Mixing
Understanding the Cartesian product is the set of functions [duplicate]
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
This question already has an answer here:
Cartesian products as sets of all functions.
3 answers
According Wikipedia, "if a tuple is defined as a function on $1, 2, ..., n$ that takes its value at $i$ to be the $i$th element of the tuple, then the Cartesian product $X_1×...×X_n$ is the set of functions
$displaystyle x(i)in X_i textfor every iin 1,ldots ,n.$"
For me, it is hard to understand the above definition. Could you give some example to illustrate the concept?
For example, let $X_1 = 1,2,X_2=a,b$. Then, $X_1×X_2= ?$.
Thank you in advance.
elementary-set-theory
edited Jul 21 at 12:29
Andrés E. Caicedo
63.2k7151235
asked Jul 21 at 6:53
Sihyun Kim
699211
marked as duplicate by Asaf Karagila♦ set-theory
Users with the  set-theory badge can single-handedly close set-theory questions as duplicates and reopen them as needed.
StackExchange.ready(function()
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function()
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function()
$hover.showInfoMessage('',
messageElement: $msg.clone().show(),
transient: false,
position: my: 'bottom left', at: 'top center', offsetTop: -7 ,
dismissable: false,
relativeToBody: true
);
,
function()
StackExchange.helpers.removeMessages();
);
);
);
Jul 21 at 7:23
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |Â
up vote
0
down vote
favorite
This question already has an answer here:
Cartesian products as sets of all functions.
3 answers
According Wikipedia, "if a tuple is defined as a function on $1, 2, ..., n$ that takes its value at $i$ to be the $i$th element of the tuple, then the Cartesian product $X_1×...×X_n$ is the set of functions
$displaystyle x(i)in X_i textfor every iin 1,ldots ,n.$"
For me, it is hard to understand the above definition. Could you give some example to illustrate the concept?
For example, let $X_1 = 1,2,X_2=a,b$. Then, $X_1×X_2= ?$.
Thank you in advance.
elementary-set-theory
edited Jul 21 at 12:29
Andrés E. Caicedo
63.2k7151235
asked Jul 21 at 6:53
Sihyun Kim
699211
marked as duplicate by Asaf Karagila♦ set-theory
Users with the  set-theory badge can single-handedly close set-theory questions as duplicates and reopen them as needed.
StackExchange.ready(function()
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function()
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function()
$hover.showInfoMessage('',
messageElement: $msg.clone().show(),
transient: false,
position: my: 'bottom left', at: 'top center', offsetTop: -7 ,
dismissable: false,
relativeToBody: true
);
,
function()
StackExchange.helpers.removeMessages();
);
);
);
Jul 21 at 7:23
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
2
$X_1 times X_2 = (1,a), (1,b), (2,a), (2,b) $
– tp1
Jul 21 at 7:00
@tp1 That is another definiton of $X_1times X_2$. I don't see the relevance.
– drhab
Jul 21 at 7:51
@drhab: normally cartesian product is defined using pair constructor function and element accessor functions: $ ct : 1 rightarrow X_1 times X_2 $ and $a_i : X_1 times X_2 rightarrow X_i $, and you'll get $x(i) in X_i $ from that.
– tp1
Jul 21 at 8:37
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
This question already has an answer here:
Cartesian products as sets of all functions.
3 answers
According Wikipedia, "if a tuple is defined as a function on $1, 2, ..., n$ that takes its value at $i$ to be the $i$th element of the tuple, then the Cartesian product $X_1×...×X_n$ is the set of functions
$displaystyle x(i)in X_i textfor every iin 1,ldots ,n.$"
For me, it is hard to understand the above definition. Could you give some example to illustrate the concept?
For example, let $X_1 = 1,2,X_2=a,b$. Then, $X_1×X_2= ?$.
Thank you in advance.
elementary-set-theory
edited Jul 21 at 12:29
Andrés E. Caicedo
63.2k7151235
asked Jul 21 at 6:53
Sihyun Kim
699211
This question already has an answer here:
Cartesian products as sets of all functions.
3 answers
According Wikipedia, "if a tuple is defined as a function on $1, 2, ..., n$ that takes its value at $i$ to be the $i$th element of the tuple, then the Cartesian product $X_1×...×X_n$ is the set of functions
$displaystyle x(i)in X_i textfor every iin 1,ldots ,n.$"
For me, it is hard to understand the above definition. Could you give some example to illustrate the concept?
For example, let $X_1 = 1,2,X_2=a,b$. Then, $X_1×X_2= ?$.
Thank you in advance.
This question already has an answer here:
Cartesian products as sets of all functions.
