Mixing
Sigma algebra respect to random variables.
$begingroup$
Let $Omega={1,2,3,4,5,6}$ and $X(omega)=omega mod 2$, which is a random variable.
Then, my first question is, what is $sigma(X)$?
In my understanding,
$sigma(X)={{omegamid X(omega)in B}mid Bsubset R}$, so $sigma(X)={emptyset,{1,2,3},{4,5,6},Omega}$.
Let $Y(omega)=1_{omegageq 4}$.
Then, my second question is, what is $sigma(X,Y)$?
In my thought, $sigma(X,Y)={{omegamid (X(omega),Y(omega))in B_1times B_2}mid B_1,B_2subseteq R}$.
$sigma(X,Y)=left{{2},{5},{1,3},{4,6},{2,5},{1,2,3},{2,4,6},{1,3,5},{4,5,6},{1,3,4,6},{1,2,3,5},{2,4,5,6},{1,2,3,4,6},{1,3,4,5,6},Omegaright}$
Is this true?
probability-theory measure-theory
asked Jan 25 at 9:44
hiratathiratat
262
$endgroup$
add a comment |
$begingroup$
Let $Omega={1,2,3,4,5,6}$ and $X(omega)=omega mod 2$, which is a random variable.
Then, my first question is, what is $sigma(X)$?
In my understanding,
$sigma(X)={{omegamid X(omega)in B}mid Bsubset R}$, so $sigma(X)={emptyset,{1,2,3},{4,5,6},Omega}$.
Let $Y(omega)=1_{omegageq 4}$.
Then, my second question is, what is $sigma(X,Y)$?
In my thought, $sigma(X,Y)={{omegamid (X(omega),Y(omega))in B_1times B_2}mid B_1,B_2subseteq R}$.
$sigma(X,Y)=left{{2},{5},{1,3},{4,6},{2,5},{1,2,3},{2,4,6},{1,3,5},{4,5,6},{1,3,4,6},{1,2,3,5},{2,4,5,6},{1,2,3,4,6},{1,3,4,5,6},Omegaright}$
Is this true?
probability-theory measure-theory
asked Jan 25 at 9:44
hiratathiratat
262
$endgroup$
add a comment |
$begingroup$
Let $Omega={1,2,3,4,5,6}$ and $X(omega)=omega mod 2$, which is a random variable.
Then, my first question is, what is $sigma(X)$?
In my understanding,
$sigma(X)={{omegamid X(omega)in B}mid Bsubset R}$, so $sigma(X)={emptyset,{1,2,3},{4,5,6},Omega}$.
Let $Y(omega)=1_{omegageq 4}$.
Then, my second question is, what is $sigma(X,Y)$?
In my thought, $sigma(X,Y)={{omegamid (X(omega),Y(omega))in B_1times B_2}mid B_1,B_2subseteq R}$.
$sigma(X,Y)=left{{2},{5},{1,3},{4,6},{2,5},{1,2,3},{2,4,6},{1,3,5},{4,5,6},{1,3,4,6},{1,2,3,5},{2,4,5,6},{1,2,3,4,6},{1,3,4,5,6},Omegaright}$
Is this true?
probability-theory measure-theory
asked Jan 25 at 9:44
hiratathiratat
262
$endgroup$
Let $Omega={1,2,3,4,5,6}$ and $X(omega)=omega mod 2$, which is a random variable.
Then, my first question is, what is $sigma(X)$?
In my understanding,
$sigma(X)={{omegamid X(omega)in B}mid Bsubset R}$, so $sigma(X)={emptyset,{1,2,3},{4,5,6},Omega}$.
Let $Y(omega)=1_{omegageq 4}$.
Then, my second question is, what is $sigma(X,Y)$?
In my thought, $sigma(X,Y)={{omegamid (X(omega),Y(omega))in B_1times B_2}mid B_1,B_2subseteq R}$.
