Why are these sets equal? From Morris H. DeGroot and Mark J. Schervish: Probability and Statistics, theorem...
"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 $bigcuplimits_{i=1}^{infty} A_i$ = $bigcuplimits_{i=1}^{n} A_i$."
My question is why are the sets $bigcuplimits_{i=1}^{infty} A_i$ and $bigcuplimits_{i=1}^{n} A_i$ equal here?
I'm confused because, suppose $A_1$ = {$a_1$}, $A_2$ = {$a_2$}, and $A_3$ = {$a_3$}. Then...
$bigcuplimits_{i=1}^{infty} A_i$ = {$a_1$, $a_2$, $emptyset$}
$bigcuplimits_{i=1}^{n} A_i$ = {$a_1$, $a_2$}
So it doesn't make sense to me that these sets are equal...
elementary-set-theory elementary-probability
asked Dec 29 '18 at 10:56
Andres KianiAndres Kiani
52
add a comment |
"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 $bigcuplimits_{i=1}^{infty} A_i$ = $bigcuplimits_{i=1}^{n} A_i$."
My question is why are the sets $bigcuplimits_{i=1}^{infty} A_i$ and $bigcuplimits_{i=1}^{n} A_i$ equal here?
I'm confused because, suppose $A_1$ = {$a_1$}, $A_2$ = {$a_2$}, and $A_3$ = {$a_3$}. Then...
$bigcuplimits_{i=1}^{infty} A_i$ = {$a_1$, $a_2$, $emptyset$}
$bigcuplimits_{i=1}^{n} A_i$ = {$a_1$, $a_2$}
So it doesn't make sense to me that these sets are equal...
elementary-set-theory elementary-probability
asked Dec 29 '18 at 10:56
Andres KianiAndres Kiani
52
1
The assumption is that $A_i=emptyset$, not that $A_i={emptyset}$.
– Lord Shark the Unknown
Dec 29 '18 at 11:00
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.
– Mauro ALLEGRANZA
Dec 29 '18 at 11:00
Oh that makes sense @LordSharktheUnknown, thank you!
– Andres Kiani
Dec 29 '18 at 11:01
add a comment |
"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 $bigcuplimits_{i=1}^{infty} A_i$ = $bigcuplimits_{i=1}^{n} A_i$."
My question is why are the sets $bigcuplimits_{i=1}^{infty} A_i$ and $bigcuplimits_{i=1}^{n} A_i$ equal here?
I'm confused because, suppose $A_1$ = {$a_1$}, $A_2$ = {$a_2$}, and $A_3$ = {$a_3$}. Then...
$bigcuplimits_{i=1}^{infty} A_i$ = {$a_1$, $a_2$, $emptyset$}
$bigcuplimits_{i=1}^{n} A_i$ = {$a_1$, $a_2$}
So it doesn't make sense to me that these sets are equal...
elementary-set-theory elementary-probability
asked Dec 29 '18 at 10:56
Andres KianiAndres Kiani
52
"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 $bigcuplimits_{i=1}^{infty} A_i$ = $bigcuplimits_{i=1}^{n} A_i$."
My question is why are the sets $bigcuplimits_{i=1}^{infty} A_i$ and $bigcuplimits_{i=1}^{n} A_i$ equal here?
I'm confused because, suppose $A_1$ = {$a_1$}, $A_2$ = {$a_2$}, and $A_3$ = {$a_3$}. Then...
$bigcuplimits_{i=1}^{infty} A_i$ = {$a_1$, $a_2$, $emptyset$}
$bigcuplimits_{i=1}^{n} A_i$ = {$a_1$, $a_2$}
So it doesn't make sense to me that these sets are equal...
elementary-set-theory elementary-probability
elementary-set-theory elementary-probability
asked Dec 29 '18 at 10:56
Andres KianiAndres Kiani
52
asked Dec 29 '18 at 10:56
Andres KianiAndres Kiani
52
asked Dec 29 '18 at 10:56
Andres KianiAndres Kiani
52
asked Dec 29 '18 at 10:56
Andres KianiAndres Kiani
52
asked Dec 29 '18 at 10:56
Andres KianiAndres Kiani
52
52
1
The assumption is that $A_i=emptyset$, not that $A_i={emptyset}$.
– Lord Shark the Unknown
Dec 29 '18 at 11:00
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.
– Mauro ALLEGRANZA
Dec 29 '18 at 11:00
Oh that makes sense @LordSharktheUnknown, thank you!