3 answers
elementary-set-theory
edited Jul 21 at 12:29
Andrés E. Caicedo
63.2k7151235
asked Jul 21 at 6:53
Sihyun Kim
699211
edited Jul 21 at 12:29
Andrés E. Caicedo
63.2k7151235
edited Jul 21 at 12:29
Andrés E. Caicedo
63.2k7151235
edited Jul 21 at 12:29
Andrés E. Caicedo
63.2k7151235
63.2k7151235
asked Jul 21 at 6:53
Sihyun Kim
699211
asked Jul 21 at 6:53
Sihyun Kim
699211
asked Jul 21 at 6:53
Sihyun Kim
699211
699211
marked as duplicate by Asaf Karagila♦ set-theory
Users with the  set-theory badge can single-handedly close set-theory questions as duplicates and reopen them as needed.
StackExchange.ready(function()
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function()
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function()
$hover.showInfoMessage('',
messageElement: $msg.clone().show(),
transient: false,
position: my: 'bottom left', at: 'top center', offsetTop: -7 ,
dismissable: false,
relativeToBody: true
);
,
function()
StackExchange.helpers.removeMessages();
);
);
);
Jul 21 at 7:23
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Asaf Karagila♦ set-theory
Users with the  set-theory badge can single-handedly close set-theory questions as duplicates and reopen them as needed.
StackExchange.ready(function()
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function()
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function()
$hover.showInfoMessage('',
messageElement: $msg.clone().show(),
transient: false,
position: my: 'bottom left', at: 'top center', offsetTop: -7 ,
dismissable: false,
relativeToBody: true
);
,
function()
StackExchange.helpers.removeMessages();
);
);
);
Jul 21 at 7:23
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
2
$X_1 times X_2 = (1,a), (1,b), (2,a), (2,b) $
– tp1
Jul 21 at 7:00
@tp1 That is another definiton of $X_1times X_2$. I don't see the relevance.
– drhab
Jul 21 at 7:51
@drhab: normally cartesian product is defined using pair constructor function and element accessor functions: $ ct : 1 rightarrow X_1 times X_2 $ and $a_i : X_1 times X_2 rightarrow X_i $, and you'll get $x(i) in X_i $ from that.
– tp1
Jul 21 at 8:37
add a comment |Â
2
$X_1 times X_2 = (1,a), (1,b), (2,a), (2,b) $
– tp1
Jul 21 at 7:00
@tp1 That is another definiton of $X_1times X_2$. I don't see the relevance.
– drhab
Jul 21 at 7:51
@drhab: normally cartesian product is defined using pair constructor function and element accessor functions: $ ct : 1 rightarrow X_1 times X_2 $ and $a_i : X_1 times X_2 rightarrow X_i $, and you'll get $x(i) in X_i $ from that.
– tp1
Jul 21 at 8:37
2
2
$X_1 times X_2 = (1,a), (1,b), (2,a), (2,b) $
– tp1
Jul 21 at 7:00
$X_1 times X_2 = (1,a), (1,b), (2,a), (2,b) $
– tp1
Jul 21 at 7:00
@tp1 That is another definiton of $X_1times X_2$. I don't see the relevance.
– drhab
Jul 21 at 7:51
@tp1 That is another definiton of $X_1times X_2$. I don't see the relevance.
– drhab
Jul 21 at 7:51
@drhab: normally cartesian product is defined using pair constructor function and element accessor functions: $ ct : 1 rightarrow X_1 times X_2 $ and $a_i : X_1 times X_2 rightarrow X_i $, and you'll get $x(i) in X_i $ from that.
– tp1
Jul 21 at 8:37
@drhab: normally cartesian product is defined using pair constructor function and element accessor functions: $ ct : 1 rightarrow X_1 times X_2 $ and $a_i : X_1 times X_2 rightarrow X_i $, and you'll get $x(i) in X_i $ from that.
– tp1
Jul 21 at 8:37
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
We are looking for functions $f:1,2to1,2,a,b$ such $f(1)in X_1=1,2$ and $f(2)in X_2=a,b$.
There are $4$ such functions and writing each function as a set of ordered pairs they are:
- $f_1,a:=langle 1,1rangle,langle 2,arangle$
- $f_1,b:=langle 1,1rangle,langle 2,brangle$
- $f_2,a:=langle 1,2rangle,langle 2,arangle$
- $f_2,b:=langle 1,2rangle,langle 2,brangle$
Then according to the mentioned definition:$$X_1times X_2=f_1,a,f_1,b,f_2,a,f_2,b=$$$$langle 1,1rangle,langle 2,arangle,langle 1,1rangle,langle 2,brangle,langle 1,2rangle,langle 2,arangle,langle 1,2rangle,langle 2,brangle$$
We can go even further by writing order pair $langle r,srangle$ as $r,r,s$ according to the definition of Kuratowski, but I don't think that that will increase your understanding.
