Proving |A|<=Aleph Null with Function f:A->B and |B|<=Aleph Null

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SUMMARY

The discussion centers on proving that the cardinality of set A, denoted |A|, is less than or equal to Aleph Null (ℵ₀) given a function f: A → B, where |B| ≤ ℵ₀ and for every b in B, the preimage |f⁻¹({b})| ≤ ℵ₀. The user concludes that A can be expressed as the union of the preimages f⁻¹({b}) for each b in B, allowing the assertion that |A| ≤ ℵ₀ to be valid. This conclusion is confirmed by another theorem referenced in the discussion.

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  • Basic grasp of union operations in set theory
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i have a function f:A->B, I am also given that |B|<=null aleph, and for every b in B, |f^-1({b})|<=null aleph, i need to prove that |A|<=null aleph.
basically i think that A equals the union of f^-1({b}) for every b in B, and by another theorem i can consequently assert that |A|<=null aleph.

but is this correct?
 
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Yes...
 
AKG, thanks.
 

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