Constructing a Set with One Element from Each Uncountably Infinite Subset

[SW VandeCarr]

Given a set of sets such that $$A_{i}\subset{C}$$. Every subset has a countable infinity of elements. I want to create a set $$W$$ such that it contains exactly one element from each subset $$A_{i}$$. I presume I can do this by describing the intersect of $$W$$ with every subset $$A_{i}$$ as containing exactly one element.

Now if, instead, I say that every subset $$A_{i}$$ is an uncountably infinite set of elements, can I still do this?

 

[Petek]

Are you familiar with the Axiom of Choice?

 

[SW VandeCarr]

Are you familiar with the Axiom of Choice?

Yes, I am (as well as Dedekind cuts). However, I wanted to know if I could describe it just this way: that every intersect of W with A_i can contain exactly one element even when there are uncountably many elements in each A_i. This means we can derive the attribute of countability from an uncountable set since W is a countable set. This would seem to obviate the difference between countable infinite sets and uncountable infinite sets.

 

[mXSCNT]

Even in the countable case, you can't always do that. For example, let
$$A_i = \{z \in \mathbb{Z} : z > i\}$$ for positive integer i.
Suppose that z is the smallest element of W (which must exist since W is a subset of the integers greater than 1). Then z appears once in all sets A_i where i < z, and zero times in all sets A_i where i >= z. So W can't consist of just z, so it must have a second-smallest element z' > z. However, z' and z are both elements of each A_i where i < z, when you specified that each element of W can only appear once in each A_i. So W can't exist.

It would be a different story if the A_i are pairwise disjoint.

 

[SW VandeCarr]

Even in the countable case, you can't always do that. For example, let
$$A_i = \{z \in \mathbb{Z} : z > i\}$$ for positive integer i.
Suppose that z is the smallest element of W (which must exist since W is a subset of the integers greater than 1). Then z appears once in all sets A_i where i < z, and zero times in all sets A_i where i >= z. So W can't consist of just z, so it must have a second-smallest element z' > z. However, z' and z are both elements of each A_i where i < z, when you specified that each element of W can only appear once in each A_i. So W can't exist.

It would be a different story if the A_i are pairwise disjoint.

Then you're saying the Axiom of Choice is "wrong" (if an axiom can be wrong) unless we specify just how W is constructed?

 

[Dragonfall]

Then you're saying the Axiom of Choice is "wrong" (if an axiom can be wrong) unless we specify just how W is constructed?

No. To use Choice, you need pairwise disjoint sets. Your W may not always exist, that's all.

 

[SW VandeCarr]

No. To use Choice, you need pairwise disjoint sets. Your W may not always exist, that's all.

OK. That means we can have countably infinite sets A_i which are pairwise disjoint sets if we define them as, for example, each set A_i consists of all powers of a particular prime unique to that set. For uncountably infinite sets A_i any unique non-overlapping interval on the real number line for each A_i will do.

However, no specifications are given by the AC. All it says (one version from the Wiki) is "For any non-empty set X there exists a choice function f defined on X." This version doesn't even seem to preclude choosing the same element more than once. My Borowski & Borwein math dictionary states "..from every family of disjoint sets, a set can be constructed containing exactly one element from each (set?) of the given family of sets." These two definitions don't seem to me to be exactly the same. Moreover, having a family of disjoint sets isn't the same as a set of pairwise disjoint sets. For example {{{1, 2},{3,4}}{{2,4},{1,3}}} is a set of pairwise disjoint sets, but is not itself a set of disjoint sets.

 

[Dragonfall]

Every set that is pairwise disjoint is disjoint. Your example is a disjoint set.

First, AC only states that the choice function exists for the power set of X. Second, the existence of that choice function does not mean the existence of your W set.