi[/itex],[itex]\pi[/itex]) then sin((-[itex]\pi[/itex],[itex]\pi[/itex])=[-1,1].

    If sin((0,[itex]\pi[/itex]) then it equals (0,1] which is neither open nor closed. I'm not sure where my reasoning is wrong.
     
  11. Apr 16, 2014 #10

    Dick

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    jbunniii's correction is correct and you are correct. I slipped up. Sorry!
     
  12. Apr 16, 2014 #11
    Thanks everyone!
     
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