Mixing
Show that the class of all finite unions of closed-open intervals on the real line is a ring of sets but is...
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From what I understood, a ring of sets is a non-empty class of sets that is closed under symmetric difference of any pair of sets of the class and under intersection of any pair of sets of the class.
Why isn't it a set of Boolean algebra and how do you show it to be a ring of sets?
I have just begun Introduction to Topology and Modern Analysis for context.
general-topology
asked Apr 20 '17 at 16:55
Mrigank Arora
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up vote
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From what I understood, a ring of sets is a non-empty class of sets that is closed under symmetric difference of any pair of sets of the class and under intersection of any pair of sets of the class.
Why isn't it a set of Boolean algebra and how do you show it to be a ring of sets?
I have just begun Introduction to Topology and Modern Analysis for context.
general-topology
asked Apr 20 '17 at 16:55
Mrigank Arora
654
There is a 'general-topology' tag. By the way, what is a closed-open interval of $mathbb{R}$?
– amrsa
Apr 20 '17 at 17:14
Like $[0,1)$. Where the first limit is included and the second not. The converse would be an open-closed interval like $(0,1]$.
– Mrigank Arora
Apr 20 '17 at 17:21
I think your definition of ring of sets is different than mine one (and I suppose, many other people's), so may be you should give that definition.
– amrsa
Apr 20 '17 at 17:24
Or maybe when you say "From what i understood", you mean that is your definition
– amrsa
Apr 20 '17 at 17:26
This is just how the book introduced it. If possible, could you tell me your definition for a ring of sets? I just started self-studying in this area, so I don't really know any conventions.
– Mrigank Arora
Apr 20 '17 at 17:30
|
show 3 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
From what I understood, a ring of sets is a non-empty class of sets that is closed under symmetric difference of any pair of sets of the class and under intersection of any pair of sets of the class.
Why isn't it a set of Boolean algebra and how do you show it to be a ring of sets?
I have just begun Introduction to Topology and Modern Analysis for context.
general-topology
asked Apr 20 '17 at 16:55
Mrigank Arora
654
From what I understood, a ring of sets is a non-empty class of sets that is closed under symmetric difference of any pair of sets of the class and under intersection of any pair of sets of the class.
Why isn't it a set of Boolean algebra and how do you show it to be a ring of sets?
I have just begun Introduction to Topology and Modern Analysis for context.
general-topology
general-topology
asked Apr 20 '17 at 16:55
Mrigank Arora
654
asked Apr 20 '17 at 16:55
Mrigank Arora
654
edited Apr 20 '17 at 17:21
asked Apr 20 '17 at 16:55
Mrigank Arora
654
asked Apr 20 '17 at 16:55
Mrigank Arora
654
asked Apr 20 '17 at 16:55
Mrigank Arora
654
654
There is a 'general-topology' tag. By the way, what is a closed-open interval of $mathbb{R}$?
– amrsa
Apr 20 '17 at 17:14
Like $[0,1)$. Where the first limit is included and the second not. The converse would be an open-closed interval like $(0,1]$.
– Mrigank Arora
Apr 20 '17 at 17:21
I think your definition of ring of sets is different than mine one (and I suppose, many other people's), so may be you should give that definition.
– amrsa
Apr 20 '17 at 17:24
Or maybe when you say "From what i understood", you mean that is your definition
– amrsa
Apr 20 '17 at 17:26
This is just how the book introduced it. If possible, could you tell me your definition for a ring of sets? I just started self-studying in this area, so I don't really know any conventions.
– Mrigank Arora
Apr 20 '17 at 17:30
|
show 3 more comments
There is a 'general-topology' tag. By the way, what is a closed-open interval of $mathbb{R}$?
– amrsa
Apr 20 '17 at 17:14
Like $[0,1)$. Where the first limit is included and the second not. The converse would be an open-closed interval like $(0,1]$.
– Mrigank Arora
Apr 20 '17 at 17:21
I think your definition of ring of sets is different than mine one (and I suppose, many other people's), so may be you should give that definition.
– amrsa
Apr 20 '17 at 17:24
Or maybe when you say "From what i understood", you mean that is your definition
– amrsa
Apr 20 '17 at 17:26
This is just how the book introduced it. If possible, could you tell me your definition for a ring of sets? I just started self-studying in this area, so I don't really know any conventions.