$sigma(X,Y)=left{{2},{5},{1,3},{4,6},{2,5},{1,2,3},{2,4,6},{1,3,5},{4,5,6},{1,3,4,6},{1,2,3,5},{2,4,5,6},{1,2,3,4,6},{1,3,4,5,6},Omegaright}$
Is this true?
probability-theory measure-theory
probability-theory measure-theory
asked Jan 25 at 9:44
hiratathiratat
262
asked Jan 25 at 9:44
hiratathiratat
262
asked Jan 25 at 9:44
hiratathiratat
262
asked Jan 25 at 9:44
hiratathiratat
262
asked Jan 25 at 9:44
hiratathiratat
262
262
add a comment |
add a comment |
2 Answers
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$begingroup$
I don't beleive you've identified $ sigmaleft(Xright) $ correctly. Consider your expression, $ X^{-1}left(Bright) = {omegamid X(omega)in B} $, for the elements of $ sigmaleft(Xright) $. This is correct. Strictly speaking, $ B $ should be restricted to be a measurable subset of $ R $, but in this case it makes no difference if you take all subsets of $ R $ to be measurable.
Since the only possible values of $ Xleft(omegaright) $ are $ 0 $ and $ 1 $, then unless $ 0in B $ or $ 1in B $, we must have $ X^{-1}left(Bright) = emptyset $. If $ 0in B $ but $ 1notin B $, then $ Xleft(omegaright)in B $ if and only if $ omega= 2, 4, mbox{ or } 6 $, so $ X^{-1}left(Bright) = left{2,4,6right} $. Similarly, if $ 1in B $ but $ 0notin B $, then $ X^{-1}left(Bright) = left{1,3,5right} $, and if both $ 0in B $ and $ 1notin B $, then $ X^{-1}left(Bright) = left{1,2,3,4,5,6right}= Omega $, thus giving $sigma(X)={emptyset,{1,3,5},{2,4,6},Omega}$.
Your definition of $ sigmaleft(X,Yright) $ is also not quite correct. It should be $ sigma(X,Y)={{omegamid (X(omega),Y(omega))in B }mid Bsubseteq Rtimes R, mbox{measurable}}$. That is, the elements of $ sigma(X,Y) $ are the inverse images not merely of product sets, but all measurable subsets of $ Rtimes R $. With that proviso, the construction of $ sigmaleft(X,Yright) $ can be carried out along the same lines as that of $ sigmaleft(Xright) $ above. The only possible values of $
left(Xleft(omega_1right), Yleft(omega_2right)right) $ are $ left(0,0 right), left(1,0 right),left(0,1 right), mbox{ and }left(1,1 right) $. If $ left(0,0right) $ is the only one of those pairs belonging to a set $ Bsubseteq Rtimes R $, then the inverse image of $ B $ with respect to the map $ left(X, Yright) $ is $$ left{left(2,1right),left(2,2right),left(2,3right),left(4,1right),left(4,2right),left(4,3right), left(6,1right),left(6,2right),left(6,3right)right} ,$$
and so this must be one of the minimal non-empty subsets of your $sigma$-algebra $ sigmaleft(X,Yright) $. The other three minimal non-empty subsets likewise contain nine members of $ OmegatimesOmega $, which I suggest you now try constructing for yourself.
answered Jan 25 at 11:28
lonza leggieralonza leggiera
1,12928
$endgroup$
add a comment |
$begingroup$
Thanks.
The definition of $sigma(X,Y)$ is
$${{(omega_1,omega_2)mid (X(omega_1),Y(omega_2))in B}mid Bsubseteq Rtimes R, mesurable}$$,
and
for the case $(0,1)in B$ and others not in $B$,
$${(2,4),(2,5),(2,6),(4,4),(4,5),(4,6),(6,4),(6,5),(6,6)}$$.
I need to consider $2^4$ cases, like $(0,0)in B$, $(0,1)in B$, $(1,0)notin B$, $(1,1)in B$..., right?
answered Jan 26 at 10:55
hiratathiratat
262
$endgroup$
$begingroup$
Yes. Although it's simpler to just do the 4 cases: $1. (0,0) in B, (0,1), (1,0), (1,1)notin B $; $2. (0,1) in B, (0,0), (1,0), (1,1)notin B $; $3. (1,0) in B, (0,0), (0,1), (1,1)notin B $; and $4. (1,1) in B, (0,0), (0,1),(1,0) notin B $. The sets in $ sigma(X) $ are then just all the unions (including the empty union) of those you get from these four cases.
$endgroup$
– lonza leggiera
Jan 27 at 20:37
$begingroup$
Thank you very much for your kind answer.
$endgroup$
– hiratat
Feb 1 at 3:19
add a comment |
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$begingroup$
I don't beleive you've identified $ sigmaleft(Xright) $ correctly. Consider your expression, $ X^{-1}left(Bright) = {omegamid X(omega)in B} $, for the elements of $ sigmaleft(Xright) $. This is correct. Strictly speaking, $ B $ should be restricted to be a measurable subset of $ R $, but in this case it makes no difference if you take all subsets of $ R $ to be measurable.