– Andres Kiani
Dec 29 '18 at 11:01
add a comment |
1
The assumption is that $A_i=emptyset$, not that $A_i={emptyset}$.
– Lord Shark the Unknown
Dec 29 '18 at 11:00
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.
– Mauro ALLEGRANZA
Dec 29 '18 at 11:00
Oh that makes sense @LordSharktheUnknown, thank you!
– Andres Kiani
Dec 29 '18 at 11:01
1
1
The assumption is that $A_i=emptyset$, not that $A_i={emptyset}$.
– Lord Shark the Unknown
Dec 29 '18 at 11:00
The assumption is that $A_i=emptyset$, not that $A_i={emptyset}$.
– Lord Shark the Unknown
Dec 29 '18 at 11:00
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.
– Mauro ALLEGRANZA
Dec 29 '18 at 11:00
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.
– Mauro ALLEGRANZA
Dec 29 '18 at 11:00
Oh that makes sense @LordSharktheUnknown, thank you!
– Andres Kiani
Dec 29 '18 at 11:01
Oh that makes sense @LordSharktheUnknown, thank you!
– Andres Kiani
Dec 29 '18 at 11:01
add a comment |
1 Answer
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You can see it this way. Take any element $x$ from $bigcuplimits_{i = 1}^{infty} A_i$. 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, $x in bigcuplimits_{i = 1}^n A_i$. Therefore, $bigcuplimits_{i = 1}^{infty} A_i subseteq bigcuplimits_{i = 1}^n A_i$.
The reverse inclusion is trivial to prove. Hence, the two sets are equal.
answered Dec 29 '18 at 11:00
Aniruddha DeshmukhAniruddha Deshmukh
934418
add a comment |
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You can see it this way. Take any element $x$ from $bigcuplimits_{i = 1}^{infty} A_i$. 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, $x in bigcuplimits_{i = 1}^n A_i$. Therefore, $bigcuplimits_{i = 1}^{infty} A_i subseteq bigcuplimits_{i = 1}^n A_i$.
The reverse inclusion is trivial to prove. Hence, the two sets are equal.
answered Dec 29 '18 at 11:00
Aniruddha DeshmukhAniruddha Deshmukh
934418
add a comment |
You can see it this way. Take any element $x$ from $bigcuplimits_{i = 1}^{infty} A_i$. 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, $x in bigcuplimits_{i = 1}^n A_i$. Therefore, $bigcuplimits_{i = 1}^{infty} A_i subseteq bigcuplimits_{i = 1}^n A_i$.
The reverse inclusion is trivial to prove. Hence, the two sets are equal.
answered Dec 29 '18 at 11:00
Aniruddha DeshmukhAniruddha Deshmukh
934418
add a comment |
You can see it this way. Take any element $x$ from $bigcuplimits_{i = 1}^{infty} A_i$. 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, $x in bigcuplimits_{i = 1}^n A_i$. Therefore, $bigcuplimits_{i = 1}^{infty} A_i subseteq bigcuplimits_{i = 1}^n A_i$.
The reverse inclusion is trivial to prove. Hence, the two sets are equal.
answered Dec 29 '18 at 11:00
Aniruddha DeshmukhAniruddha Deshmukh
934418
You can see it this way. Take any element $x$ from $bigcuplimits_{i = 1}^{infty} A_i$. 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, $x in bigcuplimits_{i = 1}^n A_i$. Therefore, $bigcuplimits_{i = 1}^{infty} A_i subseteq bigcuplimits_{i = 1}^n A_i$.
The reverse inclusion is trivial to prove. Hence, the two sets are equal.
answered Dec 29 '18 at 11:00
Aniruddha DeshmukhAniruddha Deshmukh
934418
answered Dec 29 '18 at 11:00
Aniruddha DeshmukhAniruddha Deshmukh
934418
answered Dec 29 '18 at 11:00
Aniruddha DeshmukhAniruddha Deshmukh
934418
answered Dec 29 '18 at 11:00
Aniruddha DeshmukhAniruddha Deshmukh
934418
934418
add a comment |
add a comment |
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The assumption is that $A_i=emptyset$, not that $A_i={emptyset}$.
– Lord Shark the Unknown
Dec 29 '18 at 11:00
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.
– Mauro ALLEGRANZA
Dec 29 '18 at 11:00
Oh that makes sense @LordSharktheUnknown, thank you!
– Andres Kiani
Dec 29 '18 at 11:01