Axiom of Choice to prove two propositions.

[erogard]

Hi everyone,

we recently covered some implications of the AC and are now to prove the followings statements with the help of the AC or one of its equivalent:

(1) Every uncountable set has a subset of cardinality $$\aleph_1$$ (the least initial ordinal not less or equal than $$\aleph_0$$, the latter being the cardinality of the set of natural numbers, i.e. $$N$$ itself)

(2) If B is an infinite set and A is a subset of B such that |A| $$\lneq$$ |B|, then |B - A| = |B|

I have mostly thought about (1) and to fix f as a choice function for such an uncountable set; then the image of this set under f is an element of it, of cardinality less or equal than that of the uncountable one (call it A).

(well I just realized that it is possible to edit the post so I'll be back with my full post in the proper form with my main attempts on (1) )

PS: is it possible to delete this post? figured out that the Set Theory forum might be more appropriate, my bad.

 

[mighty2000]

Hi there,

Thank you for bringing up these interesting statements about the Axiom of Choice (AC). I am familiar with the AC and its implications in mathematics, especially in set theory.

To prove statement (1), we can use the well-ordering principle, which is equivalent to the AC. This principle states that every set can be well-ordered, meaning that there exists a total order on the set such that every non-empty subset has a least element. Using this principle, we can construct a well-ordered subset of the uncountable set, which will have the same cardinality as the set itself. This subset will then have a cardinality of \aleph_1, as it is the least initial ordinal not less or equal than \aleph_0.

For statement (2), we can use the AC directly. Let A be a subset of the infinite set B such that |A| \lneq |B|. This means that there exists a bijection between A and a proper subset of B. Using the AC, we can choose an element from the complement set B - A for each element in A. This will create a bijection between B - A and B, proving that |B - A| = |B|.

I hope this helps to clarify these statements and their proofs. And don't worry about posting in the wrong forum, we all make mistakes sometimes. It's great that you realized and took the time to repost in the appropriate forum.