Mixing
Borel sets. Need to prove.
-2
I need to prove that set $(1,4)$ is Borel set. Actually, I have no idea how to do it. I was looking for theory. And find something.
Let $I={(1,4), a<b, text{ and } a,b in R}$
Than $(1,4)= cup_{n=1}^infty (1, 4-frac{1}{n}], a<b$.
and it is a proof? How I should prove it?
probability borel-sets
asked Nov 20 '18 at 8:56
Atstovas
697
add a comment |
-2
I need to prove that set $(1,4)$ is Borel set. Actually, I have no idea how to do it. I was looking for theory. And find something.
Let $I={(1,4), a<b, text{ and } a,b in R}$
Than $(1,4)= cup_{n=1}^infty (1, 4-frac{1}{n}], a<b$.
and it is a proof? How I should prove it?
probability borel-sets
asked Nov 20 '18 at 8:56
Atstovas
697
What is your definition of Borel set? It seems that you think of them as elements of $sigma$-algebra generated left-open/right-closed intervals like $(a,b]$. Then what you mention is indeed a proof that $(1,4)$ is a Borel set. That "definition" is okay, but it is better to define them as elements of $sigma$-algebra generated by open sets (as you meet in the answer of Fred). Then your definition becomes a theorem. If it is unclear which definition is practicized then it is not possible to recognize "proofs".
– drhab
Nov 20 '18 at 9:08
add a comment |
-2
-2
-2
I need to prove that set $(1,4)$ is Borel set. Actually, I have no idea how to do it. I was looking for theory. And find something.
Let $I={(1,4), a<b, text{ and } a,b in R}$
Than $(1,4)= cup_{n=1}^infty (1, 4-frac{1}{n}], a<b$.
and it is a proof? How I should prove it?
probability borel-sets
asked Nov 20 '18 at 8:56
Atstovas
697
I need to prove that set $(1,4)$ is Borel set. Actually, I have no idea how to do it. I was looking for theory. And find something.
Let $I={(1,4), a<b, text{ and } a,b in R}$
Than $(1,4)= cup_{n=1}^infty (1, 4-frac{1}{n}], a<b$.
and it is a proof? How I should prove it?
probability borel-sets
probability borel-sets
asked Nov 20 '18 at 8:56
Atstovas
697
asked Nov 20 '18 at 8:56
Atstovas
697
asked Nov 20 '18 at 8:56
Atstovas
697
asked Nov 20 '18 at 8:56
Atstovas
697
asked Nov 20 '18 at 8:56
Atstovas
697
697
What is your definition of Borel set? It seems that you think of them as elements of $sigma$-algebra generated left-open/right-closed intervals like $(a,b]$. Then what you mention is indeed a proof that $(1,4)$ is a Borel set. That "definition" is okay, but it is better to define them as elements of $sigma$-algebra generated by open sets (as you meet in the answer of Fred). Then your definition becomes a theorem. If it is unclear which definition is practicized then it is not possible to recognize "proofs".
– drhab
Nov 20 '18 at 9:08
add a comment |
What is your definition of Borel set? It seems that you think of them as elements of $sigma$-algebra generated left-open/right-closed intervals like $(a,b]$. Then what you mention is indeed a proof that $(1,4)$ is a Borel set. That "definition" is okay, but it is better to define them as elements of $sigma$-algebra generated by open sets (as you meet in the answer of Fred). Then your definition becomes a theorem. If it is unclear which definition is practicized then it is not possible to recognize "proofs".
– drhab
Nov 20 '18 at 9:08
What is your definition of Borel set? It seems that you think of them as elements of $sigma$-algebra generated left-open/right-closed intervals like $(a,b]$. Then what you mention is indeed a proof that $(1,4)$ is a Borel set. That "definition" is okay, but it is better to define them as elements of $sigma$-algebra generated by open sets (as you meet in the answer of Fred). Then your definition becomes a theorem. If it is unclear which definition is practicized then it is not possible to recognize "proofs".
– drhab
Nov 20 '18 at 9:08
What is your definition of Borel set? It seems that you think of them as elements of $sigma$-algebra generated left-open/right-closed intervals like $(a,b]$. Then what you mention is indeed a proof that $(1,4)$ is a Borel set. That "definition" is okay, but it is better to define them as elements of $sigma$-algebra generated by open sets (as you meet in the answer of Fred). Then your definition becomes a theorem. If it is unclear which definition is practicized then it is not possible to recognize "proofs".
– drhab
Nov 20 '18 at 9:08
add a comment |
1 Answer
1
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The Borel sets in $ mathbb R$ are generated by the open sets in $ mathbb R$ and $(1,4)$ is open !
answered Nov 20 '18 at 8:58
Fred
44.2k1845
Than what about set [1,4]? how to prove it is a Borel set?
– Atstovas
Nov 20 '18 at 9:00
1
The complement of $[1,4]$ is open !
– Fred
Nov 20 '18 at 9:03
and that is it? Enough to show this and nothing more to explain? I didn't think it is that easy...
– Atstovas
Nov 20 '18 at 9:03
If $A$ is a member of a $ sigma-$ algebra, then the complement of $A$ is also a member of the $ sigma-$ algebra
– Fred
Nov 20 '18 at 9:05
Ok, what about $mathbb{N}$ set. How to prove it is a Borel set?
