Why is chosen for intersection instead of union?
$begingroup$
Constructing a commutative monoid having idempotent elements (the underlying
monoid of a Boolean ring) free over a set $X$, I arrive on a very
natural way at monoid $M$ having the finite subsets of $X$ as underlying
set and equipped with multiplication $ST:=Scup T$. Any function
$f:Xrightarrowleft|Nright|$ where $left|Nright|$ denotes the
underlying set of a monoid $N$ (again commutative and having idempotent elements) induces monoidmorphism $f^{flat}:Mrightarrow N$
defined by $Smapstoprod_{sin S}fleft(sright)$.
What bothers me is that uptil now I did not encounter this in literature.
Not $cup$ is used as multiplication, but $cap$.
For instance on nlab I read under Boolean rings:
... as the free $mathbb{Z}_{2}$-vector space
$mathbb{Z}_{2}$$left[M_{n}right]$ generated from the commutative
idempotent monoid $M_{n}$ on $n$ generators. The latter can be
identified with the power set on an $n$-element set with
multiplication given by intersection, and
$mathbb{Z}_{2}$$left[M_{n}right]$ therefore has $2^{2^{n}}$
elements.
I understand that, especially if $X$ is a finite set, the monoids
$left(wpleft(Xright),cupright)$ and $left(wpleft(Xright),capright)$
are isomorphic ($Smapsto S^{c}$) but a monoid-morphism $f^{flat}:Mrightarrow N$
defined by $Smapstoprod_{sin S^{c}}fleft(sright)$ is less natural
in my eyes. If it comes to infinite sets $X$ then cofinite subsets (I prefer finite subsets) come in sight.
Is there some underlying reason for the choice for intersection here?
category-theory boolean-algebra monoid
asked May 5 '14 at 9:08
drhabdrhab
102k545136
$endgroup$
add a comment |
$begingroup$
Constructing a commutative monoid having idempotent elements (the underlying
monoid of a Boolean ring) free over a set $X$, I arrive on a very
natural way at monoid $M$ having the finite subsets of $X$ as underlying
set and equipped with multiplication $ST:=Scup T$. Any function
$f:Xrightarrowleft|Nright|$ where $left|Nright|$ denotes the
underlying set of a monoid $N$ (again commutative and having idempotent elements) induces monoidmorphism $f^{flat}:Mrightarrow N$
defined by $Smapstoprod_{sin S}fleft(sright)$.
What bothers me is that uptil now I did not encounter this in literature.
Not $cup$ is used as multiplication, but $cap$.
For instance on nlab I read under Boolean rings:
... as the free $mathbb{Z}_{2}$-vector space
$mathbb{Z}_{2}$$left[M_{n}right]$ generated from the commutative
idempotent monoid $M_{n}$ on $n$ generators. The latter can be
identified with the power set on an $n$-element set with
multiplication given by intersection, and
$mathbb{Z}_{2}$$left[M_{n}right]$ therefore has $2^{2^{n}}$
elements.
I understand that, especially if $X$ is a finite set, the monoids
$left(wpleft(Xright),cupright)$ and $left(wpleft(Xright),capright)$
are isomorphic ($Smapsto S^{c}$) but a monoid-morphism $f^{flat}:Mrightarrow N$
defined by $Smapstoprod_{sin S^{c}}fleft(sright)$ is less natural
in my eyes. If it comes to infinite sets $X$ then cofinite subsets (I prefer finite subsets) come in sight.
Is there some underlying reason for the choice for intersection here?
category-theory boolean-algebra monoid
asked May 5 '14 at 9:08
drhabdrhab
102k545136
$endgroup$
add a comment |
$begingroup$
Constructing a commutative monoid having idempotent elements (the underlying
monoid of a Boolean ring) free over a set $X$, I arrive on a very
natural way at monoid $M$ having the finite subsets of $X$ as underlying
set and equipped with multiplication $ST:=Scup T$. Any function
$f:Xrightarrowleft|Nright|$ where $left|Nright|$ denotes the
underlying set of a monoid $N$ (again commutative and having idempotent elements) induces monoidmorphism $f^{flat}:Mrightarrow N$
defined by $Smapstoprod_{sin S}fleft(sright)$.
What bothers me is that uptil now I did not encounter this in literature.
Not $cup$ is used as multiplication, but $cap$.
For instance on nlab I read under Boolean rings:
... as the free $mathbb{Z}_{2}$-vector space
$mathbb{Z}_{2}$$left[M_{n}right]$ generated from the commutative
idempotent monoid $M_{n}$ on $n$ generators. The latter can be
identified with the power set on an $n$-element set with
multiplication given by intersection, and
$mathbb{Z}_{2}$$left[M_{n}right]$ therefore has $2^{2^{n}}$
elements.
