Exploring Sets: a\b, f:a→b & Image of f

[mathshelp]

Let a and b be sets. Show that the following constructions are sets stating clearly which axioms you need

(a) a\b.


(b) A function f:a→ b.


(c) The image of f.


(d) Given that a and b have ranks α and β respectively, what are the maximum possible ranks of a\b, f:a→ b and the image of f?



I'm not sure of an answer for b and d at the moment.

For (a) I'm thinking ab= {x∈a: x is not in b} and therefore is a set by the subset axiom

and for (c) I'm thinking the image is {x∈b: there exists a y∈a, f(y)=x} which is a set by the subset axiom

What do you think?

 

[mmwave]

I would like to clarify and expand on your answers to ensure they are correct and fully supported by mathematical axioms.

(a) To show that a\b is a set, we can use the Axiom of Comprehension. This axiom states that for any set A and any property P(x), there exists a set B = {x ∈ A: P(x)} that contains all elements of A that satisfy the property P. In this case, we can define P(x) as "x is not in b". This means that a\b is a set by the Axiom of Comprehension.

(b) To show that a function f: a→b is a set, we can use the Axiom of Replacement. This axiom states that if a set A is mapped by a function f to a set B, then there exists a set C = {f(x): x∈A} that contains all the outputs of the function f. In this case, a and b are sets, and f maps elements from a to b. Therefore, f:a→b is a set by the Axiom of Replacement.

(c) The image of f is also a set by the Axiom of Replacement. This is because the image of f is defined as the set of all outputs of the function f, which is exactly what the Axiom of Replacement states.

(d) The maximum possible rank of a\b is the minimum of the ranks of a and b. This is because the elements of a\b are those that are in a but not in b, which means that the rank of a\b cannot exceed the rank of a.

The maximum possible rank of f:a→b is the maximum of the ranks of a and b. This is because the outputs of the function f are elements of b, which means that the rank of f:a→b cannot exceed the rank of b.

The maximum possible rank of the image of f is the rank of b. This is because the image of f is a subset of b, and the rank of a subset cannot exceed the rank of the original set.