Mixing
The union of a class (collection of sets) is a set if and only if the class (collection of sets) is a set
$begingroup$
Let $mathfrak{C}$ be a class. I want to show that $bigcup mathfrak{C} = {x: exists C in mathfrak{C}, x in C}$ is a set exactly when $mathfrak{C}$ is a set.
If $mathfrak{C}$ is a set, then using the Axiom of Unions, we have that $bigcup mathfrak{C}$ is a set.
But I am unsure about how to show the other direction. Any advice is appreciated.
Edit: Would it be valid to do the following? Or am I using some kind of circular logic here?
Let $mathfrak{X} = { mathfrak{C} : bigcup mathfrak{C} text{ is a set} }$.
Then $mathfrak{C} in mathfrak{X}$ whenever $bigcup mathfrak{C}$ is a set, i.e. $mathfrak{C}$ is a set whenever $bigcup mathfrak{C}$ is a set.
set-theory
edited Jan 31 at 0:33
Andrés E. Caicedo
65.9k8160252
asked Jan 30 at 19:02
TuringTester69TuringTester69
307213
$endgroup$
add a comment |
$begingroup$
Let $mathfrak{C}$ be a class. I want to show that $bigcup mathfrak{C} = {x: exists C in mathfrak{C}, x in C}$ is a set exactly when $mathfrak{C}$ is a set.
If $mathfrak{C}$ is a set, then using the Axiom of Unions, we have that $bigcup mathfrak{C}$ is a set.
But I am unsure about how to show the other direction. Any advice is appreciated.
Edit: Would it be valid to do the following? Or am I using some kind of circular logic here?
Let $mathfrak{X} = { mathfrak{C} : bigcup mathfrak{C} text{ is a set} }$.
Then $mathfrak{C} in mathfrak{X}$ whenever $bigcup mathfrak{C}$ is a set, i.e. $mathfrak{C}$ is a set whenever $bigcup mathfrak{C}$ is a set.
set-theory
edited Jan 31 at 0:33
Andrés E. Caicedo
65.9k8160252
asked Jan 30 at 19:02
TuringTester69TuringTester69
307213
$endgroup$
$begingroup$
"C $in$ X whenever UC is a set" - it follows just from the definition of X. Shouldn't it be "C $in$ X whenever C is a set, i.e. C is a set whenever UC is a set" ?
$endgroup$
– Jakub Andruszkiewicz
Feb 4 at 7:54
add a comment |
$begingroup$
Let $mathfrak{C}$ be a class. I want to show that $bigcup mathfrak{C} = {x: exists C in mathfrak{C}, x in C}$ is a set exactly when $mathfrak{C}$ is a set.
If $mathfrak{C}$ is a set, then using the Axiom of Unions, we have that $bigcup mathfrak{C}$ is a set.
But I am unsure about how to show the other direction. Any advice is appreciated.
Edit: Would it be valid to do the following? Or am I using some kind of circular logic here?
Let $mathfrak{X} = { mathfrak{C} : bigcup mathfrak{C} text{ is a set} }$.
Then $mathfrak{C} in mathfrak{X}$ whenever $bigcup mathfrak{C}$ is a set, i.e. $mathfrak{C}$ is a set whenever $bigcup mathfrak{C}$ is a set.
set-theory
edited Jan 31 at 0:33
Andrés E. Caicedo
65.9k8160252
asked Jan 30 at 19:02
TuringTester69TuringTester69
307213
$endgroup$
Let $mathfrak{C}$ be a class. I want to show that $bigcup mathfrak{C} = {x: exists C in mathfrak{C}, x in C}$ is a set exactly when $mathfrak{C}$ is a set.
If $mathfrak{C}$ is a set, then using the Axiom of Unions, we have that $bigcup mathfrak{C}$ is a set.
But I am unsure about how to show the other direction. Any advice is appreciated.
Edit: Would it be valid to do the following? Or am I using some kind of circular logic here?
Let $mathfrak{X} = { mathfrak{C} : bigcup mathfrak{C} text{ is a set} }$.
