Compact Sets and Function Pre-Image Example | Homework Help

[analysis001]

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.

The Attempt at a Solution

I'm having trouble visualizing a function that would work for this scenario. Any clues would be helpful.

 

[Mark44]

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.

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.

The Attempt at a Solution

I'm having trouble visualizing a function that would work for this scenario. Any clues would be helpful.

 

[analysis001]

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.

Ok, I might have summarized the problem wrong. I'll write it word for word here:

Consider a function f:R→R which is continuous on all of R. Find an example satisfying the following:
D\subseteqR is compact but f^-1(D) is not.

 

[Mark44]

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.

 

[analysis001]

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.

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)\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.

 

[Mark44]

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)

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.

A number d $\in$ D provided that there exists a real number x for which sin(x) = d. Draw a picture with two sets, with x in one set and d in the other set (set D). That might be helpful.

\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.