Since the only possible values of $ Xleft(omegaright) $ are $ 0 $ and $ 1 $, then unless $ 0in B $ or $ 1in B $, we must have $ X^{-1}left(Bright) = emptyset $. If $ 0in B $ but $ 1notin B $, then $ Xleft(omegaright)in B $ if and only if $ omega= 2, 4, mbox{ or } 6 $, so $ X^{-1}left(Bright) = left{2,4,6right} $. Similarly, if $ 1in B $ but $ 0notin B $, then $ X^{-1}left(Bright) = left{1,3,5right} $, and if both $ 0in B $ and $ 1notin B $, then $ X^{-1}left(Bright) = left{1,2,3,4,5,6right}= Omega $, thus giving $sigma(X)={emptyset,{1,3,5},{2,4,6},Omega}$.
Your definition of $ sigmaleft(X,Yright) $ is also not quite correct. It should be $ sigma(X,Y)={{omegamid (X(omega),Y(omega))in B }mid Bsubseteq Rtimes R, mbox{measurable}}$. That is, the elements of $ sigma(X,Y) $ are the inverse images not merely of product sets, but all measurable subsets of $ Rtimes R $. With that proviso, the construction of $ sigmaleft(X,Yright) $ can be carried out along the same lines as that of $ sigmaleft(Xright) $ above. The only possible values of $
left(Xleft(omega_1right), Yleft(omega_2right)right) $ are $ left(0,0 right), left(1,0 right),left(0,1 right), mbox{ and }left(1,1 right) $. If $ left(0,0right) $ is the only one of those pairs belonging to a set $ Bsubseteq Rtimes R $, then the inverse image of $ B $ with respect to the map $ left(X, Yright) $ is $$ left{left(2,1right),left(2,2right),left(2,3right),left(4,1right),left(4,2right),left(4,3right), left(6,1right),left(6,2right),left(6,3right)right} ,$$
and so this must be one of the minimal non-empty subsets of your $sigma$-algebra $ sigmaleft(X,Yright) $. The other three minimal non-empty subsets likewise contain nine members of $ OmegatimesOmega $, which I suggest you now try constructing for yourself.
answered Jan 25 at 11:28
lonza leggieralonza leggiera
1,12928
$endgroup$
add a comment |
$begingroup$
I don't beleive you've identified $ sigmaleft(Xright) $ correctly. Consider your expression, $ X^{-1}left(Bright) = {omegamid X(omega)in B} $, for the elements of $ sigmaleft(Xright) $. This is correct. Strictly speaking, $ B $ should be restricted to be a measurable subset of $ R $, but in this case it makes no difference if you take all subsets of $ R $ to be measurable.
Since the only possible values of $ Xleft(omegaright) $ are $ 0 $ and $ 1 $, then unless $ 0in B $ or $ 1in B $, we must have $ X^{-1}left(Bright) = emptyset $. If $ 0in B $ but $ 1notin B $, then $ Xleft(omegaright)in B $ if and only if $ omega= 2, 4, mbox{ or } 6 $, so $ X^{-1}left(Bright) = left{2,4,6right} $. Similarly, if $ 1in B $ but $ 0notin B $, then $ X^{-1}left(Bright) = left{1,3,5right} $, and if both $ 0in B $ and $ 1notin B $, then $ X^{-1}left(Bright) = left{1,2,3,4,5,6right}= Omega $, thus giving $sigma(X)={emptyset,{1,3,5},{2,4,6},Omega}$.
Your definition of $ sigmaleft(X,Yright) $ is also not quite correct. It should be $ sigma(X,Y)={{omegamid (X(omega),Y(omega))in B }mid Bsubseteq Rtimes R, mbox{measurable}}$. That is, the elements of $ sigma(X,Y) $ are the inverse images not merely of product sets, but all measurable subsets of $ Rtimes R $. With that proviso, the construction of $ sigmaleft(X,Yright) $ can be carried out along the same lines as that of $ sigmaleft(Xright) $ above. The only possible values of $
left(Xleft(omega_1right), Yleft(omega_2right)right) $ are $ left(0,0 right), left(1,0 right),left(0,1 right), mbox{ and }left(1,1 right) $. If $ left(0,0right) $ is the only one of those pairs belonging to a set $ Bsubseteq Rtimes R $, then the inverse image of $ B $ with respect to the map $ left(X, Yright) $ is $$ left{left(2,1right),left(2,2right),left(2,3right),left(4,1right),left(4,2right),left(4,3right), left(6,1right),left(6,2right),left(6,3right)right} ,$$
and so this must be one of the minimal non-empty subsets of your $sigma$-algebra $ sigmaleft(X,Yright) $. The other three minimal non-empty subsets likewise contain nine members of $ OmegatimesOmega $, which I suggest you now try constructing for yourself.