– Mrigank Arora
Apr 20 '17 at 17:30
There is a 'general-topology' tag. By the way, what is a closed-open interval of $mathbb{R}$?
– amrsa
Apr 20 '17 at 17:14
There is a 'general-topology' tag. By the way, what is a closed-open interval of $mathbb{R}$?
– amrsa
Apr 20 '17 at 17:14
Like $[0,1)$. Where the first limit is included and the second not. The converse would be an open-closed interval like $(0,1]$.
– Mrigank Arora
Apr 20 '17 at 17:21
Like $[0,1)$. Where the first limit is included and the second not. The converse would be an open-closed interval like $(0,1]$.
– Mrigank Arora
Apr 20 '17 at 17:21
I think your definition of ring of sets is different than mine one (and I suppose, many other people's), so may be you should give that definition.
– amrsa
Apr 20 '17 at 17:24
I think your definition of ring of sets is different than mine one (and I suppose, many other people's), so may be you should give that definition.
– amrsa
Apr 20 '17 at 17:24
Or maybe when you say "From what i understood", you mean that is your definition
– amrsa
Apr 20 '17 at 17:26
Or maybe when you say "From what i understood", you mean that is your definition
– amrsa
Apr 20 '17 at 17:26
This is just how the book introduced it. If possible, could you tell me your definition for a ring of sets? I just started self-studying in this area, so I don't really know any conventions.
– Mrigank Arora
Apr 20 '17 at 17:30
This is just how the book introduced it. If possible, could you tell me your definition for a ring of sets? I just started self-studying in this area, so I don't really know any conventions.
– Mrigank Arora
Apr 20 '17 at 17:30
|
show 3 more comments
1 Answer
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A ring of sets is defined in Simmons to be a non-empty class $mathcal{A}$ such that for $A,B in mathcal{A}$:
(1) $A Delta B in mathcal{A}$, where $Delta$ stands for the symmetric difference.
(2) $A cap B in mathcal{A}$.
Note that this implies $emptyset in mathcal{A}$. For this proof, you need to show the following. If $B_1, B_2 in mathcal{A}$, where $mathcal{A}$ is the class of all finite unions of closed-open intervals on the real line, then $B_1 Delta B_2 in mathcal{A}$ and $B_1 cap B_2 in mathcal{A}$.
To show that $mathcal{A}$ is not a Boolean algebra of sets, consider whether $B_1'$ is in $mathcal{A}$, where $B_1'$ stands for the complement of $B_1$. To help you get on your way, maybe define $B_1 = [a,b)$ and $B_2 = [c,d)$ where $a,b,c,d in mathbb{R}$. Then consider whether these two closed-open intervals are overlapping, containing one or the other, or disjoint.
answered 2 days ago
tucsonman101
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
A ring of sets is defined in Simmons to be a non-empty class $mathcal{A}$ such that for $A,B in mathcal{A}$:
(1) $A Delta B in mathcal{A}$, where $Delta$ stands for the symmetric difference.
(2) $A cap B in mathcal{A}$.
Note that this implies $emptyset in mathcal{A}$. For this proof, you need to show the following. If $B_1, B_2 in mathcal{A}$, where $mathcal{A}$ is the class of all finite unions of closed-open intervals on the real line, then $B_1 Delta B_2 in mathcal{A}$ and $B_1 cap B_2 in mathcal{A}$.
To show that $mathcal{A}$ is not a Boolean algebra of sets, consider whether $B_1'$ is in $mathcal{A}$, where $B_1'$ stands for the complement of $B_1$. To help you get on your way, maybe define $B_1 = [a,b)$ and $B_2 = [c,d)$ where $a,b,c,d in mathbb{R}$. Then consider whether these two closed-open intervals are overlapping, containing one or the other, or disjoint.
answered 2 days ago
tucsonman101
11
New contributor
tucsonman101 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
0
down vote
A ring of sets is defined in Simmons to be a non-empty class $mathcal{A}$ such that for $A,B in mathcal{A}$:
(1) $A Delta B in mathcal{A}$, where $Delta$ stands for the symmetric difference.
