MHB Countable Union of Countable Sets

AI Thread Summary
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
Gold Member
MHB
Messages
3,741
Reaction score
0
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?
 
Physics news on Phys.org
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)
 

Similar threads

Replies
1
Views
2K
Replies
14
Views
3K
Replies
3
Views
5K
Replies
14
Views
2K
Replies
5
Views
2K
Replies
2
Views
3K
Replies
3
Views
2K
Back
Top