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Compact Sets/Function

  • #1

Homework Statement


I need to find an example of a set D[itex]\subseteq[/itex]R 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.
 

Answers and Replies

  • #2
33,171
4,858

Homework Statement


I need to find an example of a set D[itex]\subseteq[/itex]R 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.
 
  • #3
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:RR which is continuous on all of R. Find an example satisfying the following:
D[itex]\subseteq[/itex]R is compact but f-1(D) is not.
 
  • #4
33,171
4,858
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.
 
  • #5
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)[itex]\subseteq[/itex]R 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.
 
  • #6
33,171
4,858
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.


[itex]\subseteq[/itex]R 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.
 

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