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pivoxa15
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Homework Statement
Let X be any set, f a function. Let f:X->Y
Does f(A) make sense for A in X?
I know f^(-1)(B) makes sense for B in Y.
The Attempt at a Solution
I can't see why not
HallsofIvy said:Yes, of course it does. f(A) is defined as {y| y= f(x) for some x in A}. In other words it is all values in y that you can actually "get to" using f.
HallsofIvy said:f-1(B)= {x| f(x) is in B}.
Can you answer this: if f(x)= x2 and A= [-2, 1], what is f(A)?
pivoxa15 said:Wouldn't you say for all x in A?
Since f is bijective, f(A)=[1,4]
HallsofIvy said:Yes, of course it does. f(A) is defined as {y| y= f(x) for some x in A}. In other words it is all values in y that you can actually "get to" using f.
No, I wouldn't! 4 is in f(A) for the example below because 4= f(-2). But it certainly is NOT true that 4= f(x) for ALL x in A!pivoxa15 said:Wouldn't you say for all x in A?
HallsofIvy said:f-1(B)= {x| f(x) is in B}.
Can you answer this: if f(x)= x2 and A= [-2, 1], what is f(A)?
Since f is clearly not bijective (it in neither injective nor surjective) that is not correct. To make one obvious point 0 is in [-2,1] so f(0)= 0 must be in f(A). It is NOT in [1, 4].pivoxa15 said:Since f is bijective, f(A)=[1,4]
pivoxa15 said:My bad, I should have drawn a graph. f(A)=[0,4]
With regards to 'f(A) is defined as {y| y= f(x) for some x in A}' What if I view y as a variable? So it is intepreted as all y in the codomain of f such that y=f(x) for all x in A.
Hurkyl said:It depends on how you're doing your set-building.
If we use comprehension, we are thinking about the elements of Y in the image of A.
[tex]f(A) = \{ \, y \in Y \mid \exists x \in A : f(x) = y \, \}[/tex]
If we use replacement, we are thinking about replacing every element of A with its image.
[tex]f(A) = \{ \, f(x) \mid x \in A \, \}[/tex]
What do you mean by "is not allowed"?pivoxa15 said:RIght. So I was thinking of 'replacement language' when intepreting HallsIvy's 'comprehension language' which obviously is not allowed hence the confusion.
Hurkyl said:What do you mean by "is not allowed"?
This certainly isn't grammatically correct. You have a free variable y, and the expression doesn't match any set-builder notation I know.pivoxa15 said:[tex]f(A) = \{ \, y=f(x) \mid x \in A \, \}[/tex]
HallsofIvy said:Ouch! "Replacement Language"? "Comprehesion Language"? This is getting too deep for me. Whatever happened to good old set theory?
Yes, set theory has many real-world applications, particularly in fields such as computer science, statistics, and linguistics. It is used to model relationships and structures in various systems, and to make predictions and solve problems.
A set is a collection of distinct objects, while an element is one of the objects within that set. In other words, a set is a group or category, and an element is a member or part of that group.
Set theory and logic are closely related, as set theory is often used as a foundation for mathematical logic. In set theory, statements can be expressed using logical symbols such as "and," "or," and "not," and logical principles can be used to prove theorems about sets.
Yes, set theory is a powerful and flexible framework that can be used to represent and manipulate a wide range of mathematical concepts. Many branches of mathematics, such as algebra and topology, are based on set theory.
Axioms are fundamental assumptions or principles that serve as the basis for a mathematical system. In set theory, axioms define the properties and operations of sets, and they are used to prove theorems and build more complex mathematical structures.