What is $ {square}_{mathscr A triangledown mathscr B} $?
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I'm currently working trough a set of notes on uniform spaces. I came upon a problem that I'm having some difficulties solving, so I am seeking assistance in the form of clarifications, hints or explanations.
In the notes I'm working trough, the following are the
definitions of the terms used:
• X is an arbitrary set;
• $ mathscr A $ and $ mathscr B $ are subsets of the power set of $ X $;
• $ {square}_{mathscr A} := {bigcup}_{A in mathscr A} A×A $;
• $ mathscr A triangledown mathscr B $ := {$A cup B | A in mathscr A , B in mathscr B , A cap B ne Ø$} ;
• $ circ $ is just composition of binary relations.
I am required to prove the following equality:
$ {square}_{mathscr A triangledown mathscr B} = ({square}_{mathscr A} circ {square}_{mathscr A}) cup ({square}_{mathscr A} circ {square}_{mathscr B}) cup ({square}_{mathscr B} circ {square}_{mathscr A}) cup ({square}_{mathscr B} circ {square}_{mathscr B}) $
Thus far, I have proved the inclusion
$ {square}_{mathscr A triangledown mathscr B} subseteq ({square}_{mathscr A} circ {square}_{mathscr A}) cup ({square}_{mathscr A} circ {square}_{mathscr B}) cup ({square}_{mathscr B} circ {square}_{mathscr A}) cup ({square}_{mathscr B} circ {square}_{mathscr B}) $
However, I'm having some trouble with the other direction. For instance, what if $ forall A in mathscr A, B in mathscr B (A cap B = Ø) $ ?
It seems to me thay in this case the left hand side is empty, whereas the right hand side is not, I suspect I may be missing some detail.
elementary-set-theory relations
asked Dec 1 '18 at 23:35
ToricTorusToricTorus
204
$endgroup$
add a comment |
$begingroup$
I'm currently working trough a set of notes on uniform spaces. I came upon a problem that I'm having some difficulties solving, so I am seeking assistance in the form of clarifications, hints or explanations.
In the notes I'm working trough, the following are the
definitions of the terms used:
• X is an arbitrary set;
• $ mathscr A $ and $ mathscr B $ are subsets of the power set of $ X $;
• $ {square}_{mathscr A} := {bigcup}_{A in mathscr A} A×A $;
• $ mathscr A triangledown mathscr B $ := {$A cup B | A in mathscr A , B in mathscr B , A cap B ne Ø$} ;
• $ circ $ is just composition of binary relations.
I am required to prove the following equality:
$ {square}_{mathscr A triangledown mathscr B} = ({square}_{mathscr A} circ {square}_{mathscr A}) cup ({square}_{mathscr A} circ {square}_{mathscr B}) cup ({square}_{mathscr B} circ {square}_{mathscr A}) cup ({square}_{mathscr B} circ {square}_{mathscr B}) $
Thus far, I have proved the inclusion
$ {square}_{mathscr A triangledown mathscr B} subseteq ({square}_{mathscr A} circ {square}_{mathscr A}) cup ({square}_{mathscr A} circ {square}_{mathscr B}) cup ({square}_{mathscr B} circ {square}_{mathscr A}) cup ({square}_{mathscr B} circ {square}_{mathscr B}) $
However, I'm having some trouble with the other direction. For instance, what if $ forall A in mathscr A, B in mathscr B (A cap B = Ø) $ ?
It seems to me thay in this case the left hand side is empty, whereas the right hand side is not, I suspect I may be missing some detail.
elementary-set-theory relations
asked Dec 1 '18 at 23:35
ToricTorusToricTorus
204
$endgroup$
$begingroup$
It's a notational mess. How can an element b, of A subset P(X×X) be a binary operation? If a,b are binary operations, what is their composition?
$endgroup$
– William Elliot
Dec 2 '18 at 4:34
$begingroup$
Not binary operations, binary relations.
$endgroup$
– Henno Brandsma
Dec 2 '18 at 9:23
$begingroup$
@WilliamElliot composition of relations, and it's defined as usual: if $R,S subseteq X times X$, then $R circ S = {(x,z) in X times X: exists y in X: (x,y) in R, (y,z) in S}$.
$endgroup$
– Henno Brandsma
Dec 2 '18 at 9:25
1
$begingroup$
The definition $Box_{mathscr{A}}$ is weird: $A in mathscr{A}$ means $A subseteq X times X$ and then $A times A$ is in $X^4$?
$endgroup$
– Henno Brandsma
Dec 2 '18 at 9:34
$begingroup$
Thanks for pointing that out, I had made some mistakes and typos in some of the definitions, I typed it up on my mobile phone and simply didn't notice. It should be correct now.
