Show $mathcal{B}( mathbb R)$ contains all countable subsets of $mathbb R$
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0
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We defined $mathcal{B}(mathbb R):=sigma({[a,b[}),$ where $a < b in mathbb R.$
I have been able to prove:
i) for any $b in mathbb R$, ${b} inmathcal{B}(mathbb R)$
and subsequently ii) $[a,b] in mathcal{B}(mathbb R)$ for any $a, b in mathbb R$
iii) $]a,b[ in mathcal{B}(mathbb R)$ for any $a, b in mathbb R$
Now for my problem, how can I prove that $mathcal{B}(mathbb R)$ contains all countable subsets of $mathbb R$?
My idea was to take any countable subset $C$ of $mathbb R$,
it can be written as $C = bigcup_{n in mathbb N}{a_{n}}$. Since we were able to prove (i), $C$ is immediately as the countable union $in mathcal{B}(mathbb R)$. Is this ok? Does not seem clean enough to me?
real-analysis measure-theory borel-sets borel-measures
edited Nov 21 at 20:32
Lord Shark the Unknown
99.1k958131
asked Nov 21 at 20:29
SABOY
500311
add a comment |
up vote
0
down vote
favorite
We defined $mathcal{B}(mathbb R):=sigma({[a,b[}),$ where $a < b in mathbb R.$
I have been able to prove:
i) for any $b in mathbb R$, ${b} inmathcal{B}(mathbb R)$
and subsequently ii) $[a,b] in mathcal{B}(mathbb R)$ for any $a, b in mathbb R$
iii) $]a,b[ in mathcal{B}(mathbb R)$ for any $a, b in mathbb R$
Now for my problem, how can I prove that $mathcal{B}(mathbb R)$ contains all countable subsets of $mathbb R$?
My idea was to take any countable subset $C$ of $mathbb R$,
it can be written as $C = bigcup_{n in mathbb N}{a_{n}}$. Since we were able to prove (i), $C$ is immediately as the countable union $in mathcal{B}(mathbb R)$. Is this ok? Does not seem clean enough to me?
real-analysis measure-theory borel-sets borel-measures
edited Nov 21 at 20:32
Lord Shark the Unknown
99.1k958131
asked Nov 21 at 20:29
SABOY
500311
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
We defined $mathcal{B}(mathbb R):=sigma({[a,b[}),$ where $a < b in mathbb R.$
I have been able to prove:
i) for any $b in mathbb R$, ${b} inmathcal{B}(mathbb R)$
and subsequently ii) $[a,b] in mathcal{B}(mathbb R)$ for any $a, b in mathbb R$
iii) $]a,b[ in mathcal{B}(mathbb R)$ for any $a, b in mathbb R$
Now for my problem, how can I prove that $mathcal{B}(mathbb R)$ contains all countable subsets of $mathbb R$?
My idea was to take any countable subset $C$ of $mathbb R$,
it can be written as $C = bigcup_{n in mathbb N}{a_{n}}$. Since we were able to prove (i), $C$ is immediately as the countable union $in mathcal{B}(mathbb R)$. Is this ok? Does not seem clean enough to me?
real-analysis measure-theory borel-sets borel-measures
edited Nov 21 at 20:32
Lord Shark the Unknown
99.1k958131
asked Nov 21 at 20:29
SABOY
500311
We defined $mathcal{B}(mathbb R):=sigma({[a,b[}),$ where $a < b in mathbb R.$
I have been able to prove:
i) for any $b in mathbb R$, ${b} inmathcal{B}(mathbb R)$
and subsequently ii) $[a,b] in mathcal{B}(mathbb R)$ for any $a, b in mathbb R$
iii) $]a,b[ in mathcal{B}(mathbb R)$ for any $a, b in mathbb R$
Now for my problem, how can I prove that $mathcal{B}(mathbb R)$ contains all countable subsets of $mathbb R$?
My idea was to take any countable subset $C$ of $mathbb R$,
it can be written as $C = bigcup_{n in mathbb N}{a_{n}}$. Since we were able to prove (i), $C$ is immediately as the countable union $in mathcal{B}(mathbb R)$. Is this ok? Does not seem clean enough to me?
real-analysis measure-theory borel-sets borel-measures
real-analysis measure-theory borel-sets borel-measures
edited Nov 21 at 20:32
Lord Shark the Unknown
99.1k958131
asked Nov 21 at 20:29
SABOY
500311
edited Nov 21 at 20:32
Lord Shark the Unknown
99.1k958131
asked Nov 21 at 20:29
SABOY
500311
edited Nov 21 at 20:32
Lord Shark the Unknown
99.1k958131
edited Nov 21 at 20:32
Lord Shark the Unknown
99.1k958131
edited Nov 21 at 20:32
Lord Shark the Unknown
99.1k958131
99.1k958131
asked Nov 21 at 20:29
SABOY
500311
asked Nov 21 at 20:29
SABOY
500311
asked Nov 21 at 20:29
SABOY
500311
500311
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
The usual definition of a sigma-algebra requires that it be closed under countable unions. Once that you have that every singleton is in $mathcal{B}( mathbb R)$, every countable subset of $Bbb R$ is a countable union of singletons.
answered Nov 21 at 20:35
Ross Millikan
289k23195367
So my version is correct?
