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Dictionary order relations of $BbbZ_+times [0,1)$ and $[0,1)times BbbZ_+$
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Actually, I am reading this context from the book “Topology†by James Munkres. First of all, I should recall the definition of Dictionary Order Relation, which goes as follows-
Suppose, $A, B$ are two sets with simple(or linear) order relation
$<_A$ and $<_B$ respectively. Define an order relation $<$ on $Atimes
B$ by defining $$ a_1times b_1 <a_2times b_2 $$ if $a_1<_A a_2$, or
if $a_1=a_2$ and $b_1<_B b_2$. It is called the the dictionary order
relation on $Atimes B$.
Now, then the author gives us examples of dictionary order relations on the sets $BbbZ_+times [0,1)$ and $[0,1)times BbbZ_+$. And gives us two figures to visualize these two dictionary order relations and the difference between that.
According to the book, the set $[0,1)times BbbZ_+$ in the dictionary order has a quite different order type; for example, every element of this ordered set has an immediate successor.
But I cannot visualize these two dictionary order relations in detail. If $p$ is a point on $[0,1)times BbbZ_+$, then $p$ is less than WHAT POINTS IN $[0,1)times BbbZ_+$? I cannot understand the fact "every element of this ordered set($[0,1)times BbbZ_+$) has an immediate successor".
Can anybody explain the dictionary order relation on $[0,1)times BbbZ_+$ and the differnce between $BbbZ_+times [0,1)$ and $[0,1)times BbbZ_+$ in detail?
Thanks for your assistance in advance.
relations order-theory
edited Jul 23 at 13:53
Asaf Karagila
291k31402732
asked Jul 23 at 13:23
Biswarup Saha
2298
add a comment |Â
up vote
-1
down vote
favorite
Actually, I am reading this context from the book “Topology†by James Munkres. First of all, I should recall the definition of Dictionary Order Relation, which goes as follows-
Suppose, $A, B$ are two sets with simple(or linear) order relation
$<_A$ and $<_B$ respectively. Define an order relation $<$ on $Atimes
B$ by defining $$ a_1times b_1 <a_2times b_2 $$ if $a_1<_A a_2$, or
if $a_1=a_2$ and $b_1<_B b_2$. It is called the the dictionary order
relation on $Atimes B$.
Now, then the author gives us examples of dictionary order relations on the sets $BbbZ_+times [0,1)$ and $[0,1)times BbbZ_+$. And gives us two figures to visualize these two dictionary order relations and the difference between that.
According to the book, the set $[0,1)times BbbZ_+$ in the dictionary order has a quite different order type; for example, every element of this ordered set has an immediate successor.
But I cannot visualize these two dictionary order relations in detail. If $p$ is a point on $[0,1)times BbbZ_+$, then $p$ is less than WHAT POINTS IN $[0,1)times BbbZ_+$? I cannot understand the fact "every element of this ordered set($[0,1)times BbbZ_+$) has an immediate successor".
Can anybody explain the dictionary order relation on $[0,1)times BbbZ_+$ and the differnce between $BbbZ_+times [0,1)$ and $[0,1)times BbbZ_+$ in detail?
Thanks for your assistance in advance.
relations order-theory
edited Jul 23 at 13:53
Asaf Karagila
291k31402732
asked Jul 23 at 13:23
Biswarup Saha
2298
1
Why do you need to visualize things? It's just a huge inconvenience having to visualize everything in mathematics. You have definitions. Use them.
– Asaf Karagila
Jul 23 at 13:54
add a comment |Â
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
Actually, I am reading this context from the book “Topology†by James Munkres. First of all, I should recall the definition of Dictionary Order Relation, which goes as follows-
Suppose, $A, B$ are two sets with simple(or linear) order relation
$<_A$ and $<_B$ respectively. Define an order relation $<$ on $Atimes
B$ by defining $$ a_1times b_1 <a_2times b_2 $$ if $a_1<_A a_2$, or
if $a_1=a_2$ and $b_1<_B b_2$. It is called the the dictionary order
relation on $Atimes B$.
Now, then the author gives us examples of dictionary order relations on the sets $BbbZ_+times [0,1)$ and $[0,1)times BbbZ_+$. And gives us two figures to visualize these two dictionary order relations and the difference between that.
According to the book, the set $[0,1)times BbbZ_+$ in the dictionary order has a quite different order type; for example, every element of this ordered set has an immediate successor.
But I cannot visualize these two dictionary order relations in detail. If $p$ is a point on $[0,1)times BbbZ_+$, then $p$ is less than WHAT POINTS IN $[0,1)times BbbZ_+$? I cannot understand the fact "every element of this ordered set($[0,1)times BbbZ_+$) has an immediate successor".
Can anybody explain the dictionary order relation on $[0,1)times BbbZ_+$ and the differnce between $BbbZ_+times [0,1)$ and $[0,1)times BbbZ_+$ in detail?