$endgroup$
– ToricTorus
Dec 2 '18 at 17:13
add a comment |
$begingroup$
I'm currently working trough a set of notes on uniform spaces. I came upon a problem that I'm having some difficulties solving, so I am seeking assistance in the form of clarifications, hints or explanations.
In the notes I'm working trough, the following are the
definitions of the terms used:
• X is an arbitrary set;
• $ mathscr A $ and $ mathscr B $ are subsets of the power set of $ X $;
• $ {square}_{mathscr A} := {bigcup}_{A in mathscr A} A×A $;
• $ mathscr A triangledown mathscr B $ := {$A cup B | A in mathscr A , B in mathscr B , A cap B ne Ø$} ;
• $ circ $ is just composition of binary relations.
I am required to prove the following equality:
$ {square}_{mathscr A triangledown mathscr B} = ({square}_{mathscr A} circ {square}_{mathscr A}) cup ({square}_{mathscr A} circ {square}_{mathscr B}) cup ({square}_{mathscr B} circ {square}_{mathscr A}) cup ({square}_{mathscr B} circ {square}_{mathscr B}) $
Thus far, I have proved the inclusion
$ {square}_{mathscr A triangledown mathscr B} subseteq ({square}_{mathscr A} circ {square}_{mathscr A}) cup ({square}_{mathscr A} circ {square}_{mathscr B}) cup ({square}_{mathscr B} circ {square}_{mathscr A}) cup ({square}_{mathscr B} circ {square}_{mathscr B}) $
However, I'm having some trouble with the other direction. For instance, what if $ forall A in mathscr A, B in mathscr B (A cap B = Ø) $ ?
It seems to me thay in this case the left hand side is empty, whereas the right hand side is not, I suspect I may be missing some detail.
elementary-set-theory relations
asked Dec 1 '18 at 23:35
ToricTorusToricTorus
204
$endgroup$
I'm currently working trough a set of notes on uniform spaces. I came upon a problem that I'm having some difficulties solving, so I am seeking assistance in the form of clarifications, hints or explanations.
In the notes I'm working trough, the following are the
definitions of the terms used:
• X is an arbitrary set;
• $ mathscr A $ and $ mathscr B $ are subsets of the power set of $ X $;
• $ {square}_{mathscr A} := {bigcup}_{A in mathscr A} A×A $;
• $ mathscr A triangledown mathscr B $ := {$A cup B | A in mathscr A , B in mathscr B , A cap B ne Ø$} ;
• $ circ $ is just composition of binary relations.
I am required to prove the following equality:
$ {square}_{mathscr A triangledown mathscr B} = ({square}_{mathscr A} circ {square}_{mathscr A}) cup ({square}_{mathscr A} circ {square}_{mathscr B}) cup ({square}_{mathscr B} circ {square}_{mathscr A}) cup ({square}_{mathscr B} circ {square}_{mathscr B}) $
Thus far, I have proved the inclusion
$ {square}_{mathscr A triangledown mathscr B} subseteq ({square}_{mathscr A} circ {square}_{mathscr A}) cup ({square}_{mathscr A} circ {square}_{mathscr B}) cup ({square}_{mathscr B} circ {square}_{mathscr A}) cup ({square}_{mathscr B} circ {square}_{mathscr B}) $
However, I'm having some trouble with the other direction. For instance, what if $ forall A in mathscr A, B in mathscr B (A cap B = Ø) $ ?
It seems to me thay in this case the left hand side is empty, whereas the right hand side is not, I suspect I may be missing some detail.
elementary-set-theory relations
elementary-set-theory relations
asked Dec 1 '18 at 23:35
ToricTorusToricTorus
204
asked Dec 1 '18 at 23:35
ToricTorusToricTorus
204
edited Dec 2 '18 at 17:05
ToricTorus
asked Dec 1 '18 at 23:35
ToricTorusToricTorus
204
asked Dec 1 '18 at 23:35
ToricTorusToricTorus
204
asked Dec 1 '18 at 23:35
ToricTorusToricTorus
204
204
$begingroup$
It's a notational mess. How can an element b, of A subset P(X×X) be a binary operation? If a,b are binary operations, what is their composition?
$endgroup$
– William Elliot
Dec 2 '18 at 4:34
$begingroup$
Not binary operations, binary relations.
$endgroup$
– Henno Brandsma
Dec 2 '18 at 9:23
$begingroup$
@WilliamElliot composition of relations, and it's defined as usual: if $R,S subseteq X times X$, then $R circ S = {(x,z) in X times X: exists y in X: (x,y) in R, (y,z) in S}$.
$endgroup$
– Henno Brandsma
Dec 2 '18 at 9:25
1
$begingroup$
The definition $Box_{mathscr{A}}$ is weird: $A in mathscr{A}$ means $A subseteq X times X$ and then $A times A$ is in $X^4$?
