Comparing two binary relations using Cartesian products
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Let $f$ be a binary relation between sets $A_0$ and $B_0$ and $g$ be a binary relation between sets $A_1$ and $B_1$.
It can be proved that $[fcap(A_1times B_1)=g]wedge [gcap(A_0times B_0)=f]$ implies $f=g$.
Really, it follows $gcap(A_0times B_0)cap(A_1times B_1)=g$. Thus $gsubseteq A_0times B_0$. Consequantly $g=f$.
Do the following hold?
$[fcap(A_1times B_1)supseteq g]wedge [gcap(A_0times B_0)]supseteq f$ implies $f=g$.
$[fcap(A_1times B_1)subseteq g]wedge [gcap(A_0times B_0)]subseteq f$ implies $f=g$.
elementary-set-theory relations products
edited Nov 22 at 18:07
Yanko
5,660723
asked Nov 22 at 17:44
porton
1,87911127
|
show 2 more comments
up vote
0
down vote
favorite
Let $f$ be a binary relation between sets $A_0$ and $B_0$ and $g$ be a binary relation between sets $A_1$ and $B_1$.
It can be proved that $[fcap(A_1times B_1)=g]wedge [gcap(A_0times B_0)=f]$ implies $f=g$.
Really, it follows $gcap(A_0times B_0)cap(A_1times B_1)=g$. Thus $gsubseteq A_0times B_0$. Consequantly $g=f$.
Do the following hold?
$[fcap(A_1times B_1)supseteq g]wedge [gcap(A_0times B_0)]supseteq f$ implies $f=g$.
$[fcap(A_1times B_1)subseteq g]wedge [gcap(A_0times B_0)]subseteq f$ implies $f=g$.
elementary-set-theory relations products
edited Nov 22 at 18:07
Yanko
5,660723
asked Nov 22 at 17:44
porton
1,87911127
what exactly is the difference between $land$ and $cap$ in this context?
– Yanko
Nov 22 at 17:55
@Yanko: Their meanings are standard - $cap$ is intersection of sets, $wedge$ is conjunction of statements (meaning 'and').
– Clive Newstead
Nov 22 at 17:56
@CliveNewstead I get that, but if $f,g$ are binary relations, isn't $fland g$ the same as $fcap g$ where $f$ and $g$ are being viewed as subsets of the cartesian product?
– Yanko
Nov 22 at 17:57
@Yanko: I'd presume so, yes. What's the confusion?
– Clive Newstead
Nov 22 at 17:58
2
@Yanko: Ah, you're just parsing it wrong. Read it as $$[f cap (A_1 times B_1) = g] wedge [g cap (A_0 times B_0) = f]$$ and so on.
– Clive Newstead
Nov 22 at 18:01
|
show 2 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $f$ be a binary relation between sets $A_0$ and $B_0$ and $g$ be a binary relation between sets $A_1$ and $B_1$.
It can be proved that $[fcap(A_1times B_1)=g]wedge [gcap(A_0times B_0)=f]$ implies $f=g$.
Really, it follows $gcap(A_0times B_0)cap(A_1times B_1)=g$. Thus $gsubseteq A_0times B_0$. Consequantly $g=f$.
Do the following hold?
$[fcap(A_1times B_1)supseteq g]wedge [gcap(A_0times B_0)]supseteq f$ implies $f=g$.
$[fcap(A_1times B_1)subseteq g]wedge [gcap(A_0times B_0)]subseteq f$ implies $f=g$.
elementary-set-theory relations products
edited Nov 22 at 18:07
Yanko
5,660723
asked Nov 22 at 17:44
porton
1,87911127
Let $f$ be a binary relation between sets $A_0$ and $B_0$ and $g$ be a binary relation between sets $A_1$ and $B_1$.
It can be proved that $[fcap(A_1times B_1)=g]wedge [gcap(A_0times B_0)=f]$ implies $f=g$.
Really, it follows $gcap(A_0times B_0)cap(A_1times B_1)=g$. Thus $gsubseteq A_0times B_0$. Consequantly $g=f$.
Do the following hold?
$[fcap(A_1times B_1)supseteq g]wedge [gcap(A_0times B_0)]supseteq f$ implies $f=g$.
$[fcap(A_1times B_1)subseteq g]wedge [gcap(A_0times B_0)]subseteq f$ implies $f=g$.
elementary-set-theory relations products
elementary-set-theory relations products
edited Nov 22 at 18:07
Yanko
5,660723
asked Nov 22 at 17:44
porton
1,87911127
edited Nov 22 at 18:07
Yanko
5,660723
asked Nov 22 at 17:44
porton
1,87911127
edited Nov 22 at 18:07
Yanko
5,660723
edited Nov 22 at 18:07
Yanko
5,660723
edited Nov 22 at 18:07
Yanko
5,660723
5,660723
asked Nov 22 at 17:44
porton
1,87911127
asked Nov 22 at 17:44
porton
1,87911127
asked Nov 22 at 17:44
porton
1,87911127
1,87911127
what exactly is the difference between $land$ and $cap$ in this context?
