Mixing
Measure space, definition of $C_sigma$
$begingroup$
I have read about a definition these days but found it's hard for me to understand. The definition is like this:
Let X be a set. For any set C of subsets of X, we write $C_sigma ={bigcup_{k=1}^{infty}A_k |$ each $A_kin C}$ and $C_delta ={bigcap_{k=1}^{infty}A_k |$ each $A_kin C}$. We have $C_{sigmasigma} =C_sigma $ and $C_{deltadelta}=C_delta$
Suppose X is {1, 2,3}, and C is ${ emptyset, {1}, {1,2}}$, can I know what is $C_sigma$ and $C_delta$ under this circumstance, and why $C_{sigmasigma} =C_sigma $ and $C_{deltadelta}=C_delta$
?
Thanks.
measure-theory
asked Jan 18 at 16:44
JennyJenny
83
$endgroup$
add a comment |
$begingroup$
I have read about a definition these days but found it's hard for me to understand. The definition is like this:
Let X be a set. For any set C of subsets of X, we write $C_sigma ={bigcup_{k=1}^{infty}A_k |$ each $A_kin C}$ and $C_delta ={bigcap_{k=1}^{infty}A_k |$ each $A_kin C}$. We have $C_{sigmasigma} =C_sigma $ and $C_{deltadelta}=C_delta$
Suppose X is {1, 2,3}, and C is ${ emptyset, {1}, {1,2}}$, can I know what is $C_sigma$ and $C_delta$ under this circumstance, and why $C_{sigmasigma} =C_sigma $ and $C_{deltadelta}=C_delta$
?
Thanks.
measure-theory
asked Jan 18 at 16:44
JennyJenny
83
$endgroup$
$begingroup$
You have to in addition specify the starting collection of subsets $C$
$endgroup$
– Calvin Khor
Jan 18 at 16:49
$begingroup$
Briefly put, you take any countable number of elements from $C$, which are sets, and take their union. This will give another set. We can take the union of every such choice of countable number of elements from $C$ , and that gives the set $C_{sigma}$. Since a countable union of countable unions remains countable, we get that $C_{sigma sigma} = C_{sigma}$. Something similar applies for $delta$, but with union replaced by intersection.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 18 at 16:53
$begingroup$
@астонвіллаолофмэллбэрг Thanks for the proof, but can I still get an example of what $C_sigma$ and $C_{sigmasigma}$ really is? I'm quite confused about that.
$endgroup$
– Jenny
Jan 18 at 17:01
$begingroup$
Ok, I will write an answer.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 18 at 17:02
add a comment |
$begingroup$
I have read about a definition these days but found it's hard for me to understand. The definition is like this:
Let X be a set. For any set C of subsets of X, we write $C_sigma ={bigcup_{k=1}^{infty}A_k |$ each $A_kin C}$ and $C_delta ={bigcap_{k=1}^{infty}A_k |$ each $A_kin C}$. We have $C_{sigmasigma} =C_sigma $ and $C_{deltadelta}=C_delta$
Suppose X is {1, 2,3}, and C is ${ emptyset, {1}, {1,2}}$, can I know what is $C_sigma$ and $C_delta$ under this circumstance, and why $C_{sigmasigma} =C_sigma $ and $C_{deltadelta}=C_delta$
?
Thanks.
measure-theory
asked Jan 18 at 16:44
JennyJenny
83
$endgroup$
I have read about a definition these days but found it's hard for me to understand. The definition is like this:
Let X be a set. For any set C of subsets of X, we write $C_sigma ={bigcup_{k=1}^{infty}A_k |$ each $A_kin C}$ and $C_delta ={bigcap_{k=1}^{infty}A_k |$ each $A_kin C}$. We have $C_{sigmasigma} =C_sigma $ and $C_{deltadelta}=C_delta$
Suppose X is {1, 2,3}, and C is ${ emptyset, {1}, {1,2}}$, can I know what is $C_sigma$ and $C_delta$ under this circumstance, and why $C_{sigmasigma} =C_sigma $ and $C_{deltadelta}=C_delta$
?
