When is a collection of sets closed under union?
$begingroup$
I started studying Probability, and I'm not sure if I understand what is the meaning of "closed under union".
A collection (say $F$) of subsets of a set (say $Omega$) is said to be a $sigma$-algebra if:
- $Omega in F$
$F$ is closed under complement
$F$ is closed under union
Now, consider the following example:
Given $Omega = {1, 2, 3, 4, 5}$, is $F = {emptyset, {1, 2}, {3, 4, 5}, {5}, {1, 2, 3, 4}, {1, 2, 3, 4, 5}}$ closed under union?
I'm asking because I know that $bigcup_{i = 1}^{6} X_i in F$; however it does not happen for any two elements, for example: ${1,2} cup {5} notin F$.
probability elementary-set-theory
edited Jan 6 at 9:27
Asaf Karagila♦
304k32430763
asked Jan 6 at 8:30
AndréAndré
63
$endgroup$
add a comment |
$begingroup$
I started studying Probability, and I'm not sure if I understand what is the meaning of "closed under union".
A collection (say $F$) of subsets of a set (say $Omega$) is said to be a $sigma$-algebra if:
- $Omega in F$
$F$ is closed under complement
$F$ is closed under union
Now, consider the following example:
Given $Omega = {1, 2, 3, 4, 5}$, is $F = {emptyset, {1, 2}, {3, 4, 5}, {5}, {1, 2, 3, 4}, {1, 2, 3, 4, 5}}$ closed under union?
I'm asking because I know that $bigcup_{i = 1}^{6} X_i in F$; however it does not happen for any two elements, for example: ${1,2} cup {5} notin F$.
probability elementary-set-theory
edited Jan 6 at 9:27
Asaf Karagila♦
304k32430763
asked Jan 6 at 8:30
AndréAndré
63
$endgroup$
2
$begingroup$
$sigma$-algebras are only required to be closed under countable unions.
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– Lord Shark the Unknown
Jan 6 at 8:42
$begingroup$
slight change to title; a collection of sets can be closed under union, not the set itself;
$endgroup$
– Chris
Jan 6 at 9:09
$begingroup$
You're right! Thanks, I changed it.
$endgroup$
– André
Jan 6 at 9:10
add a comment |
$begingroup$
I started studying Probability, and I'm not sure if I understand what is the meaning of "closed under union".
A collection (say $F$) of subsets of a set (say $Omega$) is said to be a $sigma$-algebra if:
- $Omega in F$
$F$ is closed under complement
$F$ is closed under union
Now, consider the following example:
Given $Omega = {1, 2, 3, 4, 5}$, is $F = {emptyset, {1, 2}, {3, 4, 5}, {5}, {1, 2, 3, 4}, {1, 2, 3, 4, 5}}$ closed under union?
I'm asking because I know that $bigcup_{i = 1}^{6} X_i in F$; however it does not happen for any two elements, for example: ${1,2} cup {5} notin F$.
probability elementary-set-theory
edited Jan 6 at 9:27
Asaf Karagila♦
304k32430763
asked Jan 6 at 8:30
AndréAndré
63
$endgroup$
I started studying Probability, and I'm not sure if I understand what is the meaning of "closed under union".
A collection (say $F$) of subsets of a set (say $Omega$) is said to be a $sigma$-algebra if:
- $Omega in F$
$F$ is closed under complement
$F$ is closed under union
Now, consider the following example:
Given $Omega = {1, 2, 3, 4, 5}$, is $F = {emptyset, {1, 2}, {3, 4, 5}, {5}, {1, 2, 3, 4}, {1, 2, 3, 4, 5}}$ closed under union?
I'm asking because I know that $bigcup_{i = 1}^{6} X_i in F$; however it does not happen for any two elements, for example: ${1,2} cup {5} notin F$.
probability elementary-set-theory
probability elementary-set-theory
edited Jan 6 at 9:27
Asaf Karagila♦
304k32430763
asked Jan 6 at 8:30
AndréAndré
63
edited Jan 6 at 9:27
Asaf Karagila♦
304k32430763
asked Jan 6 at 8:30
AndréAndré
63
edited Jan 6 at 9:27
Asaf Karagila♦
304k32430763
edited Jan 6 at 9:27
Asaf Karagila♦
304k32430763
edited Jan 6 at 9:27
Asaf Karagila♦
304k32430763
304k32430763
asked Jan 6 at 8:30
AndréAndré
63
asked Jan 6 at 8:30
AndréAndré
63
asked Jan 6 at 8:30
AndréAndré
63
63
2
$begingroup$
$sigma$-algebras are only required to be closed under countable unions.
$endgroup$
– Lord Shark the Unknown
Jan 6 at 8:42
$begingroup$
slight change to title; a collection of sets can be closed under union, not the set itself;
$endgroup$
– Chris
Jan 6 at 9:09
$begingroup$
You're right! Thanks, I changed it.
$endgroup$
– André
Jan 6 at 9:10
add a comment |
2
$begingroup$
$sigma$-algebras are only required to be closed under countable unions.
$endgroup$
– Lord Shark the Unknown
Jan 6 at 8:42
$begingroup$
slight change to title; a collection of sets can be closed under union, not the set itself;
$endgroup$
– Chris
Jan 6 at 9:09
$begingroup$
You're right! Thanks, I changed it.
