Theorem 2.12: Union of countable sets is a countable set

[@MikeRosoftJH]

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.)

[@manauwarhussain4003]

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

[@eudaimona]

Thank you for making this excellent video!

[@chumsky8754]

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.

[@soumyadasgupta7758]

Thanks for this elegant video

[@jopa1996]

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

[@hereticmindset]

You are correct in the sense that I never explicitly mention the situation you described. But if you think about it, both 1 and 2 seem trivially true. If the union is finite, well, then its even easier for us, because we can just walk each column till the end and restart on top. So x1,1 -> x2,1 -> x3,1 -> x1,2 -> and etc ... If the union is infinite, but in our sequence there are finite sets, we will exhaust these sets during our diagonal traversal. Even simpler. |infinite union of infinite sets| > |infinite union of infinite sets| |infinite union of infinite sets| > |finite union of infinite| |infinite union of infinite sets| > |infinite union of infinite and finite sets| and etc... Since we showed that the biggest one is countable, the smaller once will be also countable. That said, you are correct in saying that I made assumptions without explicitly pointing them out. Thank you!

[@elhoplita69]

That is trivial lol

[@jamesb46]

Just use prime powers lol.

[@chenweiming7639]

so good from a Chinese

[@manauwarhussain4003]

and please stop that back ground noise.

[@hereticmindset]

Thank you for your input. The truth is that I sadly was not able to record much more during the summer. But I will try to add more videos through out the year (without the background noise and with more original explanations).