Thanks.
measure-theory
measure-theory
asked Jan 18 at 16:44
JennyJenny
83
asked Jan 18 at 16:44
JennyJenny
83
edited Jan 18 at 16:51
Jenny
asked Jan 18 at 16:44
JennyJenny
83
asked Jan 18 at 16:44
JennyJenny
83
asked Jan 18 at 16:44
JennyJenny
83
83
$begingroup$
You have to in addition specify the starting collection of subsets $C$
$endgroup$
– Calvin Khor
Jan 18 at 16:49
$begingroup$
Briefly put, you take any countable number of elements from $C$, which are sets, and take their union. This will give another set. We can take the union of every such choice of countable number of elements from $C$ , and that gives the set $C_{sigma}$. Since a countable union of countable unions remains countable, we get that $C_{sigma sigma} = C_{sigma}$. Something similar applies for $delta$, but with union replaced by intersection.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 18 at 16:53
$begingroup$
@астонвіллаолофмэллбэрг Thanks for the proof, but can I still get an example of what $C_sigma$ and $C_{sigmasigma}$ really is? I'm quite confused about that.
$endgroup$
– Jenny
Jan 18 at 17:01
$begingroup$
Ok, I will write an answer.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 18 at 17:02
add a comment |
$begingroup$
You have to in addition specify the starting collection of subsets $C$
$endgroup$
– Calvin Khor
Jan 18 at 16:49
$begingroup$
Briefly put, you take any countable number of elements from $C$, which are sets, and take their union. This will give another set. We can take the union of every such choice of countable number of elements from $C$ , and that gives the set $C_{sigma}$. Since a countable union of countable unions remains countable, we get that $C_{sigma sigma} = C_{sigma}$. Something similar applies for $delta$, but with union replaced by intersection.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 18 at 16:53
$begingroup$
@астонвіллаолофмэллбэрг Thanks for the proof, but can I still get an example of what $C_sigma$ and $C_{sigmasigma}$ really is? I'm quite confused about that.
$endgroup$
– Jenny
Jan 18 at 17:01
$begingroup$
Ok, I will write an answer.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 18 at 17:02
$begingroup$
You have to in addition specify the starting collection of subsets $C$
$endgroup$
– Calvin Khor
Jan 18 at 16:49
$begingroup$
You have to in addition specify the starting collection of subsets $C$
$endgroup$
– Calvin Khor
Jan 18 at 16:49
$begingroup$
Briefly put, you take any countable number of elements from $C$, which are sets, and take their union. This will give another set. We can take the union of every such choice of countable number of elements from $C$ , and that gives the set $C_{sigma}$. Since a countable union of countable unions remains countable, we get that $C_{sigma sigma} = C_{sigma}$. Something similar applies for $delta$, but with union replaced by intersection.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 18 at 16:53
$begingroup$
Briefly put, you take any countable number of elements from $C$, which are sets, and take their union. This will give another set. We can take the union of every such choice of countable number of elements from $C$ , and that gives the set $C_{sigma}$. Since a countable union of countable unions remains countable, we get that $C_{sigma sigma} = C_{sigma}$. Something similar applies for $delta$, but with union replaced by intersection.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 18 at 16:53
$begingroup$
@астонвіллаолофмэллбэрг Thanks for the proof, but can I still get an example of what $C_sigma$ and $C_{sigmasigma}$ really is? I'm quite confused about that.
$endgroup$
– Jenny
Jan 18 at 17:01
$begingroup$
@астонвіллаолофмэллбэрг Thanks for the proof, but can I still get an example of what $C_sigma$ and $C_{sigmasigma}$ really is? I'm quite confused about that.