– SABOY
Nov 21 at 20:37
Yes, it is fine.
– Ross Millikan
Nov 21 at 20:50
add a comment |
up vote
1
down vote
Your answer is entirely correct, since $sigma$-algebras are closed under countable unions and singeltons are in this $sigma$-algebra:
Indeed, if $ain mathbb{R}$, then $${a} = bigcap_{n=1}^infty{[a,a + 1/n[} = bigcap_{n=1}^infty{[a,a + 1/n]}$$
answered Nov 21 at 20:43
Math_QED
6,88931449
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
The usual definition of a sigma-algebra requires that it be closed under countable unions. Once that you have that every singleton is in $mathcal{B}( mathbb R)$, every countable subset of $Bbb R$ is a countable union of singletons.
answered Nov 21 at 20:35
Ross Millikan
289k23195367
So my version is correct?
– SABOY
Nov 21 at 20:37
Yes, it is fine.
– Ross Millikan
Nov 21 at 20:50
add a comment |
up vote
3
down vote
accepted
The usual definition of a sigma-algebra requires that it be closed under countable unions. Once that you have that every singleton is in $mathcal{B}( mathbb R)$, every countable subset of $Bbb R$ is a countable union of singletons.
answered Nov 21 at 20:35
Ross Millikan
289k23195367
So my version is correct?
– SABOY
Nov 21 at 20:37
Yes, it is fine.
– Ross Millikan
Nov 21 at 20:50
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
The usual definition of a sigma-algebra requires that it be closed under countable unions. Once that you have that every singleton is in $mathcal{B}( mathbb R)$, every countable subset of $Bbb R$ is a countable union of singletons.
answered Nov 21 at 20:35
Ross Millikan
289k23195367
The usual definition of a sigma-algebra requires that it be closed under countable unions. Once that you have that every singleton is in $mathcal{B}( mathbb R)$, every countable subset of $Bbb R$ is a countable union of singletons.
answered Nov 21 at 20:35
Ross Millikan
289k23195367
answered Nov 21 at 20:35
Ross Millikan
289k23195367
answered Nov 21 at 20:35
Ross Millikan
289k23195367
answered Nov 21 at 20:35
Ross Millikan
289k23195367
289k23195367
So my version is correct?
– SABOY
Nov 21 at 20:37
Yes, it is fine.
– Ross Millikan
Nov 21 at 20:50
add a comment |
So my version is correct?
– SABOY
Nov 21 at 20:37
Yes, it is fine.
– Ross Millikan
Nov 21 at 20:50
So my version is correct?
– SABOY
Nov 21 at 20:37
So my version is correct?
– SABOY
Nov 21 at 20:37
Yes, it is fine.
– Ross Millikan
Nov 21 at 20:50
Yes, it is fine.
– Ross Millikan
Nov 21 at 20:50
add a comment |
up vote
1
down vote
Your answer is entirely correct, since $sigma$-algebras are closed under countable unions and singeltons are in this $sigma$-algebra:
Indeed, if $ain mathbb{R}$, then $${a} = bigcap_{n=1}^infty{[a,a + 1/n[} = bigcap_{n=1}^infty{[a,a + 1/n]}$$
answered Nov 21 at 20:43
Math_QED
6,88931449
add a comment |
up vote
1
down vote
Your answer is entirely correct, since $sigma$-algebras are closed under countable unions and singeltons are in this $sigma$-algebra:
Indeed, if $ain mathbb{R}$, then $${a} = bigcap_{n=1}^infty{[a,a + 1/n[} = bigcap_{n=1}^infty{[a,a + 1/n]}$$
answered Nov 21 at 20:43
Math_QED
6,88931449
add a comment |
up vote
1
down vote
up vote
1
down vote
Your answer is entirely correct, since $sigma$-algebras are closed under countable unions and singeltons are in this $sigma$-algebra:
Indeed, if $ain mathbb{R}$, then $${a} = bigcap_{n=1}^infty{[a,a + 1/n[} = bigcap_{n=1}^infty{[a,a + 1/n]}$$
answered Nov 21 at 20:43
Math_QED
6,88931449
Your answer is entirely correct, since $sigma$-algebras are closed under countable unions and singeltons are in this $sigma$-algebra:
Indeed, if $ain mathbb{R}$, then $${a} = bigcap_{n=1}^infty{[a,a + 1/n[} = bigcap_{n=1}^infty{[a,a + 1/n]}$$
answered Nov 21 at 20:43
Math_QED
6,88931449
answered Nov 21 at 20:43
Math_QED
6,88931449
answered Nov 21 at 20:43
Math_QED
6,88931449
answered Nov 21 at 20:43
Math_QED
6,88931449
6,88931449
add a comment |
add a comment |
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