Does this make sense in set theory?

[pivoxa15]

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]

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.

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]

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.

Wouldn't you say for all x in A?

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 bijective, f(A)=[1,4]

 

[genneth]

Wouldn't you say for all x in A?

Since f is bijective, f(A)=[1,4]

You could do that, but why waste the words when there's no confusion?

And x^2 is *not* bijective. Think again...

 

[HallsofIvy]

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.

Wouldn't you say for all x in A?

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!

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 bijective, f(A)=[1,4]

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]

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]

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.
$$f(A) = \{ \, y \in Y \mid \exists x \in A : f(x) = y \, \}$$

If we use replacement, we are thinking about replacing every element of A with its image.
$$f(A) = \{ \, f(x) \mid x \in A \, \}$$

 

[HallsofIvy]

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.

in {y| y= f(x) for some x in A}, y is a member of a set, not a variable.

 

[pivoxa15]

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.
$$f(A) = \{ \, y \in Y \mid \exists x \in A : f(x) = y \, \}$$

If we use replacement, we are thinking about replacing every element of A with its image.
$$f(A) = \{ \, f(x) \mid x \in A \, \}$$

RIght. So I was thinking of 'replacement language' when intepreting HallsIvy's 'comprehension language' which obviously is not allowed hence the confusion.

 

[Hurkyl]

RIght. So I was thinking of 'replacement language' when intepreting HallsIvy's 'comprehension language' which obviously is not allowed hence the confusion.

What do you mean by "is not allowed"?

 

[pivoxa15]

What do you mean by "is not allowed"?

Adding y in replacement language
$$f(A) = \{ \, y=f(x) \mid x \in A \, \}$$

and intepreting that y as the same y in comprehension language.

 

[Hurkyl]

$$f(A) = \{ \, y=f(x) \mid x \in A \, \}$$

This certainly isn't grammatically correct. You have a free variable y, and the expression doesn't match any set-builder notation I know.

Anyways, it strikes me that I should have made a more positive response, and I'll do that now.

The axiom of comprehension (a.k.a. axiom of subsets) says that if you have some set S and some predicate P, there exists a set containing exactly those elements of S that satisfy P. A standard way to denote that set is with the syntax
$$\{ \, s \in S \mid P(s) \, \}.$$
Sometimes, when S can be inferred from the context, you will see the shorthand notation
$$\{ \, s \mid P(s) \, \}.$$

So, the expression
$$f(A) = \{ \, y \in Y \mid \exists x \in A : f(x) = y \, \}$$
is certainly a grammatically correct mathematical statement.

 

[HallsofIvy]

Ouch! "Replacement Language"? "Comprehesion Language"? This is getting too deep for me. Whatever happened to good old set theory?

 

[genneth]

Ouch! "Replacement Language"? "Comprehesion Language"? This is getting too deep for me. Whatever happened to good old set theory?

Left with Russell when he got his barber to shave him.