answered Jan 25 at 11:28
lonza leggieralonza leggiera
1,12928
$endgroup$
add a comment |
$begingroup$
I don't beleive you've identified $ sigmaleft(Xright) $ correctly. Consider your expression, $ X^{-1}left(Bright) = {omegamid X(omega)in B} $, for the elements of $ sigmaleft(Xright) $. This is correct. Strictly speaking, $ B $ should be restricted to be a measurable subset of $ R $, but in this case it makes no difference if you take all subsets of $ R $ to be measurable.
Since the only possible values of $ Xleft(omegaright) $ are $ 0 $ and $ 1 $, then unless $ 0in B $ or $ 1in B $, we must have $ X^{-1}left(Bright) = emptyset $. If $ 0in B $ but $ 1notin B $, then $ Xleft(omegaright)in B $ if and only if $ omega= 2, 4, mbox{ or } 6 $, so $ X^{-1}left(Bright) = left{2,4,6right} $. Similarly, if $ 1in B $ but $ 0notin B $, then $ X^{-1}left(Bright) = left{1,3,5right} $, and if both $ 0in B $ and $ 1notin B $, then $ X^{-1}left(Bright) = left{1,2,3,4,5,6right}= Omega $, thus giving $sigma(X)={emptyset,{1,3,5},{2,4,6},Omega}$.
Your definition of $ sigmaleft(X,Yright) $ is also not quite correct. It should be $ sigma(X,Y)={{omegamid (X(omega),Y(omega))in B }mid Bsubseteq Rtimes R, mbox{measurable}}$. That is, the elements of $ sigma(X,Y) $ are the inverse images not merely of product sets, but all measurable subsets of $ Rtimes R $. With that proviso, the construction of $ sigmaleft(X,Yright) $ can be carried out along the same lines as that of $ sigmaleft(Xright) $ above. The only possible values of $
left(Xleft(omega_1right), Yleft(omega_2right)right) $ are $ left(0,0 right), left(1,0 right),left(0,1 right), mbox{ and }left(1,1 right) $. If $ left(0,0right) $ is the only one of those pairs belonging to a set $ Bsubseteq Rtimes R $, then the inverse image of $ B $ with respect to the map $ left(X, Yright) $ is $$ left{left(2,1right),left(2,2right),left(2,3right),left(4,1right),left(4,2right),left(4,3right), left(6,1right),left(6,2right),left(6,3right)right} ,$$
and so this must be one of the minimal non-empty subsets of your $sigma$-algebra $ sigmaleft(X,Yright) $. The other three minimal non-empty subsets likewise contain nine members of $ OmegatimesOmega $, which I suggest you now try constructing for yourself.
answered Jan 25 at 11:28
lonza leggieralonza leggiera
1,12928
$endgroup$
I don't beleive you've identified $ sigmaleft(Xright) $ correctly. Consider your expression, $ X^{-1}left(Bright) = {omegamid X(omega)in B} $, for the elements of $ sigmaleft(Xright) $. This is correct. Strictly speaking, $ B $ should be restricted to be a measurable subset of $ R $, but in this case it makes no difference if you take all subsets of $ R $ to be measurable.
Since the only possible values of $ Xleft(omegaright) $ are $ 0 $ and $ 1 $, then unless $ 0in B $ or $ 1in B $, we must have $ X^{-1}left(Bright) = emptyset $. If $ 0in B $ but $ 1notin B $, then $ Xleft(omegaright)in B $ if and only if $ omega= 2, 4, mbox{ or } 6 $, so $ X^{-1}left(Bright) = left{2,4,6right} $. Similarly, if $ 1in B $ but $ 0notin B $, then $ X^{-1}left(Bright) = left{1,3,5right} $, and if both $ 0in B $ and $ 1notin B $, then $ X^{-1}left(Bright) = left{1,2,3,4,5,6right}= Omega $, thus giving $sigma(X)={emptyset,{1,3,5},{2,4,6},Omega}$.
