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)
 
Greetings, I am studying probability theory [non-measure theory] from a textbook. I stumbled to the topic stating that Cauchy Distribution has no moments. It was not proved, and I tried working it via direct calculation of the improper integral of E[X^n] for the case n=1. Anyhow, I wanted to generalize this without success. I stumbled upon this thread here: https://www.physicsforums.com/threads/how-to-prove-the-cauchy-distribution-has-no-moments.992416/ I really enjoyed the proof...

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