The discussion centers on finding a continuous function f: R→R such that for an open set A ⊆ R, the image f(A) is not open. A suggested function is f(x) = sin(x), with the open interval A = (0, π) resulting in f(A) = (0, 1], which is neither open nor closed. The participants clarify that while sin((-π, π)) results in the open interval (-1, 1), the function's behavior on specific intervals can yield non-open images, demonstrating the nuances of continuity and image sets in real analysis.
PREREQUISITESMathematics students, particularly those studying real analysis, topology, and anyone interested in the properties of continuous functions and their images.
analysis001 said:Homework Statement
I'm trying to do a problem, and in order to do it I need to find a function f:R→R which is continuous on all of R, where A\subseteqR 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.
analysis001 said:Oh ok would f(A)=sin(A) work?
Dick said:It would. Now you have to give an example of an open set A such that f(A) isn't open.
analysis001 said:If I take A to be the interval (-\pi,\pi) which is open, then sin(A) would be on the interval [-1,1], which is closed. Am I understanding this correctly?
That's not true. ##\sin(\pi/2) = 1##, ##\sin(-\pi/2) = -1##Dick said:Almost, but sin((-pi,pi))=(-1,1) which is an open interval. The endpoints aren't included. Try sin((0,pi)).
Dick said:Almost, but sin((-pi,pi))=(-1,1) which is an open interval. The endpoints aren't included. Try sin((0,pi)).
analysis001 said:I don't get why sin((-\pi,\pi)=(-1,1). I thought because sin(\pi/2)=1 and since \pi/2\in (-\pi,\pi) and also because sin(-\pi/2)=-1 and since -\pi/2\in (-\pi,\pi) then sin((-\pi,\pi)=[-1,1].
If sin((0,\pi) then it equals (0,1] which is neither open nor closed. I'm not sure where my reasoning is wrong.