– Atstovas
Nov 20 '18 at 9:05
|
show 1 more comment
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1 Answer
1
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
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2
The Borel sets in $ mathbb R$ are generated by the open sets in $ mathbb R$ and $(1,4)$ is open !
answered Nov 20 '18 at 8:58
Fred
44.2k1845
Than what about set [1,4]? how to prove it is a Borel set?
– Atstovas
Nov 20 '18 at 9:00
1
The complement of $[1,4]$ is open !
– Fred
Nov 20 '18 at 9:03
and that is it? Enough to show this and nothing more to explain? I didn't think it is that easy...
– Atstovas
Nov 20 '18 at 9:03
If $A$ is a member of a $ sigma-$ algebra, then the complement of $A$ is also a member of the $ sigma-$ algebra
– Fred
Nov 20 '18 at 9:05
Ok, what about $mathbb{N}$ set. How to prove it is a Borel set?
– Atstovas
Nov 20 '18 at 9:05
|
show 1 more comment
2
The Borel sets in $ mathbb R$ are generated by the open sets in $ mathbb R$ and $(1,4)$ is open !
answered Nov 20 '18 at 8:58
Fred
44.2k1845
Than what about set [1,4]? how to prove it is a Borel set?
– Atstovas
Nov 20 '18 at 9:00
1
The complement of $[1,4]$ is open !
– Fred
Nov 20 '18 at 9:03
and that is it? Enough to show this and nothing more to explain? I didn't think it is that easy...
– Atstovas
Nov 20 '18 at 9:03
If $A$ is a member of a $ sigma-$ algebra, then the complement of $A$ is also a member of the $ sigma-$ algebra
– Fred
Nov 20 '18 at 9:05
Ok, what about $mathbb{N}$ set. How to prove it is a Borel set?
– Atstovas
Nov 20 '18 at 9:05
|
show 1 more comment
2
2
2
The Borel sets in $ mathbb R$ are generated by the open sets in $ mathbb R$ and $(1,4)$ is open !
answered Nov 20 '18 at 8:58
Fred
44.2k1845
The Borel sets in $ mathbb R$ are generated by the open sets in $ mathbb R$ and $(1,4)$ is open !
answered Nov 20 '18 at 8:58
Fred
44.2k1845
answered Nov 20 '18 at 8:58
Fred
44.2k1845
answered Nov 20 '18 at 8:58
Fred
44.2k1845
answered Nov 20 '18 at 8:58
Fred
44.2k1845
44.2k1845
Than what about set [1,4]? how to prove it is a Borel set?
– Atstovas
Nov 20 '18 at 9:00
1
The complement of $[1,4]$ is open !
– Fred
Nov 20 '18 at 9:03
and that is it? Enough to show this and nothing more to explain? I didn't think it is that easy...
– Atstovas
Nov 20 '18 at 9:03
If $A$ is a member of a $ sigma-$ algebra, then the complement of $A$ is also a member of the $ sigma-$ algebra
– Fred
Nov 20 '18 at 9:05
Ok, what about $mathbb{N}$ set. How to prove it is a Borel set?
– Atstovas
Nov 20 '18 at 9:05
|
show 1 more comment
Than what about set [1,4]? how to prove it is a Borel set?
– Atstovas
Nov 20 '18 at 9:00
1
The complement of $[1,4]$ is open !
– Fred
Nov 20 '18 at 9:03
and that is it? Enough to show this and nothing more to explain? I didn't think it is that easy...
– Atstovas
Nov 20 '18 at 9:03
If $A$ is a member of a $ sigma-$ algebra, then the complement of $A$ is also a member of the $ sigma-$ algebra
– Fred
Nov 20 '18 at 9:05
Ok, what about $mathbb{N}$ set. How to prove it is a Borel set?
– Atstovas
Nov 20 '18 at 9:05
Than what about set [1,4]? how to prove it is a Borel set?
– Atstovas
Nov 20 '18 at 9:00
Than what about set [1,4]? how to prove it is a Borel set?
– Atstovas
Nov 20 '18 at 9:00
1
1
The complement of $[1,4]$ is open !
– Fred
Nov 20 '18 at 9:03
The complement of $[1,4]$ is open !
– Fred
Nov 20 '18 at 9:03
and that is it? Enough to show this and nothing more to explain? I didn't think it is that easy...
– Atstovas
Nov 20 '18 at 9:03
and that is it? Enough to show this and nothing more to explain? I didn't think it is that easy...
– Atstovas
Nov 20 '18 at 9:03
If $A$ is a member of a $ sigma-$ algebra, then the complement of $A$ is also a member of the $ sigma-$ algebra
– Fred
Nov 20 '18 at 9:05
If $A$ is a member of a $ sigma-$ algebra, then the complement of $A$ is also a member of the $ sigma-$ algebra
– Fred
Nov 20 '18 at 9:05
Ok, what about $mathbb{N}$ set. How to prove it is a Borel set?
– Atstovas
Nov 20 '18 at 9:05
Ok, what about $mathbb{N}$ set. How to prove it is a Borel set?
– Atstovas
Nov 20 '18 at 9:05
|
show 1 more comment
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What is your definition of Borel set? It seems that you think of them as elements of $sigma$-algebra generated left-open/right-closed intervals like $(a,b]$. Then what you mention is indeed a proof that $(1,4)$ is a Borel set. That "definition" is okay, but it is better to define them as elements of $sigma$-algebra generated by open sets (as you meet in the answer of Fred). Then your definition becomes a theorem. If it is unclear which definition is practicized then it is not possible to recognize "proofs".
– drhab
Nov 20 '18 at 9:08