$endgroup$
– Jenny
Jan 18 at 17:01
$begingroup$
Ok, I will write an answer.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 18 at 17:02
$begingroup$
Ok, I will write an answer.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 18 at 17:02
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Given the $X$ and $C$, the set $C_{sigma}$, in words, consists of all possible countable unions of elements of $C$ (which are sets).
That is, we take countably many elements of $C$ (which includes taking finitely many elements), then we take their union. A new set will come out of this. We lump all that together, and create $C_{sigma}$.
For the sake of illustration I will take a different $C$. You may compute $C_{sigma}$ in your case.
Let us take $C = {{1},{2,3},{2},{1,3}}$.
Take the elements ${1} in C$ and ${1,3} in C$. Their union is ${1,3}$.
Take the elements ${1},{2,3},{2} in C$. Their union is ${1,2,3}$.
Take just the element ${2,3} in C$. Its union (the union involving just one set) is just ${2,3}$.
Take absolutely no element of $C$. This forms the empty union, which is equal to the empty set.
Take elements of $C$ with repetition. For example, take ${1},{2,3},{1},{2},{2,3},{1,3}in C$ , their union is ${1,2,3}$.
The set of all possible unions that we have obtained on the right, are the elements of $C_{sigma}$. So, for example, in our case, $C_{sigma}$ contains the set ${1,3}$ as an element. It contains ${1,2,3}$ as an element. It contains ${2,3}$ as an element. It contains the empty set as an element.
Can you show that ${1} in C_{sigma}$? How about ${1,2}$? To show that something is an element of $C_{sigma}$, you have to exhibit it as a union of sets which are in $C$.
Slightly more involved, but try to show that ${3} notin C_{sigma}$.
Now, $delta$ is the same thing , but with intersection. We need to take a better example to understand this.
Take $C={{1,2,3},{1,2},{3},{2,3}}$.
Now, take the elements ${1,2}, {1,2,3}in C$. Their intersection is ${1,2}$.
Take the elements ${1,2,3},{3},{2,3} in C$. Their intersection is ${3}$.
Take just the element ${2,3}in C$. Its intersection (the intersection involving just one set) is ${2,3}$.
Take absolutely no element of $C$. This is the empty intersection, which is equal to the universal set , in this case $ X= {1,2,3}$.
Take elements of $C$ with repetition : for example, ${1,2},{3},{1,2},{1,2,3},{1,2},{3} in C$, their intersection is the empty set.
The set of all possible intersections that we have obtained on the right, are the elements of $C_{delta}$. For example, in our case $C_{delta}$ contains the set ${1,2}$ as an element. It contains ${3}$ as an element. It contains ${2,3}$ as an element. It contains ${1,2,3}$ as an element. It contains the empty set as an element.
Can you show that ${2}in C_{delta}$?
Try to show that ${1} notin C_{delta}$.
Now, there is the most important thing that we must mention at this point : infinite unions and intersections are permitted. We took only like $2,3,$ maybe at most $6$ elements of $C$ above and took their union/intersection to get an element of $C_{sigma}$. In general, however, we may take infinite intersections as well. However, for this , $C$ will at least have to be infinite to bring some more clarity.
For example, consider the set $X = [0,1]$, and consider the collection of subsets $C = {[0,frac 1n)}$. Then,the set ${0}$ is an infinite intersection $cap_{k=0}^{infty} [0,frac 1k)$ of subsets of $C$, but not a finite intersection. That is, ${0} in C_{delta}$ as an infinite intersection, not a finite one.
The fact that $C_{sigmasigma} = C_{sigma}$ and $C_{deltadelta} =C_{delta}$ follows like this : $C_{sigmasigma}$ consists of the countable union of elements of $C_{sigma}$ which consists of the countable union of elements of $C$. Combining thee unions, we retain countability from a well known fact that a countable union of countable unions remains countable, and therefore $C_{sigma sigma}$ is a subset of $C_{sigma}$. The reverse inclusion is near obvious. Similarly for $delta$.