Your definition of $ sigmaleft(X,Yright) $ is also not quite correct. It should be $ sigma(X,Y)={{omegamid (X(omega),Y(omega))in B }mid Bsubseteq Rtimes R, mbox{measurable}}$. That is, the elements of $ sigma(X,Y) $ are the inverse images not merely of product sets, but all measurable subsets of $ Rtimes R $. With that proviso, the construction of $ sigmaleft(X,Yright) $ can be carried out along the same lines as that of $ sigmaleft(Xright) $ above. The only possible values of $
left(Xleft(omega_1right), Yleft(omega_2right)right) $ are $ left(0,0 right), left(1,0 right),left(0,1 right), mbox{ and }left(1,1 right) $. If $ left(0,0right) $ is the only one of those pairs belonging to a set $ Bsubseteq Rtimes R $, then the inverse image of $ B $ with respect to the map $ left(X, Yright) $ is $$ left{left(2,1right),left(2,2right),left(2,3right),left(4,1right),left(4,2right),left(4,3right), left(6,1right),left(6,2right),left(6,3right)right} ,$$
and so this must be one of the minimal non-empty subsets of your $sigma$-algebra $ sigmaleft(X,Yright) $. The other three minimal non-empty subsets likewise contain nine members of $ OmegatimesOmega $, which I suggest you now try constructing for yourself.
answered Jan 25 at 11:28
lonza leggieralonza leggiera
1,12928
answered Jan 25 at 11:28
lonza leggieralonza leggiera
1,12928
answered Jan 25 at 11:28
lonza leggieralonza leggiera
1,12928
answered Jan 25 at 11:28
lonza leggieralonza leggiera
1,12928
1,12928
add a comment |
add a comment |
$begingroup$
Thanks.
The definition of $sigma(X,Y)$ is
$${{(omega_1,omega_2)mid (X(omega_1),Y(omega_2))in B}mid Bsubseteq Rtimes R, mesurable}$$,
and
for the case $(0,1)in B$ and others not in $B$,
$${(2,4),(2,5),(2,6),(4,4),(4,5),(4,6),(6,4),(6,5),(6,6)}$$.
I need to consider $2^4$ cases, like $(0,0)in B$, $(0,1)in B$, $(1,0)notin B$, $(1,1)in B$..., right?
answered Jan 26 at 10:55
hiratathiratat
262
$endgroup$
$begingroup$
Yes. Although it's simpler to just do the 4 cases: $1. (0,0) in B, (0,1), (1,0), (1,1)notin B $; $2. (0,1) in B, (0,0), (1,0), (1,1)notin B $; $3. (1,0) in B, (0,0), (0,1), (1,1)notin B $; and $4. (1,1) in B, (0,0), (0,1),(1,0) notin B $. The sets in $ sigma(X) $ are then just all the unions (including the empty union) of those you get from these four cases.
$endgroup$
– lonza leggiera
Jan 27 at 20:37
$begingroup$
Thank you very much for your kind answer.
$endgroup$
– hiratat
Feb 1 at 3:19
add a comment |
$begingroup$
Thanks.
The definition of $sigma(X,Y)$ is
$${{(omega_1,omega_2)mid (X(omega_1),Y(omega_2))in B}mid Bsubseteq Rtimes R, mesurable}$$,
and
for the case $(0,1)in B$ and others not in $B$,
$${(2,4),(2,5),(2,6),(4,4),(4,5),(4,6),(6,4),(6,5),(6,6)}$$.
I need to consider $2^4$ cases, like $(0,0)in B$, $(0,1)in B$, $(1,0)notin B$, $(1,1)in B$..., right?
answered Jan 26 at 10:55
hiratathiratat
262
$endgroup$
$begingroup$
Yes. Although it's simpler to just do the 4 cases: $1. (0,0) in B, (0,1), (1,0), (1,1)notin B $; $2. (0,1) in B, (0,0), (1,0), (1,1)notin B $; $3. (1,0) in B, (0,0), (0,1), (1,1)notin B $; and $4. (1,1) in B, (0,0), (0,1),(1,0) notin B $. The sets in $ sigma(X) $ are then just all the unions (including the empty union) of those you get from these four cases.