I understand that, especially if $X$ is a finite set, the monoids
$left(wpleft(Xright),cupright)$ and $left(wpleft(Xright),capright)$
are isomorphic ($Smapsto S^{c}$) but a monoid-morphism $f^{flat}:Mrightarrow N$
defined by $Smapstoprod_{sin S^{c}}fleft(sright)$ is less natural
in my eyes. If it comes to infinite sets $X$ then cofinite subsets (I prefer finite subsets) come in sight.
Is there some underlying reason for the choice for intersection here?
category-theory boolean-algebra monoid
asked May 5 '14 at 9:08
drhabdrhab
102k545136
$endgroup$
Constructing a commutative monoid having idempotent elements (the underlying
monoid of a Boolean ring) free over a set $X$, I arrive on a very
natural way at monoid $M$ having the finite subsets of $X$ as underlying
set and equipped with multiplication $ST:=Scup T$. Any function
$f:Xrightarrowleft|Nright|$ where $left|Nright|$ denotes the
underlying set of a monoid $N$ (again commutative and having idempotent elements) induces monoidmorphism $f^{flat}:Mrightarrow N$
defined by $Smapstoprod_{sin S}fleft(sright)$.
What bothers me is that uptil now I did not encounter this in literature.
Not $cup$ is used as multiplication, but $cap$.
For instance on nlab I read under Boolean rings:
... as the free $mathbb{Z}_{2}$-vector space
$mathbb{Z}_{2}$$left[M_{n}right]$ generated from the commutative
idempotent monoid $M_{n}$ on $n$ generators. The latter can be
identified with the power set on an $n$-element set with
multiplication given by intersection, and
$mathbb{Z}_{2}$$left[M_{n}right]$ therefore has $2^{2^{n}}$
elements.
I understand that, especially if $X$ is a finite set, the monoids
$left(wpleft(Xright),cupright)$ and $left(wpleft(Xright),capright)$
are isomorphic ($Smapsto S^{c}$) but a monoid-morphism $f^{flat}:Mrightarrow N$
defined by $Smapstoprod_{sin S^{c}}fleft(sright)$ is less natural
in my eyes. If it comes to infinite sets $X$ then cofinite subsets (I prefer finite subsets) come in sight.
Is there some underlying reason for the choice for intersection here?
category-theory boolean-algebra monoid
category-theory boolean-algebra monoid
asked May 5 '14 at 9:08
drhabdrhab
102k545136
asked May 5 '14 at 9:08
drhabdrhab
102k545136
edited Dec 20 '18 at 14:25
drhab
asked May 5 '14 at 9:08
drhabdrhab
102k545136
asked May 5 '14 at 9:08
drhabdrhab
102k545136
asked May 5 '14 at 9:08
drhabdrhab
102k545136
102k545136
add a comment |
add a comment |
1 Answer
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$begingroup$
As you note, the distinction disappears after switching to complements (and hence from finite to cofinite subsets). If one considers characteristic maps $f_Scolon Xto{0,1}$ with $$f(x)=begin{cases}1&text{if }xin S\0&text{if }xnotin Send{cases}$$ Then $cap$ corresponds to pointwise multiplication, i.e. $f_{Scap T}(x)=f_S(x)f_T(x)$.
answered May 5 '14 at 9:24
Hagen von EitzenHagen von Eitzen
281k23272505
$endgroup$
$begingroup$
Hmm... So the reason for preferring intersection is that $f_{Scap T}left(xright)=f_{S}left(xright)f_{T}left(xright)$ is more 'convenient' than $f_{Scup T}left(xright)=1-left(1-f_{S}left(xright)right)left(1-f_{T}left(xright)right)$?
$endgroup$
– drhab
May 5 '14 at 9:30
$begingroup$
Yes. For the same reason some write $cdot$ and $+$ instead of $land$ and $lor$
$endgroup$
– Hagen von Eitzen
May 5 '14 at 9:56
add a comment |
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1 Answer
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1 Answer
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$begingroup$
As you note, the distinction disappears after switching to complements (and hence from finite to cofinite subsets). If one considers characteristic maps $f_Scolon Xto{0,1}$ with $$f(x)=begin{cases}1&text{if }xin S\0&text{if }xnotin Send{cases}$$ Then $cap$ corresponds to pointwise multiplication, i.e. $f_{Scap T}(x)=f_S(x)f_T(x)$.
answered May 5 '14 at 9:24
Hagen von EitzenHagen von Eitzen
281k23272505
$endgroup$
$begingroup$
Hmm... So the reason for preferring intersection is that $f_{Scap T}left(xright)=f_{S}left(xright)f_{T}left(xright)$ is more 'convenient' than $f_{Scup T}left(xright)=1-left(1-f_{S}left(xright)right)left(1-f_{T}left(xright)right)$?
