Theorem 2.12: Union of countable sets is a countable set

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There's one non-trivial step here: We know that each set is countable, meaning that for each set there exists some function: N -> En that covers all the elements. But to prove that the union of countably many countable sets is countable, you need to choose this function infinitely many times at the same time; there are uncountably many one-to-one
functions N->N. (To be exact, you need a function: N -> fn, where fn is a function from N to En which covers all the elements of En.)

Turns out, to prove that a union of countably many countable sets is countable requires some form of axiom of choice. (On the other hand, that a Cartesian product of two countable sets is countable doesn't require the axiom of choice.)

MikeRosoftJH

What do you think of the following : Start with odd Naturals. Then make the set formed by multiplying those by 2 (singly even) . Then form the set by multiplying the odds by 2^2 (the doubly evens) and continue for all 2^i. You've just created a countable collection of countable sets that is a partition of N.

chumsky

You are explaing well but the fact is the we need a proof other than the books proof...other wise...great...thanx for this .

manauwarhussain

Thank you for making this excellent video!

eudaimona

The idea is very good but you have to consider that 1. the union might not be infinite and also that 2.your countable sets might not be actually infinite also (u explicitly said a countable set is infinite...not(true) ) so you have to like find a rigorous way to say like if this xij is not defined (either because theres not so many countable sets in your union or that the specific set in which your in has not infitely many elements) well skip it. which is not easy
bottomline this proof is not complete but its a strong argument

jopa

and please stop that back ground noise.

manauwarhussain

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