Isn't a semialgebra an algebra?
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I was solving this exercise in a book, "A First Look at Rigorous Probability Theory", by Jeffrey Rosenthal.
Exercise 2.7.3. Suppose $mathcal{F}$ is a
collection of subsets of $Omega$, such that
$Omega in mathcal{F}$.
b) Assume $mathcal{F}$ is a semialgebra.
Prove that $mathcal{F}$ is an algebra.
However, it seems to me this exercise is wrong.
A counterexample suffices to show it.
Suppose a set $mathcal{J}$ of all intervals in $[0,1]$. Then, it is easy
to show that $mathcal{J}$ is a semi-algebra. Suppose two different
intervals $A = [0,frac{1}{3}]$ and $B = [frac{2}{3},1]$ are present in
$mathcal{J}$. $A cup B$ does not belong to $mathcal{J}$ because it is not an interval. $mathcal{J}$ is not an algebra because
it is not closed under finite unions.
Am I missing something?
measure-theory examples-counterexamples
asked Dec 24 '18 at 22:33
DunhoClarkDunhoClark
505
$endgroup$
|
show 3 more comments
$begingroup$
I was solving this exercise in a book, "A First Look at Rigorous Probability Theory", by Jeffrey Rosenthal.
Exercise 2.7.3. Suppose $mathcal{F}$ is a
collection of subsets of $Omega$, such that
$Omega in mathcal{F}$.
b) Assume $mathcal{F}$ is a semialgebra.
Prove that $mathcal{F}$ is an algebra.
However, it seems to me this exercise is wrong.
A counterexample suffices to show it.
Suppose a set $mathcal{J}$ of all intervals in $[0,1]$. Then, it is easy
to show that $mathcal{J}$ is a semi-algebra. Suppose two different
intervals $A = [0,frac{1}{3}]$ and $B = [frac{2}{3},1]$ are present in
$mathcal{J}$. $A cup B$ does not belong to $mathcal{J}$ because it is not an interval. $mathcal{J}$ is not an algebra because
it is not closed under finite unions.
Am I missing something?
measure-theory examples-counterexamples
asked Dec 24 '18 at 22:33
DunhoClarkDunhoClark
505
$endgroup$
$begingroup$
Which book are you referring to? It might be helpful for others to know. Please include the details in an edit.
$endgroup$
– Shaun
Dec 24 '18 at 22:44
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What is your definition of "semialgebra"?
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– Eric Wofsey
Dec 24 '18 at 22:46
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@EricWofsey algebra would be the same, but accepting it is closed under complement and under finite union.
$endgroup$
– DunhoClark
Dec 24 '18 at 22:50
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That's not the standard definition of semi-algebra. en.wikibooks.org/wiki/Measure_Theory/…
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– Joel Pereira
Dec 24 '18 at 22:58
$begingroup$
@JoelPereira I made a mistake, the book uses the same definition. But the question remains.
$endgroup$
– DunhoClark
Dec 24 '18 at 22:59
|
show 3 more comments
$begingroup$
I was solving this exercise in a book, "A First Look at Rigorous Probability Theory", by Jeffrey Rosenthal.
Exercise 2.7.3. Suppose $mathcal{F}$ is a
collection of subsets of $Omega$, such that
$Omega in mathcal{F}$.
b) Assume $mathcal{F}$ is a semialgebra.
Prove that $mathcal{F}$ is an algebra.
However, it seems to me this exercise is wrong.
A counterexample suffices to show it.
Suppose a set $mathcal{J}$ of all intervals in $[0,1]$. Then, it is easy
to show that $mathcal{J}$ is a semi-algebra. Suppose two different
intervals $A = [0,frac{1}{3}]$ and $B = [frac{2}{3},1]$ are present in
$mathcal{J}$. $A cup B$ does not belong to $mathcal{J}$ because it is not an interval. $mathcal{J}$ is not an algebra because
it is not closed under finite unions.
Am I missing something?
measure-theory examples-counterexamples
asked Dec 24 '18 at 22:33
DunhoClarkDunhoClark
505
$endgroup$
I was solving this exercise in a book, "A First Look at Rigorous Probability Theory", by Jeffrey Rosenthal.
Exercise 2.7.3. Suppose $mathcal{F}$ is a
collection of subsets of $Omega$, such that
$Omega in mathcal{F}$.
b) Assume $mathcal{F}$ is a semialgebra.
Prove that $mathcal{F}$ is an algebra.
However, it seems to me this exercise is wrong.
A counterexample suffices to show it.
Suppose a set $mathcal{J}$ of all intervals in $[0,1]$. Then, it is easy
to show that $mathcal{J}$ is a semi-algebra. Suppose two different
intervals $A = [0,frac{1}{3}]$ and $B = [frac{2}{3},1]$ are present in
$mathcal{J}$. $A cup B$ does not belong to $mathcal{J}$ because it is not an interval. $mathcal{J}$ is not an algebra because
it is not closed under finite unions.