– Yanko
Nov 22 at 17:55
@Yanko: Their meanings are standard - $cap$ is intersection of sets, $wedge$ is conjunction of statements (meaning 'and').
– Clive Newstead
Nov 22 at 17:56
@CliveNewstead I get that, but if $f,g$ are binary relations, isn't $fland g$ the same as $fcap g$ where $f$ and $g$ are being viewed as subsets of the cartesian product?
– Yanko
Nov 22 at 17:57
@Yanko: I'd presume so, yes. What's the confusion?
– Clive Newstead
Nov 22 at 17:58
2
@Yanko: Ah, you're just parsing it wrong. Read it as $$[f cap (A_1 times B_1) = g] wedge [g cap (A_0 times B_0) = f]$$ and so on.
– Clive Newstead
Nov 22 at 18:01
|
show 2 more comments
what exactly is the difference between $land$ and $cap$ in this context?
– Yanko
Nov 22 at 17:55
@Yanko: Their meanings are standard - $cap$ is intersection of sets, $wedge$ is conjunction of statements (meaning 'and').
– Clive Newstead
Nov 22 at 17:56
@CliveNewstead I get that, but if $f,g$ are binary relations, isn't $fland g$ the same as $fcap g$ where $f$ and $g$ are being viewed as subsets of the cartesian product?
– Yanko
Nov 22 at 17:57
@Yanko: I'd presume so, yes. What's the confusion?
– Clive Newstead
Nov 22 at 17:58
2
@Yanko: Ah, you're just parsing it wrong. Read it as $$[f cap (A_1 times B_1) = g] wedge [g cap (A_0 times B_0) = f]$$ and so on.
– Clive Newstead
Nov 22 at 18:01
what exactly is the difference between $land$ and $cap$ in this context?
– Yanko
Nov 22 at 17:55
what exactly is the difference between $land$ and $cap$ in this context?
– Yanko
Nov 22 at 17:55
@Yanko: Their meanings are standard - $cap$ is intersection of sets, $wedge$ is conjunction of statements (meaning 'and').
– Clive Newstead
Nov 22 at 17:56
@Yanko: Their meanings are standard - $cap$ is intersection of sets, $wedge$ is conjunction of statements (meaning 'and').
– Clive Newstead
Nov 22 at 17:56
@CliveNewstead I get that, but if $f,g$ are binary relations, isn't $fland g$ the same as $fcap g$ where $f$ and $g$ are being viewed as subsets of the cartesian product?
– Yanko
Nov 22 at 17:57
@CliveNewstead I get that, but if $f,g$ are binary relations, isn't $fland g$ the same as $fcap g$ where $f$ and $g$ are being viewed as subsets of the cartesian product?
– Yanko
Nov 22 at 17:57
@Yanko: I'd presume so, yes. What's the confusion?
– Clive Newstead
Nov 22 at 17:58
@Yanko: I'd presume so, yes. What's the confusion?
– Clive Newstead
Nov 22 at 17:58
2
2
@Yanko: Ah, you're just parsing it wrong. Read it as $$[f cap (A_1 times B_1) = g] wedge [g cap (A_0 times B_0) = f]$$ and so on.
– Clive Newstead
Nov 22 at 18:01
@Yanko: Ah, you're just parsing it wrong. Read it as $$[f cap (A_1 times B_1) = g] wedge [g cap (A_0 times B_0) = f]$$ and so on.
– Clive Newstead
Nov 22 at 18:01
|
show 2 more comments
1 Answer
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votes
up vote
2
down vote
accepted
It is true in general that if $X subseteq Y cap Z$, then $X subseteq Y$ and $X subseteq Z$. The hypothesis of statement (1) implies that $f subseteq g$ and $g subseteq f$, and so $f = g$, as required. So statement (1) is true.
The hypothesis of statement (2) holds for all $f,g$, provided that $A_0 cap A_1 = varnothing$ or $B_0 cap B_1 = varnothing$, so it doesn't follow in general that $f=g$. So statement (2) is false in general.
In fact, more generally, we have $(A_0 times B_0) cap (A_1 times B_1) = (A_0 cap A_1) times (B_0 cap B_1)$, and so the hypothesis of (2) holds whenever the relations $f$ and $g$ agree on $(A_0 cap A_1) times (B_0 cap B_1)$; but then $f$ and $g$ can say what they like about the elements not in these intersections, so they need not be equal.
answered Nov 22 at 18:05
Clive Newstead
49.9k473132
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
It is true in general that if $X subseteq Y cap Z$, then $X subseteq Y$ and $X subseteq Z$. The hypothesis of statement (1) implies that $f subseteq g$ and $g subseteq f$, and so $f = g$, as required. So statement (1) is true.
