"Proof. Consider the infinite sequence of events $A_1, A_2, \dots,$ in which $A_1, \dots, A_n$ are the $n$ given disjoint events and $A_i$ = $\emptyset$ for $i > n$. Then the events in this infinite sequence are disjoint and $\bigcup\limits_{i=1}^{\infty} A_i$ = $\bigcup\limits_{i=1}^{n} A_i$."

My question is why are the sets $\bigcup\limits_{i=1}^{\infty} A_i$ and $\bigcup\limits_{i=1}^{n} A_i$ equal here?

I'm confused because, suppose $A_1$ = {$a_1$}, $A_2$ = {$a_2$}, and $A_3$ = {$a_3$}. Then...

  • $\bigcup\limits_{i=1}^{\infty} A_i$ = {$a_1$, $a_2$, $\emptyset$}

  • $\bigcup\limits_{i=1}^{n} A_i$ = {$a_1$, $a_2$}

So it doesn't make sense to me that these sets are equal...

  • 1
    $\begingroup$ The assumption is that $A_i=\emptyset$, not that $A_i=\{\emptyset\}$. $\endgroup$ – Lord Shark the Unknown Dec 29 '18 at 11:00
  • $\begingroup$ If all the $A_i$ are empty, for $i > n$, then "adding" them to $\bigcup_{i=1}^n A_i$ does not add new elements. $\endgroup$ – Mauro ALLEGRANZA Dec 29 '18 at 11:00
  • $\begingroup$ Oh that makes sense @LordSharktheUnknown, thank you! $\endgroup$ – Andres Kiani Dec 29 '18 at 11:01

You can see it this way. Take any element $x$ from $\bigcup\limits_{i = 1}^{\infty} A_i$. Then, it is in one of the $A_i$ for some $i$. Since $A_i = \emptyset$ for $i > n$, that $i$ has to be $\leq n$. Hence, $x \in \bigcup\limits_{i = 1}^n A_i$. Therefore, $\bigcup\limits_{i = 1}^{\infty} A_i \subseteq \bigcup\limits_{i = 1}^n A_i$.

The reverse inclusion is trivial to prove. Hence, the two sets are equal.


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