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Axiom of Choice to prove two propositions.

  1. Dec 5, 2009 #1
    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 [tex]\aleph_1[/tex] (the least initial ordinal not less or equal than [tex]\aleph_0[/tex], the latter being the cardinality of the set of natural numbers, i.e. [tex]N[/tex] itself)

    (2) If B is an infinite set and A is a subset of B such that |A| [tex] \lneq [/tex] |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.
     
    Last edited: Dec 5, 2009
  2. jcsd
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