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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.

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.

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