If you have NOT understood the infinite case yet, then do not worry : just remember what finite unions/intersections look like, and this will give you a good idea of what $C_{sigma / delta}$ looks like.
answered Jan 18 at 17:43
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.9k33477
$endgroup$
$begingroup$
I don't have the reputation to upvote for this answer, but this is definitely a more than 5 star one!
$endgroup$
– Jenny
Jan 18 at 19:58
$begingroup$
Thank you for the compliment!
$endgroup$
– астон вілла олоф мэллбэрг
Jan 19 at 3:32
add a comment |
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1 Answer
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$begingroup$
Given the $X$ and $C$, the set $C_{sigma}$, in words, consists of all possible countable unions of elements of $C$ (which are sets).
That is, we take countably many elements of $C$ (which includes taking finitely many elements), then we take their union. A new set will come out of this. We lump all that together, and create $C_{sigma}$.
For the sake of illustration I will take a different $C$. You may compute $C_{sigma}$ in your case.
Let us take $C = {{1},{2,3},{2},{1,3}}$.
Take the elements ${1} in C$ and ${1,3} in C$. Their union is ${1,3}$.
Take the elements ${1},{2,3},{2} in C$. Their union is ${1,2,3}$.
Take just the element ${2,3} in C$. Its union (the union involving just one set) is just ${2,3}$.
Take absolutely no element of $C$. This forms the empty union, which is equal to the empty set.
Take elements of $C$ with repetition. For example, take ${1},{2,3},{1},{2},{2,3},{1,3}in C$ , their union is ${1,2,3}$.
The set of all possible unions that we have obtained on the right, are the elements of $C_{sigma}$. So, for example, in our case, $C_{sigma}$ contains the set ${1,3}$ as an element. It contains ${1,2,3}$ as an element. It contains ${2,3}$ as an element. It contains the empty set as an element.
Can you show that ${1} in C_{sigma}$? How about ${1,2}$? To show that something is an element of $C_{sigma}$, you have to exhibit it as a union of sets which are in $C$.
Slightly more involved, but try to show that ${3} notin C_{sigma}$.
Now, $delta$ is the same thing , but with intersection. We need to take a better example to understand this.
Take $C={{1,2,3},{1,2},{3},{2,3}}$.
Now, take the elements ${1,2}, {1,2,3}in C$. Their intersection is ${1,2}$.
Take the elements ${1,2,3},{3},{2,3} in C$. Their intersection is ${3}$.
Take just the element ${2,3}in C$. Its intersection (the intersection involving just one set) is ${2,3}$.
Take absolutely no element of $C$. This is the empty intersection, which is equal to the universal set , in this case $ X= {1,2,3}$.
Take elements of $C$ with repetition : for example, ${1,2},{3},{1,2},{1,2,3},{1,2},{3} in C$, their intersection is the empty set.
The set of all possible intersections that we have obtained on the right, are the elements of $C_{delta}$. For example, in our case $C_{delta}$ contains the set ${1,2}$ as an element. It contains ${3}$ as an element. It contains ${2,3}$ as an element. It contains ${1,2,3}$ as an element. It contains the empty set as an element.
Can you show that ${2}in C_{delta}$?
Try to show that ${1} notin C_{delta}$.
Now, there is the most important thing that we must mention at this point : infinite unions and intersections are permitted. We took only like $2,3,$ maybe at most $6$ elements of $C$ above and took their union/intersection to get an element of $C_{sigma}$. In general, however, we may take infinite intersections as well. However, for this , $C$ will at least have to be infinite to bring some more clarity.
For example, consider the set $X = [0,1]$, and consider the collection of subsets $C = {[0,frac 1n)}$. Then,the set ${0}$ is an infinite intersection $cap_{k=0}^{infty} [0,frac 1k)$ of subsets of $C$, but not a finite intersection. That is, ${0} in C_{delta}$ as an infinite intersection, not a finite one.
