The discussion revolves around finding an example of a compact set \( D \subseteq \mathbb{R} \) such that the pre-image \( f^{-1}(D) \) is not compact. The context involves concepts from topology and real analysis, particularly focusing on the properties of continuous functions and their images.
The discussion is ongoing, with participants questioning the notation and attempting to clarify the problem statement. Some guidance has been offered regarding visualizing the relationship between sets and functions, but no consensus has been reached on a specific example.
There is a noted confusion regarding the notation used in the problem, particularly distinguishing between the pre-image and the function inverse. Participants are also grappling with the implications of continuity and compactness in the context of the sine function.
Unless f is a map from D to itself, your notation doesn't make sense. The symbol f(D) implies that D is the domain, and f(D) is the image under f. The symbol f-1(D) implies that D is now the range.analysis001 said:Homework Statement
I need to find an example of a set D\subseteqR is compact but f-1(D) is not.
Homework Equations
f-1(D) is the pre-image of f(D), not the inverse.
analysis001 said:The Attempt at a Solution
I'm having trouble visualizing a function that would work for this scenario. Any clues would be helpful.
Mark44 said:Unless f is a map from D to itself, your notation doesn't make sense. The symbol f(D) implies that D is the domain, and f(D) is the image under f. The symbol f-1(D) implies that D is now the range.
Mark44 said:The image of the sine function is the interval [-1, 1]. The inverse image of [-1, 1] is R. If we're talking about inverse image as opposed to function inverse (f-1), this should work. If you really do mean f-1 as the function inverse, then no, it won't work, as the sine function isn't one-to-one, so doesn't have an inverse that is a function.
No. You're getting all balled up in the notation and not understanding what it's supposed to mean. The problem is to find a function and a set D (NOT f(D)) that is compact, but the inverse image of D is not compact.analysis001 said:Yes, it's talking about the inverse image, not the function inverse. I don't really see how f(D)=sin(D) would work though. If the question was to find a f(D)
analysis001 said:\subseteqR where f(D) is compact but f-1(D) is not then I see how f(D)=sin(D) would work, because f(D)=[-1,1] is compact but f-1(D)=R is not (I think). Maybe I'm just understanding it wrong because I don't see how f(D)=sin(D) works.