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I have two countable families of sets $A_i$ and $B_i$ indexed by $\mathbb{N}$. I want to prove that:

$$ \bigcup_{i \in \mathbb{N}} A_i = \bigcup_{i \in \mathbb{N}} Bi $$

And I already have proved that they are equal for the finite union, that is:

$$ \bigcup_{i = 1}^n A_i = \bigcup_{i = 1}^n Bi $$

for all $n \in \mathbb{N}$.

Is that sufficient to complete the proof? If yes, I'm missing the formal argument to do so. I can't take a "limit" on $n$ or similar. What should I do then?

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2 Answers 2

up vote 10 down vote accepted

If $x\in\bigcup_{j\in\mathbb N}A_j$, then there exists $n$ with $x\in A_n$. So $$ x\in A_n\subset\bigcup_{j=1}^nA_j=\bigcup_{j=1}^nB_j\subset \bigcup_{j\in\mathbb N}B_j. $$ This shows that $\bigcup_{j\in\mathbb N}A_j\subset \bigcup_{j\in\mathbb N}B_j$. Now you can exchange the roles of that $A_j$ and $B_j$ to obtain the reverse inclusion.

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If finite unions of sets in the two families are equal, then define $A'_n = \bigcup_{i=1}^n A_i$ and sim for $B_n'$. Then clearly $B_n' = A_n'$, and we have $\bigcup^{\infty} A_i = \bigcup^{\infty} A_n' = \bigcup^{\infty} B_n' = \bigcup^{\infty} B_i$, QED. But I like the above answer better.

That trick of defining new sets to be the finite union up to $n$ of the set sequence is used all the time, for instance a lot in measure theory.

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Funny, I like your answer better :) –  Martin Argerami May 17 at 2:58
    
@MartinArgerami we should be answer buddies then. Lol –  Enjoys Math May 17 at 3:03
    
:) The only thing about your answer is that it should be $A_n'$ and not $A_i'$. –  Martin Argerami May 17 at 3:04
    
@MartinArgerami Thanks. Fixed it. –  Enjoys Math May 17 at 3:05
    
Thank's for the trick then, I came up with this studying Measure Theory –  victorsouza May 17 at 19:14

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