The fact that $C_{sigmasigma} = C_{sigma}$ and $C_{deltadelta} =C_{delta}$ follows like this : $C_{sigmasigma}$ consists of the countable union of elements of $C_{sigma}$ which consists of the countable union of elements of $C$. Combining thee unions, we retain countability from a well known fact that a countable union of countable unions remains countable, and therefore $C_{sigma sigma}$ is a subset of $C_{sigma}$. The reverse inclusion is near obvious. Similarly for $delta$.
If you have NOT understood the infinite case yet, then do not worry : just remember what finite unions/intersections look like, and this will give you a good idea of what $C_{sigma / delta}$ looks like.
answered Jan 18 at 17:43
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.9k33477
$endgroup$
$begingroup$
I don't have the reputation to upvote for this answer, but this is definitely a more than 5 star one!
$endgroup$
– Jenny
Jan 18 at 19:58
$begingroup$
Thank you for the compliment!
$endgroup$
– астон вілла олоф мэллбэрг
Jan 19 at 3:32
add a comment |
$begingroup$
Given the $X$ and $C$, the set $C_{sigma}$, in words, consists of all possible countable unions of elements of $C$ (which are sets).
That is, we take countably many elements of $C$ (which includes taking finitely many elements), then we take their union. A new set will come out of this. We lump all that together, and create $C_{sigma}$.
For the sake of illustration I will take a different $C$. You may compute $C_{sigma}$ in your case.
Let us take $C = {{1},{2,3},{2},{1,3}}$.
Take the elements ${1} in C$ and ${1,3} in C$. Their union is ${1,3}$.
Take the elements ${1},{2,3},{2} in C$. Their union is ${1,2,3}$.
Take just the element ${2,3} in C$. Its union (the union involving just one set) is just ${2,3}$.
Take absolutely no element of $C$. This forms the empty union, which is equal to the empty set.
Take elements of $C$ with repetition. For example, take ${1},{2,3},{1},{2},{2,3},{1,3}in C$ , their union is ${1,2,3}$.
The set of all possible unions that we have obtained on the right, are the elements of $C_{sigma}$. So, for example, in our case, $C_{sigma}$ contains the set ${1,3}$ as an element. It contains ${1,2,3}$ as an element. It contains ${2,3}$ as an element. It contains the empty set as an element.
Can you show that ${1} in C_{sigma}$? How about ${1,2}$? To show that something is an element of $C_{sigma}$, you have to exhibit it as a union of sets which are in $C$.
Slightly more involved, but try to show that ${3} notin C_{sigma}$.
Now, $delta$ is the same thing , but with intersection. We need to take a better example to understand this.
Take $C={{1,2,3},{1,2},{3},{2,3}}$.
Now, take the elements ${1,2}, {1,2,3}in C$. Their intersection is ${1,2}$.
Take the elements ${1,2,3},{3},{2,3} in C$. Their intersection is ${3}$.
Take just the element ${2,3}in C$. Its intersection (the intersection involving just one set) is ${2,3}$.
Take absolutely no element of $C$. This is the empty intersection, which is equal to the universal set , in this case $ X= {1,2,3}$.
Take elements of $C$ with repetition : for example, ${1,2},{3},{1,2},{1,2,3},{1,2},{3} in C$, their intersection is the empty set.
The set of all possible intersections that we have obtained on the right, are the elements of $C_{delta}$. For example, in our case $C_{delta}$ contains the set ${1,2}$ as an element. It contains ${3}$ as an element. It contains ${2,3}$ as an element. It contains ${1,2,3}$ as an element. It contains the empty set as an element.
Can you show that ${2}in C_{delta}$?
Try to show that ${1} notin C_{delta}$.
Now, there is the most important thing that we must mention at this point : infinite unions and intersections are permitted. We took only like $2,3,$ maybe at most $6$ elements of $C$ above and took their union/intersection to get an element of $C_{sigma}$. In general, however, we may take infinite intersections as well. However, for this , $C$ will at least have to be infinite to bring some more clarity.