Am I missing something?
measure-theory examples-counterexamples
measure-theory examples-counterexamples
asked Dec 24 '18 at 22:33
DunhoClarkDunhoClark
505
asked Dec 24 '18 at 22:33
DunhoClarkDunhoClark
505
edited Dec 24 '18 at 22:48
DunhoClark
asked Dec 24 '18 at 22:33
DunhoClarkDunhoClark
505
asked Dec 24 '18 at 22:33
DunhoClarkDunhoClark
505
asked Dec 24 '18 at 22:33
DunhoClarkDunhoClark
505
505
$begingroup$
Which book are you referring to? It might be helpful for others to know. Please include the details in an edit.
$endgroup$
– Shaun
Dec 24 '18 at 22:44
$begingroup$
What is your definition of "semialgebra"?
$endgroup$
– Eric Wofsey
Dec 24 '18 at 22:46
$begingroup$
@EricWofsey algebra would be the same, but accepting it is closed under complement and under finite union.
$endgroup$
– DunhoClark
Dec 24 '18 at 22:50
$begingroup$
That's not the standard definition of semi-algebra. en.wikibooks.org/wiki/Measure_Theory/…
$endgroup$
– Joel Pereira
Dec 24 '18 at 22:58
$begingroup$
@JoelPereira I made a mistake, the book uses the same definition. But the question remains.
$endgroup$
– DunhoClark
Dec 24 '18 at 22:59
|
show 3 more comments
$begingroup$
Which book are you referring to? It might be helpful for others to know. Please include the details in an edit.
$endgroup$
– Shaun
Dec 24 '18 at 22:44
$begingroup$
What is your definition of "semialgebra"?
$endgroup$
– Eric Wofsey
Dec 24 '18 at 22:46
$begingroup$
@EricWofsey algebra would be the same, but accepting it is closed under complement and under finite union.
$endgroup$
– DunhoClark
Dec 24 '18 at 22:50
$begingroup$
That's not the standard definition of semi-algebra. en.wikibooks.org/wiki/Measure_Theory/…
$endgroup$
– Joel Pereira
Dec 24 '18 at 22:58
$begingroup$
@JoelPereira I made a mistake, the book uses the same definition. But the question remains.
$endgroup$
– DunhoClark
Dec 24 '18 at 22:59
$begingroup$
Which book are you referring to? It might be helpful for others to know. Please include the details in an edit.
$endgroup$
– Shaun
Dec 24 '18 at 22:44
$begingroup$
Which book are you referring to? It might be helpful for others to know. Please include the details in an edit.
$endgroup$
– Shaun
Dec 24 '18 at 22:44
$begingroup$
What is your definition of "semialgebra"?
$endgroup$
– Eric Wofsey
Dec 24 '18 at 22:46
$begingroup$
What is your definition of "semialgebra"?
$endgroup$
– Eric Wofsey
Dec 24 '18 at 22:46
$begingroup$
@EricWofsey algebra would be the same, but accepting it is closed under complement and under finite union.
$endgroup$
– DunhoClark
Dec 24 '18 at 22:50
$begingroup$
@EricWofsey algebra would be the same, but accepting it is closed under complement and under finite union.
$endgroup$
– DunhoClark
Dec 24 '18 at 22:50
$begingroup$
That's not the standard definition of semi-algebra. en.wikibooks.org/wiki/Measure_Theory/…
$endgroup$
– Joel Pereira
Dec 24 '18 at 22:58
$begingroup$
That's not the standard definition of semi-algebra. en.wikibooks.org/wiki/Measure_Theory/…
$endgroup$
– Joel Pereira
Dec 24 '18 at 22:58
$begingroup$
@JoelPereira I made a mistake, the book uses the same definition. But the question remains.
$endgroup$
– DunhoClark
Dec 24 '18 at 22:59
$begingroup$
@JoelPereira I made a mistake, the book uses the same definition. But the question remains.
$endgroup$
– DunhoClark
Dec 24 '18 at 22:59
|
show 3 more comments
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$begingroup$
Which book are you referring to? It might be helpful for others to know. Please include the details in an edit.
$endgroup$
– Shaun
Dec 24 '18 at 22:44
$begingroup$
What is your definition of "semialgebra"?
$endgroup$
– Eric Wofsey
Dec 24 '18 at 22:46
$begingroup$
@EricWofsey algebra would be the same, but accepting it is closed under complement and under finite union.
$endgroup$
– DunhoClark
Dec 24 '18 at 22:50
$begingroup$
That's not the standard definition of semi-algebra. en.wikibooks.org/wiki/Measure_Theory/…
$endgroup$
– Joel Pereira
Dec 24 '18 at 22:58
$begingroup$
@JoelPereira I made a mistake, the book uses the same definition. But the question remains.
$endgroup$
– DunhoClark
Dec 24 '18 at 22:59