Note that we get back the "familiar" definition of $X_1times X_2$ just by replacing $f_1,a,f_1,b,f_2,a,f_2,b$ by $langle 1,arangle,langle 1,brangle,langle 2,arangle,langle 2,brangle$ respectively.
answered Jul 21 at 7:11
drhab
86.5k541118
add a comment |Â
up vote
1
down vote
In terms of the definition you've quoted, in your example $X_1times X_2$ is the set of all functions $x$ from $,1,2,$ to $,1,2,a,b,$ such that $x(1)$ is in $,1,2,$ and $x(2)$ is in $,a,b,,$. There are four such functions, which I'll call $x_1,x_2,x_3,x_4$, given by
$x_1(1)=1,x_1(2)=a$
$x_2(1)=1,x_2(2)=b$
$x_3(1)=2,x_3(2)=a$
$x_4(1)=2,x_4(2)=b$
answered Jul 21 at 7:07
Gerry Myerson
143k7144294
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
We are looking for functions $f:1,2to1,2,a,b$ such $f(1)in X_1=1,2$ and $f(2)in X_2=a,b$.
There are $4$ such functions and writing each function as a set of ordered pairs they are:
- $f_1,a:=langle 1,1rangle,langle 2,arangle$
- $f_1,b:=langle 1,1rangle,langle 2,brangle$
- $f_2,a:=langle 1,2rangle,langle 2,arangle$
- $f_2,b:=langle 1,2rangle,langle 2,brangle$
Then according to the mentioned definition:$$X_1times X_2=f_1,a,f_1,b,f_2,a,f_2,b=$$$$langle 1,1rangle,langle 2,arangle,langle 1,1rangle,langle 2,brangle,langle 1,2rangle,langle 2,arangle,langle 1,2rangle,langle 2,brangle$$
We can go even further by writing order pair $langle r,srangle$ as $r,r,s$ according to the definition of Kuratowski, but I don't think that that will increase your understanding.
Note that we get back the "familiar" definition of $X_1times X_2$ just by replacing $f_1,a,f_1,b,f_2,a,f_2,b$ by $langle 1,arangle,langle 1,brangle,langle 2,arangle,langle 2,brangle$ respectively.
answered Jul 21 at 7:11
drhab
86.5k541118
add a comment |Â
up vote
2
down vote
accepted
We are looking for functions $f:1,2to1,2,a,b$ such $f(1)in X_1=1,2$ and $f(2)in X_2=a,b$.
There are $4$ such functions and writing each function as a set of ordered pairs they are:
- $f_1,a:=langle 1,1rangle,langle 2,arangle$
- $f_1,b:=langle 1,1rangle,langle 2,brangle$
- $f_2,a:=langle 1,2rangle,langle 2,arangle$
- $f_2,b:=langle 1,2rangle,langle 2,brangle$
Then according to the mentioned definition:$$X_1times X_2=f_1,a,f_1,b,f_2,a,f_2,b=$$$$langle 1,1rangle,langle 2,arangle,langle 1,1rangle,langle 2,brangle,langle 1,2rangle,langle 2,arangle,langle 1,2rangle,langle 2,brangle$$
We can go even further by writing order pair $langle r,srangle$ as $r,r,s$ according to the definition of Kuratowski, but I don't think that that will increase your understanding.
Note that we get back the "familiar" definition of $X_1times X_2$ just by replacing $f_1,a,f_1,b,f_2,a,f_2,b$ by $langle 1,arangle,langle 1,brangle,langle 2,arangle,langle 2,brangle$ respectively.
answered Jul 21 at 7:11
drhab
86.5k541118
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
We are looking for functions $f:1,2to1,2,a,b$ such $f(1)in X_1=1,2$ and $f(2)in X_2=a,b$.
There are $4$ such functions and writing each function as a set of ordered pairs they are:
- $f_1,a:=langle 1,1rangle,langle 2,arangle$
- $f_1,b:=langle 1,1rangle,langle 2,brangle$
- $f_2,a:=langle 1,2rangle,langle 2,arangle$
- $f_2,b:=langle 1,2rangle,langle 2,brangle$
Then according to the mentioned definition:$$X_1times X_2=f_1,a,f_1,b,f_2,a,f_2,b=$$$$langle 1,1rangle,langle 2,arangle,langle 1,1rangle,langle 2,brangle,langle 1,2rangle,langle 2,arangle,langle 1,2rangle,langle 2,brangle$$
We can go even further by writing order pair $langle r,srangle$ as $r,r,s$ according to the definition of Kuratowski, but I don't think that that will increase your understanding.
Note that we get back the "familiar" definition of $X_1times X_2$ just by replacing $f_1,a,f_1,b,f_2,a,f_2,b$ by $langle 1,arangle,langle 1,brangle,langle 2,arangle,langle 2,brangle$ respectively.
answered Jul 21 at 7:11
drhab
86.5k541118
We are looking for functions $f:1,2to1,2,a,b$ such $f(1)in X_1=1,2$ and $f(2)in X_2=a,b$.