Then $mathfrak{C} in mathfrak{X}$ whenever $bigcup mathfrak{C}$ is a set, i.e. $mathfrak{C}$ is a set whenever $bigcup mathfrak{C}$ is a set.
set-theory
set-theory
edited Jan 31 at 0:33
Andrés E. Caicedo
65.9k8160252
asked Jan 30 at 19:02
TuringTester69TuringTester69
307213
edited Jan 31 at 0:33
Andrés E. Caicedo
65.9k8160252
asked Jan 30 at 19:02
TuringTester69TuringTester69
307213
edited Jan 31 at 0:33
Andrés E. Caicedo
65.9k8160252
edited Jan 31 at 0:33
Andrés E. Caicedo
65.9k8160252
edited Jan 31 at 0:33
Andrés E. Caicedo
65.9k8160252
65.9k8160252
asked Jan 30 at 19:02
TuringTester69TuringTester69
307213
asked Jan 30 at 19:02
TuringTester69TuringTester69
307213
asked Jan 30 at 19:02
TuringTester69TuringTester69
307213
307213
$begingroup$
"C $in$ X whenever UC is a set" - it follows just from the definition of X. Shouldn't it be "C $in$ X whenever C is a set, i.e. C is a set whenever UC is a set" ?
$endgroup$
– Jakub Andruszkiewicz
Feb 4 at 7:54
add a comment |
$begingroup$
"C $in$ X whenever UC is a set" - it follows just from the definition of X. Shouldn't it be "C $in$ X whenever C is a set, i.e. C is a set whenever UC is a set" ?
$endgroup$
– Jakub Andruszkiewicz
Feb 4 at 7:54
$begingroup$
"C $in$ X whenever UC is a set" - it follows just from the definition of X. Shouldn't it be "C $in$ X whenever C is a set, i.e. C is a set whenever UC is a set" ?
$endgroup$
– Jakub Andruszkiewicz
Feb 4 at 7:54
$begingroup$
"C $in$ X whenever UC is a set" - it follows just from the definition of X. Shouldn't it be "C $in$ X whenever C is a set, i.e. C is a set whenever UC is a set" ?
$endgroup$
– Jakub Andruszkiewicz
Feb 4 at 7:54
add a comment |
1 Answer
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$begingroup$
If $X$ = $bigcup C$ is a set, then $Csubseteq P(X)$, so $C$ is a set.
answered Jan 30 at 19:11
Jakub AndruszkiewiczJakub Andruszkiewicz
2116
$endgroup$
add a comment |
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1 Answer
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$begingroup$
If $X$ = $bigcup C$ is a set, then $Csubseteq P(X)$, so $C$ is a set.
answered Jan 30 at 19:11
Jakub AndruszkiewiczJakub Andruszkiewicz
2116
$endgroup$
add a comment |
$begingroup$
If $X$ = $bigcup C$ is a set, then $Csubseteq P(X)$, so $C$ is a set.
answered Jan 30 at 19:11
Jakub AndruszkiewiczJakub Andruszkiewicz
2116
$endgroup$
add a comment |
$begingroup$
If $X$ = $bigcup C$ is a set, then $Csubseteq P(X)$, so $C$ is a set.
answered Jan 30 at 19:11
Jakub AndruszkiewiczJakub Andruszkiewicz
2116
$endgroup$
If $X$ = $bigcup C$ is a set, then $Csubseteq P(X)$, so $C$ is a set.
answered Jan 30 at 19:11
Jakub AndruszkiewiczJakub Andruszkiewicz
2116
answered Jan 30 at 19:11
Jakub AndruszkiewiczJakub Andruszkiewicz
2116
answered Jan 30 at 19:11
Jakub AndruszkiewiczJakub Andruszkiewicz
2116
answered Jan 30 at 19:11
Jakub AndruszkiewiczJakub Andruszkiewicz
2116
2116
add a comment |
add a comment |
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$begingroup$
"C $in$ X whenever UC is a set" - it follows just from the definition of X. Shouldn't it be "C $in$ X whenever C is a set, i.e. C is a set whenever UC is a set" ?
$endgroup$
– Jakub Andruszkiewicz
Feb 4 at 7:54