Example of semi-algebras that are not algebras
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I know that every algebra is as semi-algebra, and the book (A course in Real Analysis, McDonald and Weiss) tells me that the opposite is not true: not every semi-algebra is an algebra. Why not?
A semi-algebra contains all finite intersections of its members. This satisfies the condition that an algebra must contain all finite intersection (or unions).
So I think the problem is with the complement-condition:
- If the complement to a member in the semi-algebra is $emptyset$,
then this satisfies some of the condition to be an algebra (that
every complement must be in the algebra). - So I guess that something goes wrong with the complement being a
finite union of pairwise disjoint sets? But I don´t know why?
Do anyone have an example of semi-algebras that are not algebras?
measure-theory
asked Oct 8 '14 at 8:24
user7919user7919
255
$endgroup$
add a comment |
$begingroup$
I know that every algebra is as semi-algebra, and the book (A course in Real Analysis, McDonald and Weiss) tells me that the opposite is not true: not every semi-algebra is an algebra. Why not?
A semi-algebra contains all finite intersections of its members. This satisfies the condition that an algebra must contain all finite intersection (or unions).
So I think the problem is with the complement-condition:
- If the complement to a member in the semi-algebra is $emptyset$,
then this satisfies some of the condition to be an algebra (that
every complement must be in the algebra). - So I guess that something goes wrong with the complement being a
finite union of pairwise disjoint sets? But I don´t know why?
Do anyone have an example of semi-algebras that are not algebras?
measure-theory
asked Oct 8 '14 at 8:24
user7919user7919
255
$endgroup$
$begingroup$
For discussion and examples, see also physicsforums.com/threads/definition-of-a-semialgebra.334682.
$endgroup$
– Dietrich Burde
Oct 8 '14 at 8:31
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statement "every semi-algebra is not an algebra" is not true, and "not every semi-algebra is an algebra" is a true statement.
$endgroup$
– drhab
Oct 8 '14 at 8:55
add a comment |
$begingroup$
I know that every algebra is as semi-algebra, and the book (A course in Real Analysis, McDonald and Weiss) tells me that the opposite is not true: not every semi-algebra is an algebra. Why not?
A semi-algebra contains all finite intersections of its members. This satisfies the condition that an algebra must contain all finite intersection (or unions).
So I think the problem is with the complement-condition:
- If the complement to a member in the semi-algebra is $emptyset$,
then this satisfies some of the condition to be an algebra (that
every complement must be in the algebra). - So I guess that something goes wrong with the complement being a
finite union of pairwise disjoint sets? But I don´t know why?
Do anyone have an example of semi-algebras that are not algebras?
measure-theory
asked Oct 8 '14 at 8:24
user7919user7919
255
$endgroup$
I know that every algebra is as semi-algebra, and the book (A course in Real Analysis, McDonald and Weiss) tells me that the opposite is not true: not every semi-algebra is an algebra. Why not?
A semi-algebra contains all finite intersections of its members. This satisfies the condition that an algebra must contain all finite intersection (or unions).
So I think the problem is with the complement-condition:
- If the complement to a member in the semi-algebra is $emptyset$,
then this satisfies some of the condition to be an algebra (that
every complement must be in the algebra). - So I guess that something goes wrong with the complement being a
finite union of pairwise disjoint sets? But I don´t know why?
Do anyone have an example of semi-algebras that are not algebras?
measure-theory
measure-theory
asked Oct 8 '14 at 8:24
user7919user7919
255
asked Oct 8 '14 at 8:24
user7919user7919
255
edited Oct 8 '14 at 8:57
user7919
asked Oct 8 '14 at 8:24
user7919user7919
255
asked Oct 8 '14 at 8:24
user7919user7919
255
asked Oct 8 '14 at 8:24
user7919user7919
255
255
$begingroup$
For discussion and examples, see also physicsforums.com/threads/definition-of-a-semialgebra.334682.
$endgroup$
– Dietrich Burde
Oct 8 '14 at 8:31
$begingroup$
statement "every semi-algebra is not an algebra" is not true, and "not every semi-algebra is an algebra" is a true statement.
$endgroup$
– drhab
Oct 8 '14 at 8:55
add a comment |
$begingroup$
For discussion and examples, see also physicsforums.com/threads/definition-of-a-semialgebra.334682.
$endgroup$
– Dietrich Burde
Oct 8 '14 at 8:31
$begingroup$
statement "every semi-algebra is not an algebra" is not true, and "not every semi-algebra is an algebra" is a true statement.
$endgroup$
– drhab
Oct 8 '14 at 8:55
$begingroup$
For discussion and examples, see also physicsforums.com/threads/definition-of-a-semialgebra.334682.
$endgroup$
– Dietrich Burde
Oct 8 '14 at 8:31
$begingroup$
For discussion and examples, see also physicsforums.com/threads/definition-of-a-semialgebra.334682.
$endgroup$
– Dietrich Burde
Oct 8 '14 at 8:31
$begingroup$
statement "every semi-algebra is not an algebra" is not true, and "not every semi-algebra is an algebra" is a true statement.
