Set theory: Axioms of Construction

oliphant
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Homework Statement



We're asked to prove that a few constructions of the sets a,b are themselves sets, stating which axioms we use to do so.

a) a\b
b)the function f:a->b
c)the image of f

Homework Equations



The following standard definitions of axioms of construction: Extensionality, Pair Set, Power Set, Union, Subset

The Attempt at a Solution



I think for a) we could define a\b = \{\bigcup (x \in P(a)) | x \notin b \} so we would be using the axioms of power set, union, subset.

Now I have no idea how to do b), and I'd need to define that set first before I could do c). So if someone could point me in the right direction that would be great.
 
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As for a), if you mean the set A \backslash B, I cannot understand why you use the Power Set of A; this is the set of subsets of A, while A \backslash B is the set of elements of A that are not in B. If A and B are sets, then A \backslash B is defined by the formula:

x \in A\backslash B \equiv x\in A\wedge x\notin B

As for b) (and c)), remember that a function is just a set of ordered pairs, with a certain property.
 
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Oops, that's what I wrote down first of all (well a\b = \{x \in a | x \notin b\} for a) but then decided to over complicate it and confuse matters.

So if we think of the function as a set of ordered pairs (s,t) (where the same a cannot give different bs). I actually have a definition for the set of ordered pairs a \times b = \{x \in P P \bigcup \{a,b\} | \thereis s \in a, t \in b, x = (s,t)\}, so would it just be a case of shoehorning in a condition that when f(a_1) = b and f(a_2) = b, then a_1 = a_2?
 
That's pretty much it; you know that the class of ordered pairs is set, then you pick the subset representing the function. The image is a subset of the function set.
 
JSuarez said:
That's pretty much it; you know that the class of ordered pairs is set, then you pick the subset representing the function. The image is a subset of the function set.
Brilliant, thank you very much!
 
There question then asks what the maximum rank for each of the 3 sets we defined would be (taking alpha, beta as ranks for a, b). for a) would it simply be alpha, as in when beta is the empty set, right? What sort of reasoning could I use for b?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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