For example, consider the set $X = [0,1]$, and consider the collection of subsets $C = {[0,frac 1n)}$. Then,the set ${0}$ is an infinite intersection $cap_{k=0}^{infty} [0,frac 1k)$ of subsets of $C$, but not a finite intersection. That is, ${0} in C_{delta}$ as an infinite intersection, not a finite one.
The fact that $C_{sigmasigma} = C_{sigma}$ and $C_{deltadelta} =C_{delta}$ follows like this : $C_{sigmasigma}$ consists of the countable union of elements of $C_{sigma}$ which consists of the countable union of elements of $C$. Combining thee unions, we retain countability from a well known fact that a countable union of countable unions remains countable, and therefore $C_{sigma sigma}$ is a subset of $C_{sigma}$. The reverse inclusion is near obvious. Similarly for $delta$.
If you have NOT understood the infinite case yet, then do not worry : just remember what finite unions/intersections look like, and this will give you a good idea of what $C_{sigma / delta}$ looks like.
answered Jan 18 at 17:43
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.9k33477
$endgroup$
$begingroup$
I don't have the reputation to upvote for this answer, but this is definitely a more than 5 star one!
$endgroup$
– Jenny
Jan 18 at 19:58
$begingroup$
Thank you for the compliment!
$endgroup$
– астон вілла олоф мэллбэрг
Jan 19 at 3:32
add a comment |
$begingroup$
Given the $X$ and $C$, the set $C_{sigma}$, in words, consists of all possible countable unions of elements of $C$ (which are sets).
That is, we take countably many elements of $C$ (which includes taking finitely many elements), then we take their union. A new set will come out of this. We lump all that together, and create $C_{sigma}$.
For the sake of illustration I will take a different $C$. You may compute $C_{sigma}$ in your case.
Let us take $C = {{1},{2,3},{2},{1,3}}$.
Take the elements ${1} in C$ and ${1,3} in C$. Their union is ${1,3}$.
Take the elements ${1},{2,3},{2} in C$. Their union is ${1,2,3}$.
Take just the element ${2,3} in C$. Its union (the union involving just one set) is just ${2,3}$.
Take absolutely no element of $C$. This forms the empty union, which is equal to the empty set.
Take elements of $C$ with repetition. For example, take ${1},{2,3},{1},{2},{2,3},{1,3}in C$ , their union is ${1,2,3}$.
The set of all possible unions that we have obtained on the right, are the elements of $C_{sigma}$. So, for example, in our case, $C_{sigma}$ contains the set ${1,3}$ as an element. It contains ${1,2,3}$ as an element. It contains ${2,3}$ as an element. It contains the empty set as an element.
Can you show that ${1} in C_{sigma}$? How about ${1,2}$? To show that something is an element of $C_{sigma}$, you have to exhibit it as a union of sets which are in $C$.
Slightly more involved, but try to show that ${3} notin C_{sigma}$.
Now, $delta$ is the same thing , but with intersection. We need to take a better example to understand this.
Take $C={{1,2,3},{1,2},{3},{2,3}}$.
Now, take the elements ${1,2}, {1,2,3}in C$. Their intersection is ${1,2}$.
Take the elements ${1,2,3},{3},{2,3} in C$. Their intersection is ${3}$.
Take just the element ${2,3}in C$. Its intersection (the intersection involving just one set) is ${2,3}$.
Take absolutely no element of $C$. This is the empty intersection, which is equal to the universal set , in this case $ X= {1,2,3}$.
Take elements of $C$ with repetition : for example, ${1,2},{3},{1,2},{1,2,3},{1,2},{3} in C$, their intersection is the empty set.
The set of all possible intersections that we have obtained on the right, are the elements of $C_{delta}$. For example, in our case $C_{delta}$ contains the set ${1,2}$ as an element. It contains ${3}$ as an element. It contains ${2,3}$ as an element. It contains ${1,2,3}$ as an element. It contains the empty set as an element.