$endgroup$
– drhab
Oct 8 '14 at 8:55
$begingroup$
statement "every semi-algebra is not an algebra" is not true, and "not every semi-algebra is an algebra" is a true statement.
$endgroup$
– drhab
Oct 8 '14 at 8:55
add a comment |
2 Answers
2
active
oldest
votes
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The collection of half-open intervals $[a,b)$ in $mathbb{R}$ is a semialgebra, but not an algebra, since the complement of $[a,b)$ is $(infty,a)cup [b,infty)$, which is not a half open interval.
Edit: In order for what I said to be a semialgebra, you need to allow $a,b$ to be $infty$, and $[infty,b)$ is interpreted to be $(infty, b)$.
answered Oct 8 '14 at 8:42
Mike EarnestMike Earnest
21.8k12051
$endgroup$
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Hmm, I started to think about this. Why is the complement in the semi-algebra then?
$endgroup$
– user7919
Oct 8 '14 at 8:53
$begingroup$
The complement is not in the semi-algebra. A semi-algebra is not closed under complements, but if $S$ in the semi-algebra, then $S^c$ is a finite disjoint union of elements in the semi-algebra. The half open intervals work because the complement of $[a,b)$ is the disjoint union of $(infty,a)$ and $[b,infty)$, which are both in the semi-algebra (I've edited my post to clarify you need to allow $a,b$ to be $infty$ for this to works).
$endgroup$
– Mike Earnest
Oct 8 '14 at 17:55
add a comment |
$begingroup$
Let $Omega={a,b,c}$ and $mathcal{C}={Omega,phi,{a},{b},{c}}$ then $mathcal{C}$ is a semi-algebra but NOT an algebra.
answered Dec 9 '18 at 3:31
GMaGMa
2111
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add a comment |
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2 Answers
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2 Answers
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active
oldest
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votes
$begingroup$
The collection of half-open intervals $[a,b)$ in $mathbb{R}$ is a semialgebra, but not an algebra, since the complement of $[a,b)$ is $(infty,a)cup [b,infty)$, which is not a half open interval.
Edit: In order for what I said to be a semialgebra, you need to allow $a,b$ to be $infty$, and $[infty,b)$ is interpreted to be $(infty, b)$.
answered Oct 8 '14 at 8:42
Mike EarnestMike Earnest
21.8k12051
$endgroup$
$begingroup$
Hmm, I started to think about this. Why is the complement in the semi-algebra then?
$endgroup$
– user7919
Oct 8 '14 at 8:53
$begingroup$
The complement is not in the semi-algebra. A semi-algebra is not closed under complements, but if $S$ in the semi-algebra, then $S^c$ is a finite disjoint union of elements in the semi-algebra. The half open intervals work because the complement of $[a,b)$ is the disjoint union of $(infty,a)$ and $[b,infty)$, which are both in the semi-algebra (I've edited my post to clarify you need to allow $a,b$ to be $infty$ for this to works).
$endgroup$
– Mike Earnest
Oct 8 '14 at 17:55
add a comment |
$begingroup$
The collection of half-open intervals $[a,b)$ in $mathbb{R}$ is a semialgebra, but not an algebra, since the complement of $[a,b)$ is $(infty,a)cup [b,infty)$, which is not a half open interval.
Edit: In order for what I said to be a semialgebra, you need to allow $a,b$ to be $infty$, and $[infty,b)$ is interpreted to be $(infty, b)$.
answered Oct 8 '14 at 8:42
Mike EarnestMike Earnest
21.8k12051
$endgroup$
$begingroup$
Hmm, I started to think about this. Why is the complement in the semi-algebra then?
$endgroup$
– user7919
Oct 8 '14 at 8:53
$begingroup$
The complement is not in the semi-algebra. A semi-algebra is not closed under complements, but if $S$ in the semi-algebra, then $S^c$ is a finite disjoint union of elements in the semi-algebra. The half open intervals work because the complement of $[a,b)$ is the disjoint union of $(infty,a)$ and $[b,infty)$, which are both in the semi-algebra (I've edited my post to clarify you need to allow $a,b$ to be $infty$ for this to works).
$endgroup$
– Mike Earnest
Oct 8 '14 at 17:55
add a comment |
$begingroup$
The collection of half-open intervals $[a,b)$ in $mathbb{R}$ is a semialgebra, but not an algebra, since the complement of $[a,b)$ is $(infty,a)cup [b,infty)$, which is not a half open interval.
Edit: In order for what I said to be a semialgebra, you need to allow $a,b$ to be $infty$, and $[infty,b)$ is interpreted to be $(infty, b)$.
answered Oct 8 '14 at 8:42
Mike EarnestMike Earnest
21.8k12051
$endgroup$
The collection of half-open intervals $[a,b)$ in $mathbb{R}$ is a semialgebra, but not an algebra, since the complement of $[a,b)$ is $(infty,a)cup [b,infty)$, which is not a half open interval.