$endgroup$
– lonza leggiera
Jan 27 at 20:37
$begingroup$
Thank you very much for your kind answer.
$endgroup$
– hiratat
Feb 1 at 3:19
add a comment |
$begingroup$
Thanks.
The definition of $sigma(X,Y)$ is
$${{(omega_1,omega_2)mid (X(omega_1),Y(omega_2))in B}mid Bsubseteq Rtimes R, mesurable}$$,
and
for the case $(0,1)in B$ and others not in $B$,
$${(2,4),(2,5),(2,6),(4,4),(4,5),(4,6),(6,4),(6,5),(6,6)}$$.
I need to consider $2^4$ cases, like $(0,0)in B$, $(0,1)in B$, $(1,0)notin B$, $(1,1)in B$..., right?
answered Jan 26 at 10:55
hiratathiratat
262
$endgroup$
Thanks.
The definition of $sigma(X,Y)$ is
$${{(omega_1,omega_2)mid (X(omega_1),Y(omega_2))in B}mid Bsubseteq Rtimes R, mesurable}$$,
and
for the case $(0,1)in B$ and others not in $B$,
$${(2,4),(2,5),(2,6),(4,4),(4,5),(4,6),(6,4),(6,5),(6,6)}$$.
I need to consider $2^4$ cases, like $(0,0)in B$, $(0,1)in B$, $(1,0)notin B$, $(1,1)in B$..., right?
answered Jan 26 at 10:55
hiratathiratat
262
answered Jan 26 at 10:55
hiratathiratat
262
answered Jan 26 at 10:55
hiratathiratat
262
answered Jan 26 at 10:55
hiratathiratat
262
262
$begingroup$
Yes. Although it's simpler to just do the 4 cases: $1. (0,0) in B, (0,1), (1,0), (1,1)notin B $; $2. (0,1) in B, (0,0), (1,0), (1,1)notin B $; $3. (1,0) in B, (0,0), (0,1), (1,1)notin B $; and $4. (1,1) in B, (0,0), (0,1),(1,0) notin B $. The sets in $ sigma(X) $ are then just all the unions (including the empty union) of those you get from these four cases.
$endgroup$
– lonza leggiera
Jan 27 at 20:37
$begingroup$
Thank you very much for your kind answer.
$endgroup$
– hiratat
Feb 1 at 3:19
add a comment |
$begingroup$
Yes. Although it's simpler to just do the 4 cases: $1. (0,0) in B, (0,1), (1,0), (1,1)notin B $; $2. (0,1) in B, (0,0), (1,0), (1,1)notin B $; $3. (1,0) in B, (0,0), (0,1), (1,1)notin B $; and $4. (1,1) in B, (0,0), (0,1),(1,0) notin B $. The sets in $ sigma(X) $ are then just all the unions (including the empty union) of those you get from these four cases.
$endgroup$
– lonza leggiera
Jan 27 at 20:37
$begingroup$
Thank you very much for your kind answer.
$endgroup$
– hiratat
Feb 1 at 3:19
$begingroup$
Yes. Although it's simpler to just do the 4 cases: $1. (0,0) in B, (0,1), (1,0), (1,1)notin B $; $2. (0,1) in B, (0,0), (1,0), (1,1)notin B $; $3. (1,0) in B, (0,0), (0,1), (1,1)notin B $; and $4. (1,1) in B, (0,0), (0,1),(1,0) notin B $. The sets in $ sigma(X) $ are then just all the unions (including the empty union) of those you get from these four cases.
$endgroup$
– lonza leggiera
Jan 27 at 20:37
$begingroup$
Yes. Although it's simpler to just do the 4 cases: $1. (0,0) in B, (0,1), (1,0), (1,1)notin B $; $2. (0,1) in B, (0,0), (1,0), (1,1)notin B $; $3. (1,0) in B, (0,0), (0,1), (1,1)notin B $; and $4. (1,1) in B, (0,0), (0,1),(1,0) notin B $. The sets in $ sigma(X) $ are then just all the unions (including the empty union) of those you get from these four cases.
$endgroup$
– lonza leggiera
Jan 27 at 20:37
$begingroup$
Thank you very much for your kind answer.
$endgroup$
– hiratat
Feb 1 at 3:19
$begingroup$
Thank you very much for your kind answer.
$endgroup$
– hiratat
Feb 1 at 3:19
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