(2) $A cap B in mathcal{A}$.
Note that this implies $emptyset in mathcal{A}$. For this proof, you need to show the following. If $B_1, B_2 in mathcal{A}$, where $mathcal{A}$ is the class of all finite unions of closed-open intervals on the real line, then $B_1 Delta B_2 in mathcal{A}$ and $B_1 cap B_2 in mathcal{A}$.
To show that $mathcal{A}$ is not a Boolean algebra of sets, consider whether $B_1'$ is in $mathcal{A}$, where $B_1'$ stands for the complement of $B_1$. To help you get on your way, maybe define $B_1 = [a,b)$ and $B_2 = [c,d)$ where $a,b,c,d in mathbb{R}$. Then consider whether these two closed-open intervals are overlapping, containing one or the other, or disjoint.
answered 2 days ago
tucsonman101
11
New contributor
tucsonman101 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
0
down vote
up vote
0
down vote
A ring of sets is defined in Simmons to be a non-empty class $mathcal{A}$ such that for $A,B in mathcal{A}$:
(1) $A Delta B in mathcal{A}$, where $Delta$ stands for the symmetric difference.
(2) $A cap B in mathcal{A}$.
Note that this implies $emptyset in mathcal{A}$. For this proof, you need to show the following. If $B_1, B_2 in mathcal{A}$, where $mathcal{A}$ is the class of all finite unions of closed-open intervals on the real line, then $B_1 Delta B_2 in mathcal{A}$ and $B_1 cap B_2 in mathcal{A}$.
To show that $mathcal{A}$ is not a Boolean algebra of sets, consider whether $B_1'$ is in $mathcal{A}$, where $B_1'$ stands for the complement of $B_1$. To help you get on your way, maybe define $B_1 = [a,b)$ and $B_2 = [c,d)$ where $a,b,c,d in mathbb{R}$. Then consider whether these two closed-open intervals are overlapping, containing one or the other, or disjoint.
answered 2 days ago
tucsonman101
11
New contributor
tucsonman101 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
A ring of sets is defined in Simmons to be a non-empty class $mathcal{A}$ such that for $A,B in mathcal{A}$:
(1) $A Delta B in mathcal{A}$, where $Delta$ stands for the symmetric difference.
(2) $A cap B in mathcal{A}$.
Note that this implies $emptyset in mathcal{A}$. For this proof, you need to show the following. If $B_1, B_2 in mathcal{A}$, where $mathcal{A}$ is the class of all finite unions of closed-open intervals on the real line, then $B_1 Delta B_2 in mathcal{A}$ and $B_1 cap B_2 in mathcal{A}$.
To show that $mathcal{A}$ is not a Boolean algebra of sets, consider whether $B_1'$ is in $mathcal{A}$, where $B_1'$ stands for the complement of $B_1$. To help you get on your way, maybe define $B_1 = [a,b)$ and $B_2 = [c,d)$ where $a,b,c,d in mathbb{R}$. Then consider whether these two closed-open intervals are overlapping, containing one or the other, or disjoint.
answered 2 days ago
tucsonman101
11
New contributor
tucsonman101 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 2 days ago
tucsonman101
11
New contributor
tucsonman101 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 2 days ago
tucsonman101
11
answered 2 days ago
tucsonman101
11
11
New contributor
tucsonman101 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
tucsonman101 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
tucsonman101 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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There is a 'general-topology' tag. By the way, what is a closed-open interval of $mathbb{R}$?
– amrsa
Apr 20 '17 at 17:14
Like $[0,1)$. Where the first limit is included and the second not. The converse would be an open-closed interval like $(0,1]$.
– Mrigank Arora
Apr 20 '17 at 17:21
I think your definition of ring of sets is different than mine one (and I suppose, many other people's), so may be you should give that definition.
– amrsa
Apr 20 '17 at 17:24
Or maybe when you say "From what i understood", you mean that is your definition
– amrsa
Apr 20 '17 at 17:26
This is just how the book introduced it. If possible, could you tell me your definition for a ring of sets? I just started self-studying in this area, so I don't really know any conventions.
– Mrigank Arora
Apr 20 '17 at 17:30