There are $4$ such functions and writing each function as a set of ordered pairs they are:
- $f_1,a:=langle 1,1rangle,langle 2,arangle$
- $f_1,b:=langle 1,1rangle,langle 2,brangle$
- $f_2,a:=langle 1,2rangle,langle 2,arangle$
- $f_2,b:=langle 1,2rangle,langle 2,brangle$
Then according to the mentioned definition:$$X_1times X_2=f_1,a,f_1,b,f_2,a,f_2,b=$$$$langle 1,1rangle,langle 2,arangle,langle 1,1rangle,langle 2,brangle,langle 1,2rangle,langle 2,arangle,langle 1,2rangle,langle 2,brangle$$
We can go even further by writing order pair $langle r,srangle$ as $r,r,s$ according to the definition of Kuratowski, but I don't think that that will increase your understanding.
Note that we get back the "familiar" definition of $X_1times X_2$ just by replacing $f_1,a,f_1,b,f_2,a,f_2,b$ by $langle 1,arangle,langle 1,brangle,langle 2,arangle,langle 2,brangle$ respectively.
answered Jul 21 at 7:11
drhab
86.5k541118
edited Jul 21 at 7:26
answered Jul 21 at 7:11
drhab
86.5k541118
answered Jul 21 at 7:11
drhab
86.5k541118
answered Jul 21 at 7:11
drhab
86.5k541118
86.5k541118
add a comment |Â
add a comment |Â
up vote
1
down vote
In terms of the definition you've quoted, in your example $X_1times X_2$ is the set of all functions $x$ from $,1,2,$ to $,1,2,a,b,$ such that $x(1)$ is in $,1,2,$ and $x(2)$ is in $,a,b,,$. There are four such functions, which I'll call $x_1,x_2,x_3,x_4$, given by
$x_1(1)=1,x_1(2)=a$
$x_2(1)=1,x_2(2)=b$
$x_3(1)=2,x_3(2)=a$
$x_4(1)=2,x_4(2)=b$
answered Jul 21 at 7:07
Gerry Myerson
143k7144294
add a comment |Â
up vote
1
down vote
In terms of the definition you've quoted, in your example $X_1times X_2$ is the set of all functions $x$ from $,1,2,$ to $,1,2,a,b,$ such that $x(1)$ is in $,1,2,$ and $x(2)$ is in $,a,b,,$. There are four such functions, which I'll call $x_1,x_2,x_3,x_4$, given by
$x_1(1)=1,x_1(2)=a$
$x_2(1)=1,x_2(2)=b$
$x_3(1)=2,x_3(2)=a$
$x_4(1)=2,x_4(2)=b$
answered Jul 21 at 7:07
Gerry Myerson
143k7144294
add a comment |Â
up vote
1
down vote
up vote
1
down vote
In terms of the definition you've quoted, in your example $X_1times X_2$ is the set of all functions $x$ from $,1,2,$ to $,1,2,a,b,$ such that $x(1)$ is in $,1,2,$ and $x(2)$ is in $,a,b,,$. There are four such functions, which I'll call $x_1,x_2,x_3,x_4$, given by
$x_1(1)=1,x_1(2)=a$
$x_2(1)=1,x_2(2)=b$
$x_3(1)=2,x_3(2)=a$
$x_4(1)=2,x_4(2)=b$
answered Jul 21 at 7:07
Gerry Myerson
143k7144294
In terms of the definition you've quoted, in your example $X_1times X_2$ is the set of all functions $x$ from $,1,2,$ to $,1,2,a,b,$ such that $x(1)$ is in $,1,2,$ and $x(2)$ is in $,a,b,,$. There are four such functions, which I'll call $x_1,x_2,x_3,x_4$, given by
$x_1(1)=1,x_1(2)=a$
$x_2(1)=1,x_2(2)=b$
$x_3(1)=2,x_3(2)=a$
$x_4(1)=2,x_4(2)=b$
answered Jul 21 at 7:07
Gerry Myerson
143k7144294
answered Jul 21 at 7:07
Gerry Myerson
143k7144294
answered Jul 21 at 7:07
Gerry Myerson
143k7144294
answered Jul 21 at 7:07
Gerry Myerson
143k7144294
143k7144294
add a comment |Â
add a comment |Â
2
$X_1 times X_2 = (1,a), (1,b), (2,a), (2,b) $
– tp1
Jul 21 at 7:00
@tp1 That is another definiton of $X_1times X_2$. I don't see the relevance.
– drhab
Jul 21 at 7:51
@drhab: normally cartesian product is defined using pair constructor function and element accessor functions: $ ct : 1 rightarrow X_1 times X_2 $ and $a_i : X_1 times X_2 rightarrow X_i $, and you'll get $x(i) in X_i $ from that.
– tp1
Jul 21 at 8:37