Can you show that ${2}in C_{delta}$?
Try to show that ${1} notin C_{delta}$.
Now, there is the most important thing that we must mention at this point : infinite unions and intersections are permitted. We took only like $2,3,$ maybe at most $6$ elements of $C$ above and took their union/intersection to get an element of $C_{sigma}$. In general, however, we may take infinite intersections as well. However, for this , $C$ will at least have to be infinite to bring some more clarity.
For example, consider the set $X = [0,1]$, and consider the collection of subsets $C = {[0,frac 1n)}$. Then,the set ${0}$ is an infinite intersection $cap_{k=0}^{infty} [0,frac 1k)$ of subsets of $C$, but not a finite intersection. That is, ${0} in C_{delta}$ as an infinite intersection, not a finite one.
The fact that $C_{sigmasigma} = C_{sigma}$ and $C_{deltadelta} =C_{delta}$ follows like this : $C_{sigmasigma}$ consists of the countable union of elements of $C_{sigma}$ which consists of the countable union of elements of $C$. Combining thee unions, we retain countability from a well known fact that a countable union of countable unions remains countable, and therefore $C_{sigma sigma}$ is a subset of $C_{sigma}$. The reverse inclusion is near obvious. Similarly for $delta$.
If you have NOT understood the infinite case yet, then do not worry : just remember what finite unions/intersections look like, and this will give you a good idea of what $C_{sigma / delta}$ looks like.
answered Jan 18 at 17:43
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.9k33477
$endgroup$
Given the $X$ and $C$, the set $C_{sigma}$, in words, consists of all possible countable unions of elements of $C$ (which are sets).
That is, we take countably many elements of $C$ (which includes taking finitely many elements), then we take their union. A new set will come out of this. We lump all that together, and create $C_{sigma}$.
For the sake of illustration I will take a different $C$. You may compute $C_{sigma}$ in your case.
Let us take $C = {{1},{2,3},{2},{1,3}}$.
Take the elements ${1} in C$ and ${1,3} in C$. Their union is ${1,3}$.
Take the elements ${1},{2,3},{2} in C$. Their union is ${1,2,3}$.
Take just the element ${2,3} in C$. Its union (the union involving just one set) is just ${2,3}$.
Take absolutely no element of $C$. This forms the empty union, which is equal to the empty set.
Take elements of $C$ with repetition. For example, take ${1},{2,3},{1},{2},{2,3},{1,3}in C$ , their union is ${1,2,3}$.
The set of all possible unions that we have obtained on the right, are the elements of $C_{sigma}$. So, for example, in our case, $C_{sigma}$ contains the set ${1,3}$ as an element. It contains ${1,2,3}$ as an element. It contains ${2,3}$ as an element. It contains the empty set as an element.
Can you show that ${1} in C_{sigma}$? How about ${1,2}$? To show that something is an element of $C_{sigma}$, you have to exhibit it as a union of sets which are in $C$.
Slightly more involved, but try to show that ${3} notin C_{sigma}$.
Now, $delta$ is the same thing , but with intersection. We need to take a better example to understand this.
Take $C={{1,2,3},{1,2},{3},{2,3}}$.
Now, take the elements ${1,2}, {1,2,3}in C$. Their intersection is ${1,2}$.
Take the elements ${1,2,3},{3},{2,3} in C$. Their intersection is ${3}$.
Take just the element ${2,3}in C$. Its intersection (the intersection involving just one set) is ${2,3}$.
Take absolutely no element of $C$. This is the empty intersection, which is equal to the universal set , in this case $ X= {1,2,3}$.
Take elements of $C$ with repetition : for example, ${1,2},{3},{1,2},{1,2,3},{1,2},{3} in C$, their intersection is the empty set.