Edit: In order for what I said to be a semialgebra, you need to allow $a,b$ to be $infty$, and $[infty,b)$ is interpreted to be $(infty, b)$.
answered Oct 8 '14 at 8:42
Mike EarnestMike Earnest
21.8k12051
edited Oct 8 '14 at 17:53
answered Oct 8 '14 at 8:42
Mike EarnestMike Earnest
21.8k12051
answered Oct 8 '14 at 8:42
Mike EarnestMike Earnest
21.8k12051
answered Oct 8 '14 at 8:42
Mike EarnestMike Earnest
21.8k12051
21.8k12051
$begingroup$
Hmm, I started to think about this. Why is the complement in the semi-algebra then?
$endgroup$
– user7919
Oct 8 '14 at 8:53
$begingroup$
The complement is not in the semi-algebra. A semi-algebra is not closed under complements, but if $S$ in the semi-algebra, then $S^c$ is a finite disjoint union of elements in the semi-algebra. The half open intervals work because the complement of $[a,b)$ is the disjoint union of $(infty,a)$ and $[b,infty)$, which are both in the semi-algebra (I've edited my post to clarify you need to allow $a,b$ to be $infty$ for this to works).
$endgroup$
– Mike Earnest
Oct 8 '14 at 17:55
add a comment |
$begingroup$
Hmm, I started to think about this. Why is the complement in the semi-algebra then?
$endgroup$
– user7919
Oct 8 '14 at 8:53
$begingroup$
The complement is not in the semi-algebra. A semi-algebra is not closed under complements, but if $S$ in the semi-algebra, then $S^c$ is a finite disjoint union of elements in the semi-algebra. The half open intervals work because the complement of $[a,b)$ is the disjoint union of $(infty,a)$ and $[b,infty)$, which are both in the semi-algebra (I've edited my post to clarify you need to allow $a,b$ to be $infty$ for this to works).
$endgroup$
– Mike Earnest
Oct 8 '14 at 17:55
$begingroup$
Hmm, I started to think about this. Why is the complement in the semi-algebra then?
$endgroup$
– user7919
Oct 8 '14 at 8:53
$begingroup$
Hmm, I started to think about this. Why is the complement in the semi-algebra then?
$endgroup$
– user7919
Oct 8 '14 at 8:53
$begingroup$
The complement is not in the semi-algebra. A semi-algebra is not closed under complements, but if $S$ in the semi-algebra, then $S^c$ is a finite disjoint union of elements in the semi-algebra. The half open intervals work because the complement of $[a,b)$ is the disjoint union of $(infty,a)$ and $[b,infty)$, which are both in the semi-algebra (I've edited my post to clarify you need to allow $a,b$ to be $infty$ for this to works).
$endgroup$
– Mike Earnest
Oct 8 '14 at 17:55
$begingroup$
The complement is not in the semi-algebra. A semi-algebra is not closed under complements, but if $S$ in the semi-algebra, then $S^c$ is a finite disjoint union of elements in the semi-algebra. The half open intervals work because the complement of $[a,b)$ is the disjoint union of $(infty,a)$ and $[b,infty)$, which are both in the semi-algebra (I've edited my post to clarify you need to allow $a,b$ to be $infty$ for this to works).
$endgroup$
– Mike Earnest
Oct 8 '14 at 17:55
add a comment |
$begingroup$
Let $Omega={a,b,c}$ and $mathcal{C}={Omega,phi,{a},{b},{c}}$ then $mathcal{C}$ is a semi-algebra but NOT an algebra.
answered Dec 9 '18 at 3:31
GMaGMa
2111
$endgroup$
add a comment |
$begingroup$
Let $Omega={a,b,c}$ and $mathcal{C}={Omega,phi,{a},{b},{c}}$ then $mathcal{C}$ is a semi-algebra but NOT an algebra.
answered Dec 9 '18 at 3:31
GMaGMa
2111
$endgroup$
add a comment |
$begingroup$
Let $Omega={a,b,c}$ and $mathcal{C}={Omega,phi,{a},{b},{c}}$ then $mathcal{C}$ is a semi-algebra but NOT an algebra.
answered Dec 9 '18 at 3:31
GMaGMa
2111
$endgroup$
Let $Omega={a,b,c}$ and $mathcal{C}={Omega,phi,{a},{b},{c}}$ then $mathcal{C}$ is a semi-algebra but NOT an algebra.
answered Dec 9 '18 at 3:31
GMaGMa
2111
answered Dec 9 '18 at 3:31
GMaGMa
2111
answered Dec 9 '18 at 3:31
GMaGMa
2111
answered Dec 9 '18 at 3:31
GMaGMa
2111
2111
add a comment |
add a comment |
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$begingroup$
For discussion and examples, see also physicsforums.com/threads/definition-of-a-semialgebra.334682.
$endgroup$
– Dietrich Burde
Oct 8 '14 at 8:31
$begingroup$
statement "every semi-algebra is not an algebra" is not true, and "not every semi-algebra is an algebra" is a true statement.
$endgroup$
– drhab
Oct 8 '14 at 8:55