The hypothesis of statement (2) holds for all $f,g$, provided that $A_0 cap A_1 = varnothing$ or $B_0 cap B_1 = varnothing$, so it doesn't follow in general that $f=g$. So statement (2) is false in general.
In fact, more generally, we have $(A_0 times B_0) cap (A_1 times B_1) = (A_0 cap A_1) times (B_0 cap B_1)$, and so the hypothesis of (2) holds whenever the relations $f$ and $g$ agree on $(A_0 cap A_1) times (B_0 cap B_1)$; but then $f$ and $g$ can say what they like about the elements not in these intersections, so they need not be equal.
answered Nov 22 at 18:05
Clive Newstead
49.9k473132
add a comment |
up vote
2
down vote
accepted
It is true in general that if $X subseteq Y cap Z$, then $X subseteq Y$ and $X subseteq Z$. The hypothesis of statement (1) implies that $f subseteq g$ and $g subseteq f$, and so $f = g$, as required. So statement (1) is true.
The hypothesis of statement (2) holds for all $f,g$, provided that $A_0 cap A_1 = varnothing$ or $B_0 cap B_1 = varnothing$, so it doesn't follow in general that $f=g$. So statement (2) is false in general.
In fact, more generally, we have $(A_0 times B_0) cap (A_1 times B_1) = (A_0 cap A_1) times (B_0 cap B_1)$, and so the hypothesis of (2) holds whenever the relations $f$ and $g$ agree on $(A_0 cap A_1) times (B_0 cap B_1)$; but then $f$ and $g$ can say what they like about the elements not in these intersections, so they need not be equal.
answered Nov 22 at 18:05
Clive Newstead
49.9k473132
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
It is true in general that if $X subseteq Y cap Z$, then $X subseteq Y$ and $X subseteq Z$. The hypothesis of statement (1) implies that $f subseteq g$ and $g subseteq f$, and so $f = g$, as required. So statement (1) is true.
The hypothesis of statement (2) holds for all $f,g$, provided that $A_0 cap A_1 = varnothing$ or $B_0 cap B_1 = varnothing$, so it doesn't follow in general that $f=g$. So statement (2) is false in general.
In fact, more generally, we have $(A_0 times B_0) cap (A_1 times B_1) = (A_0 cap A_1) times (B_0 cap B_1)$, and so the hypothesis of (2) holds whenever the relations $f$ and $g$ agree on $(A_0 cap A_1) times (B_0 cap B_1)$; but then $f$ and $g$ can say what they like about the elements not in these intersections, so they need not be equal.
answered Nov 22 at 18:05
Clive Newstead
49.9k473132
It is true in general that if $X subseteq Y cap Z$, then $X subseteq Y$ and $X subseteq Z$. The hypothesis of statement (1) implies that $f subseteq g$ and $g subseteq f$, and so $f = g$, as required. So statement (1) is true.
The hypothesis of statement (2) holds for all $f,g$, provided that $A_0 cap A_1 = varnothing$ or $B_0 cap B_1 = varnothing$, so it doesn't follow in general that $f=g$. So statement (2) is false in general.
In fact, more generally, we have $(A_0 times B_0) cap (A_1 times B_1) = (A_0 cap A_1) times (B_0 cap B_1)$, and so the hypothesis of (2) holds whenever the relations $f$ and $g$ agree on $(A_0 cap A_1) times (B_0 cap B_1)$; but then $f$ and $g$ can say what they like about the elements not in these intersections, so they need not be equal.
answered Nov 22 at 18:05
Clive Newstead
49.9k473132
answered Nov 22 at 18:05
Clive Newstead
49.9k473132
answered Nov 22 at 18:05
Clive Newstead
49.9k473132
answered Nov 22 at 18:05
Clive Newstead
49.9k473132
49.9k473132
add a comment |
add a comment |
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what exactly is the difference between $land$ and $cap$ in this context?
– Yanko
Nov 22 at 17:55
@Yanko: Their meanings are standard - $cap$ is intersection of sets, $wedge$ is conjunction of statements (meaning 'and').
– Clive Newstead
Nov 22 at 17:56
@CliveNewstead I get that, but if $f,g$ are binary relations, isn't $fland g$ the same as $fcap g$ where $f$ and $g$ are being viewed as subsets of the cartesian product?
– Yanko
Nov 22 at 17:57
@Yanko: I'd presume so, yes. What's the confusion?
– Clive Newstead
Nov 22 at 17:58
2
@Yanko: Ah, you're just parsing it wrong. Read it as $$[f cap (A_1 times B_1) = g] wedge [g cap (A_0 times B_0) = f]$$ and so on.
– Clive Newstead
Nov 22 at 18:01