$endgroup$
– André
Jan 6 at 9:10
2
2
$begingroup$
$sigma$-algebras are only required to be closed under countable unions.
$endgroup$
– Lord Shark the Unknown
Jan 6 at 8:42
$begingroup$
$sigma$-algebras are only required to be closed under countable unions.
$endgroup$
– Lord Shark the Unknown
Jan 6 at 8:42
$begingroup$
slight change to title; a collection of sets can be closed under union, not the set itself;
$endgroup$
– Chris
Jan 6 at 9:09
$begingroup$
slight change to title; a collection of sets can be closed under union, not the set itself;
$endgroup$
– Chris
Jan 6 at 9:09
$begingroup$
You're right! Thanks, I changed it.
$endgroup$
– André
Jan 6 at 9:10
$begingroup$
You're right! Thanks, I changed it.
$endgroup$
– André
Jan 6 at 9:10
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
"$F$ is closed under union" means that for all $A,B in F$, $A cup B in F$.
So here ${1,2}$ and ${5}$ are elements of $F$, but their union is not, so $F$ is not closed by union.
edited Jan 6 at 9:20
Henno Brandsma
109k347114
answered Jan 6 at 8:37
MindlackMindlack
3,86018
$endgroup$
add a comment |
$begingroup$
“Closed under union” means that the union of any set of members of $F$ is also a member of $F$.
In your example $F$ is not closed under union (although it is closed under complement).
answered Jan 6 at 8:39
gandalf61gandalf61
8,689725
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add a comment |
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2 Answers
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$begingroup$
"$F$ is closed under union" means that for all $A,B in F$, $A cup B in F$.
So here ${1,2}$ and ${5}$ are elements of $F$, but their union is not, so $F$ is not closed by union.
edited Jan 6 at 9:20
Henno Brandsma
109k347114
answered Jan 6 at 8:37
MindlackMindlack
3,86018
$endgroup$
add a comment |
$begingroup$
"$F$ is closed under union" means that for all $A,B in F$, $A cup B in F$.
So here ${1,2}$ and ${5}$ are elements of $F$, but their union is not, so $F$ is not closed by union.
edited Jan 6 at 9:20
Henno Brandsma
109k347114
answered Jan 6 at 8:37
MindlackMindlack
3,86018
$endgroup$
add a comment |
$begingroup$
"$F$ is closed under union" means that for all $A,B in F$, $A cup B in F$.
So here ${1,2}$ and ${5}$ are elements of $F$, but their union is not, so $F$ is not closed by union.
edited Jan 6 at 9:20
Henno Brandsma
109k347114
answered Jan 6 at 8:37
MindlackMindlack
3,86018
$endgroup$
"$F$ is closed under union" means that for all $A,B in F$, $A cup B in F$.
So here ${1,2}$ and ${5}$ are elements of $F$, but their union is not, so $F$ is not closed by union.
edited Jan 6 at 9:20
Henno Brandsma
109k347114
answered Jan 6 at 8:37
MindlackMindlack
3,86018
edited Jan 6 at 9:20
Henno Brandsma
109k347114
edited Jan 6 at 9:20
Henno Brandsma
109k347114
edited Jan 6 at 9:20
Henno Brandsma
109k347114
109k347114
answered Jan 6 at 8:37
MindlackMindlack
3,86018
answered Jan 6 at 8:37
MindlackMindlack
3,86018
answered Jan 6 at 8:37
MindlackMindlack
3,86018
3,86018
add a comment |
add a comment |
$begingroup$
“Closed under union” means that the union of any set of members of $F$ is also a member of $F$.
In your example $F$ is not closed under union (although it is closed under complement).
answered Jan 6 at 8:39
gandalf61gandalf61
8,689725
$endgroup$
add a comment |
$begingroup$
“Closed under union” means that the union of any set of members of $F$ is also a member of $F$.
In your example $F$ is not closed under union (although it is closed under complement).
answered Jan 6 at 8:39
gandalf61gandalf61
8,689725
$endgroup$
add a comment |
$begingroup$
“Closed under union” means that the union of any set of members of $F$ is also a member of $F$.
In your example $F$ is not closed under union (although it is closed under complement).
answered Jan 6 at 8:39
gandalf61gandalf61
8,689725
$endgroup$
“Closed under union” means that the union of any set of members of $F$ is also a member of $F$.
In your example $F$ is not closed under union (although it is closed under complement).
answered Jan 6 at 8:39
gandalf61gandalf61
8,689725
answered Jan 6 at 8:39
gandalf61gandalf61
8,689725
answered Jan 6 at 8:39
gandalf61gandalf61
8,689725
answered Jan 6 at 8:39
gandalf61gandalf61
8,689725
8,689725
add a comment |
add a comment |
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2
$begingroup$
$sigma$-algebras are only required to be closed under countable unions.
$endgroup$
– Lord Shark the Unknown
Jan 6 at 8:42
$begingroup$
slight change to title; a collection of sets can be closed under union, not the set itself;
$endgroup$
– Chris
Jan 6 at 9:09
$begingroup$
You're right! Thanks, I changed it.
$endgroup$
– André
Jan 6 at 9:10