$endgroup$
– drhab
May 5 '14 at 9:30
$begingroup$
Yes. For the same reason some write $cdot$ and $+$ instead of $land$ and $lor$
$endgroup$
– Hagen von Eitzen
May 5 '14 at 9:56
add a comment |
$begingroup$
As you note, the distinction disappears after switching to complements (and hence from finite to cofinite subsets). If one considers characteristic maps $f_Scolon Xto{0,1}$ with $$f(x)=begin{cases}1&text{if }xin S\0&text{if }xnotin Send{cases}$$ Then $cap$ corresponds to pointwise multiplication, i.e. $f_{Scap T}(x)=f_S(x)f_T(x)$.
answered May 5 '14 at 9:24
Hagen von EitzenHagen von Eitzen
281k23272505
$endgroup$
$begingroup$
Hmm... So the reason for preferring intersection is that $f_{Scap T}left(xright)=f_{S}left(xright)f_{T}left(xright)$ is more 'convenient' than $f_{Scup T}left(xright)=1-left(1-f_{S}left(xright)right)left(1-f_{T}left(xright)right)$?
$endgroup$
– drhab
May 5 '14 at 9:30
$begingroup$
Yes. For the same reason some write $cdot$ and $+$ instead of $land$ and $lor$
$endgroup$
– Hagen von Eitzen
May 5 '14 at 9:56
add a comment |
$begingroup$
As you note, the distinction disappears after switching to complements (and hence from finite to cofinite subsets). If one considers characteristic maps $f_Scolon Xto{0,1}$ with $$f(x)=begin{cases}1&text{if }xin S\0&text{if }xnotin Send{cases}$$ Then $cap$ corresponds to pointwise multiplication, i.e. $f_{Scap T}(x)=f_S(x)f_T(x)$.
answered May 5 '14 at 9:24
Hagen von EitzenHagen von Eitzen
281k23272505
$endgroup$
As you note, the distinction disappears after switching to complements (and hence from finite to cofinite subsets). If one considers characteristic maps $f_Scolon Xto{0,1}$ with $$f(x)=begin{cases}1&text{if }xin S\0&text{if }xnotin Send{cases}$$ Then $cap$ corresponds to pointwise multiplication, i.e. $f_{Scap T}(x)=f_S(x)f_T(x)$.
answered May 5 '14 at 9:24
Hagen von EitzenHagen von Eitzen
281k23272505
answered May 5 '14 at 9:24
Hagen von EitzenHagen von Eitzen
281k23272505
answered May 5 '14 at 9:24
Hagen von EitzenHagen von Eitzen
281k23272505
answered May 5 '14 at 9:24
Hagen von EitzenHagen von Eitzen
281k23272505
281k23272505
$begingroup$
Hmm... So the reason for preferring intersection is that $f_{Scap T}left(xright)=f_{S}left(xright)f_{T}left(xright)$ is more 'convenient' than $f_{Scup T}left(xright)=1-left(1-f_{S}left(xright)right)left(1-f_{T}left(xright)right)$?
$endgroup$
– drhab
May 5 '14 at 9:30
$begingroup$
Yes. For the same reason some write $cdot$ and $+$ instead of $land$ and $lor$
$endgroup$
– Hagen von Eitzen
May 5 '14 at 9:56
add a comment |
$begingroup$
Hmm... So the reason for preferring intersection is that $f_{Scap T}left(xright)=f_{S}left(xright)f_{T}left(xright)$ is more 'convenient' than $f_{Scup T}left(xright)=1-left(1-f_{S}left(xright)right)left(1-f_{T}left(xright)right)$?
$endgroup$
– drhab
May 5 '14 at 9:30
$begingroup$
Yes. For the same reason some write $cdot$ and $+$ instead of $land$ and $lor$
$endgroup$
– Hagen von Eitzen
May 5 '14 at 9:56
$begingroup$
Hmm... So the reason for preferring intersection is that $f_{Scap T}left(xright)=f_{S}left(xright)f_{T}left(xright)$ is more 'convenient' than $f_{Scup T}left(xright)=1-left(1-f_{S}left(xright)right)left(1-f_{T}left(xright)right)$?
$endgroup$
– drhab
May 5 '14 at 9:30
$begingroup$
Hmm... So the reason for preferring intersection is that $f_{Scap T}left(xright)=f_{S}left(xright)f_{T}left(xright)$ is more 'convenient' than $f_{Scup T}left(xright)=1-left(1-f_{S}left(xright)right)left(1-f_{T}left(xright)right)$?
$endgroup$
– drhab
May 5 '14 at 9:30
$begingroup$
Yes. For the same reason some write $cdot$ and $+$ instead of $land$ and $lor$
$endgroup$
– Hagen von Eitzen
May 5 '14 at 9:56
$begingroup$
Yes. For the same reason some write $cdot$ and $+$ instead of $land$ and $lor$
$endgroup$
– Hagen von Eitzen
May 5 '14 at 9:56
add a comment |
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