MHB Countable Union of Countable Sets

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The discussion centers on the proposition that if there is a sequence of surjective functions from the natural numbers to a sequence of sets, then there exists a surjective function from the natural numbers to the union of those sets. This implies that the union of countably many sets, each of which is at most countable, remains at most countable. The user seeks an example to illustrate this concept. The conversation confirms the understanding of the proposition's implications regarding countable sets. The topic highlights fundamental principles in set theory regarding countable unions.
evinda
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Hello! (Wave)

I am looking at the proposition:

[m] If $(A_n)_{n \in \omega}$ is a sequence of sets and $(f_n)_{n \in \omega}$ is a sequence of functions then:

for all $n \in \omega, f_n: \omega \overset{\text{ surjective }}{\rightarrow} A_n$ then there is a function $f: \omega \overset{\text{ surjective }}{\rightarrow} \bigcup_{n \in \omega} A_n$. [/m]Could you give me an example of such a case?
 
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This simply says that the union of countably many sets that are at most countable is at most countable.
 
Evgeny.Makarov said:
This simply says that the union of countably many sets that are at most countable is at most countable.

I see... Thanks a lot! (Smile)
 
There is a nice little variation of the problem. The host says, after you have chosen the door, that you can change your guess, but to sweeten the deal, he says you can choose the two other doors, if you wish. This proposition is a no brainer, however before you are quick enough to accept it, the host opens one of the two doors and it is empty. In this version you really want to change your pick, but at the same time ask yourself is the host impartial and does that change anything. The host...

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