Continuous Function- Open Sets

  • #1

Homework Statement


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.
 

Answers and Replies

  • #2
Dick
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Homework Statement


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.

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

It would. Now you have to give an example of an open set A such that f(A) isn't open.
 
  • #5
It would. Now you have to give an example of an open set A such that f(A) isn't open.

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?
 
  • #6
Dick
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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?

Almost, but sin((-pi,pi))=(-1,1) which is an open interval. The endpoints aren't included. Try sin((0,pi)).
 
  • #7
jbunniii
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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)).
That's not true. ##\sin(\pi/2) = 1##, ##\sin(-\pi/2) = -1##

Any open interval containing ##[-\pi/2,\pi/2]## will work.
 
  • #8
STEMucator
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I believe something like:

##f: x → a^2 - (x-a)^2## should fit those requirements.
 
  • #9
Almost, but sin((-pi,pi))=(-1,1) which is an open interval. The endpoints aren't included. Try sin((0,pi)).

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.
 
  • #10
Dick
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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.

jbunniii's correction is correct and you are correct. I slipped up. Sorry!
 
  • #11
Thanks everyone!
 

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