The set of all possible intersections that we have obtained on the right, are the elements of $C_{delta}$. For example, in our case $C_{delta}$ contains the set ${1,2}$ as an element. It contains ${3}$ as an element. It contains ${2,3}$ as an element. It contains ${1,2,3}$ as an element. It contains the empty set as an element.
Can you show that ${2}in C_{delta}$?
Try to show that ${1} notin C_{delta}$.
Now, there is the most important thing that we must mention at this point : infinite unions and intersections are permitted. We took only like $2,3,$ maybe at most $6$ elements of $C$ above and took their union/intersection to get an element of $C_{sigma}$. In general, however, we may take infinite intersections as well. However, for this , $C$ will at least have to be infinite to bring some more clarity.
For example, consider the set $X = [0,1]$, and consider the collection of subsets $C = {[0,frac 1n)}$. Then,the set ${0}$ is an infinite intersection $cap_{k=0}^{infty} [0,frac 1k)$ of subsets of $C$, but not a finite intersection. That is, ${0} in C_{delta}$ as an infinite intersection, not a finite one.
The fact that $C_{sigmasigma} = C_{sigma}$ and $C_{deltadelta} =C_{delta}$ follows like this : $C_{sigmasigma}$ consists of the countable union of elements of $C_{sigma}$ which consists of the countable union of elements of $C$. Combining thee unions, we retain countability from a well known fact that a countable union of countable unions remains countable, and therefore $C_{sigma sigma}$ is a subset of $C_{sigma}$. The reverse inclusion is near obvious. Similarly for $delta$.
If you have NOT understood the infinite case yet, then do not worry : just remember what finite unions/intersections look like, and this will give you a good idea of what $C_{sigma / delta}$ looks like.
answered Jan 18 at 17:43
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.9k33477
edited Jan 19 at 3:32
answered Jan 18 at 17:43
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.9k33477
answered Jan 18 at 17:43
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.9k33477
answered Jan 18 at 17:43
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.9k33477
38.9k33477
$begingroup$
I don't have the reputation to upvote for this answer, but this is definitely a more than 5 star one!
$endgroup$
– Jenny
Jan 18 at 19:58
$begingroup$
Thank you for the compliment!
$endgroup$
– астон вілла олоф мэллбэрг
Jan 19 at 3:32
add a comment |
$begingroup$
I don't have the reputation to upvote for this answer, but this is definitely a more than 5 star one!
$endgroup$
– Jenny
Jan 18 at 19:58
$begingroup$
Thank you for the compliment!
$endgroup$
– астон вілла олоф мэллбэрг
Jan 19 at 3:32
$begingroup$
I don't have the reputation to upvote for this answer, but this is definitely a more than 5 star one!
$endgroup$
– Jenny
Jan 18 at 19:58
$begingroup$
I don't have the reputation to upvote for this answer, but this is definitely a more than 5 star one!
$endgroup$
– Jenny
Jan 18 at 19:58
$begingroup$
Thank you for the compliment!
$endgroup$
– астон вілла олоф мэллбэрг
Jan 19 at 3:32
$begingroup$
Thank you for the compliment!
$endgroup$
– астон вілла олоф мэллбэрг
Jan 19 at 3:32
add a comment |
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$begingroup$
You have to in addition specify the starting collection of subsets $C$
$endgroup$
– Calvin Khor
Jan 18 at 16:49
$begingroup$
Briefly put, you take any countable number of elements from $C$, which are sets, and take their union. This will give another set. We can take the union of every such choice of countable number of elements from $C$ , and that gives the set $C_{sigma}$. Since a countable union of countable unions remains countable, we get that $C_{sigma sigma} = C_{sigma}$. Something similar applies for $delta$, but with union replaced by intersection.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 18 at 16:53
$begingroup$
@астонвіллаолофмэллбэрг Thanks for the proof, but can I still get an example of what $C_sigma$ and $C_{sigmasigma}$ really is? I'm quite confused about that.
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– Jenny
Jan 18 at 17:01
$begingroup$
Ok, I will write an answer.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 18 at 17:02