$endgroup$
– Henno Brandsma
Dec 2 '18 at 9:34
$begingroup$
Thanks for pointing that out, I had made some mistakes and typos in some of the definitions, I typed it up on my mobile phone and simply didn't notice. It should be correct now.
$endgroup$
– ToricTorus
Dec 2 '18 at 17:13
add a comment |
$begingroup$
It's a notational mess. How can an element b, of A subset P(X×X) be a binary operation? If a,b are binary operations, what is their composition?
$endgroup$
– William Elliot
Dec 2 '18 at 4:34
$begingroup$
Not binary operations, binary relations.
$endgroup$
– Henno Brandsma
Dec 2 '18 at 9:23
$begingroup$
@WilliamElliot composition of relations, and it's defined as usual: if $R,S subseteq X times X$, then $R circ S = {(x,z) in X times X: exists y in X: (x,y) in R, (y,z) in S}$.
$endgroup$
– Henno Brandsma
Dec 2 '18 at 9:25
1
$begingroup$
The definition $Box_{mathscr{A}}$ is weird: $A in mathscr{A}$ means $A subseteq X times X$ and then $A times A$ is in $X^4$?
$endgroup$
– Henno Brandsma
Dec 2 '18 at 9:34
$begingroup$
Thanks for pointing that out, I had made some mistakes and typos in some of the definitions, I typed it up on my mobile phone and simply didn't notice. It should be correct now.
$endgroup$
– ToricTorus
Dec 2 '18 at 17:13
$begingroup$
It's a notational mess. How can an element b, of A subset P(X×X) be a binary operation? If a,b are binary operations, what is their composition?
$endgroup$
– William Elliot
Dec 2 '18 at 4:34
$begingroup$
It's a notational mess. How can an element b, of A subset P(X×X) be a binary operation? If a,b are binary operations, what is their composition?
$endgroup$
– William Elliot
Dec 2 '18 at 4:34
$begingroup$
Not binary operations, binary relations.
$endgroup$
– Henno Brandsma
Dec 2 '18 at 9:23
$begingroup$
Not binary operations, binary relations.
$endgroup$
– Henno Brandsma
Dec 2 '18 at 9:23
$begingroup$
@WilliamElliot composition of relations, and it's defined as usual: if $R,S subseteq X times X$, then $R circ S = {(x,z) in X times X: exists y in X: (x,y) in R, (y,z) in S}$.
$endgroup$
– Henno Brandsma
Dec 2 '18 at 9:25
$begingroup$
@WilliamElliot composition of relations, and it's defined as usual: if $R,S subseteq X times X$, then $R circ S = {(x,z) in X times X: exists y in X: (x,y) in R, (y,z) in S}$.
$endgroup$
– Henno Brandsma
Dec 2 '18 at 9:25
1
1
$begingroup$
The definition $Box_{mathscr{A}}$ is weird: $A in mathscr{A}$ means $A subseteq X times X$ and then $A times A$ is in $X^4$?
$endgroup$
– Henno Brandsma
Dec 2 '18 at 9:34
$begingroup$
The definition $Box_{mathscr{A}}$ is weird: $A in mathscr{A}$ means $A subseteq X times X$ and then $A times A$ is in $X^4$?
$endgroup$
– Henno Brandsma
Dec 2 '18 at 9:34
$begingroup$
Thanks for pointing that out, I had made some mistakes and typos in some of the definitions, I typed it up on my mobile phone and simply didn't notice. It should be correct now.
$endgroup$
– ToricTorus
Dec 2 '18 at 17:13
$begingroup$
Thanks for pointing that out, I had made some mistakes and typos in some of the definitions, I typed it up on my mobile phone and simply didn't notice. It should be correct now.
$endgroup$
– ToricTorus
Dec 2 '18 at 17:13
add a comment |
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$begingroup$
It's a notational mess. How can an element b, of A subset P(X×X) be a binary operation? If a,b are binary operations, what is their composition?
$endgroup$
– William Elliot
Dec 2 '18 at 4:34
$begingroup$
Not binary operations, binary relations.
$endgroup$
– Henno Brandsma
Dec 2 '18 at 9:23
$begingroup$
@WilliamElliot composition of relations, and it's defined as usual: if $R,S subseteq X times X$, then $R circ S = {(x,z) in X times X: exists y in X: (x,y) in R, (y,z) in S}$.
$endgroup$
– Henno Brandsma
Dec 2 '18 at 9:25
1
$begingroup$
The definition $Box_{mathscr{A}}$ is weird: $A in mathscr{A}$ means $A subseteq X times X$ and then $A times A$ is in $X^4$?
$endgroup$
– Henno Brandsma
Dec 2 '18 at 9:34
$begingroup$
Thanks for pointing that out, I had made some mistakes and typos in some of the definitions, I typed it up on my mobile phone and simply didn't notice. It should be correct now.
$endgroup$
– ToricTorus
Dec 2 '18 at 17:13