Thanks for your assistance in advance.
relations order-theory
edited Jul 23 at 13:53
Asaf Karagila
291k31402732
asked Jul 23 at 13:23
Biswarup Saha
2298
Actually, I am reading this context from the book “Topology†by James Munkres. First of all, I should recall the definition of Dictionary Order Relation, which goes as follows-
Suppose, $A, B$ are two sets with simple(or linear) order relation
$<_A$ and $<_B$ respectively. Define an order relation $<$ on $Atimes
B$ by defining $$ a_1times b_1 <a_2times b_2 $$ if $a_1<_A a_2$, or
if $a_1=a_2$ and $b_1<_B b_2$. It is called the the dictionary order
relation on $Atimes B$.
Now, then the author gives us examples of dictionary order relations on the sets $BbbZ_+times [0,1)$ and $[0,1)times BbbZ_+$. And gives us two figures to visualize these two dictionary order relations and the difference between that.
According to the book, the set $[0,1)times BbbZ_+$ in the dictionary order has a quite different order type; for example, every element of this ordered set has an immediate successor.
But I cannot visualize these two dictionary order relations in detail. If $p$ is a point on $[0,1)times BbbZ_+$, then $p$ is less than WHAT POINTS IN $[0,1)times BbbZ_+$? I cannot understand the fact "every element of this ordered set($[0,1)times BbbZ_+$) has an immediate successor".
Can anybody explain the dictionary order relation on $[0,1)times BbbZ_+$ and the differnce between $BbbZ_+times [0,1)$ and $[0,1)times BbbZ_+$ in detail?
Thanks for your assistance in advance.
relations order-theory
edited Jul 23 at 13:53
Asaf Karagila
291k31402732
asked Jul 23 at 13:23
Biswarup Saha
2298
edited Jul 23 at 13:53
Asaf Karagila
291k31402732
edited Jul 23 at 13:53
Asaf Karagila
291k31402732
edited Jul 23 at 13:53
Asaf Karagila
291k31402732
291k31402732
asked Jul 23 at 13:23
Biswarup Saha
2298
asked Jul 23 at 13:23
Biswarup Saha
2298
asked Jul 23 at 13:23
Biswarup Saha
2298
2298
1
Why do you need to visualize things? It's just a huge inconvenience having to visualize everything in mathematics. You have definitions. Use them.
– Asaf Karagila
Jul 23 at 13:54
add a comment |Â
1
Why do you need to visualize things? It's just a huge inconvenience having to visualize everything in mathematics. You have definitions. Use them.
– Asaf Karagila
Jul 23 at 13:54
1
1
Why do you need to visualize things? It's just a huge inconvenience having to visualize everything in mathematics. You have definitions. Use them.
– Asaf Karagila
Jul 23 at 13:54
Why do you need to visualize things? It's just a huge inconvenience having to visualize everything in mathematics. You have definitions. Use them.
– Asaf Karagila
Jul 23 at 13:54
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
0
down vote
accepted
When you want to visualize the dictionary order (or as it is probably best known, the lexicographic order) one good trick is to think that the order is obtained by first "looking up2 and then to the right. So, in the order $mathbbNtimes mathbbZ$, the elements greater than $(-1,0)$ will be $$(-1,1),(-1,2),ldots,(0,0),(0,1),(0,2),ldots,(1,0),(1,1),(1,2),ldots.$$
In the structure, $(0,1]times mathbbZ_+$, the succesor of a given point $p=(r,n)$ will be obtained looking up, and it will be $(r,n+1)$. On the contrary, in the order $mathbbZ_+times (0,1]$, the successor of a point $p=(n,r)$ would be given by $(n,s)$ where $s$ is the next element after $r$ in the order $(0,1]$. Since $(0,1]$ is a dense linear order, such element $s$ cannot exists, and therefore $p=(n,r)$ does not have a successor in $mathbbZ_+times (0,1]$.
answered Jul 23 at 13:56
DarÃo G
3,942613
Dario G, okay, you cleared my confusion.
– Biswarup Saha
Jul 23 at 14:13
I think the last $(0,1]times Bbb Z_+$ should be $Bbb Z_+times (0,1]$ instead
– Alessandro Codenotti
Jul 23 at 16:39
Corrected! Thanks
– DarÃo G
Jul 23 at 16:42
add a comment |Â
up vote
0
down vote
A simpler example may help your intuition regards dictionary
order. N is the positive integers.
(0,1,2) × N looks like 1,2,3,... 1,2,3,... 1,2,3,...
N × (0,1,2) looks like 0,1,2, 0,1,2, 0,1,2, ...
answered Jul 24 at 8:52
William Elliot
5,0942416
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
When you want to visualize the dictionary order (or as it is probably best known, the lexicographic order) one good trick is to think that the order is obtained by first "looking up2 and then to the right. So, in the order $mathbbNtimes mathbbZ$, the elements greater than $(-1,0)$ will be $$(-1,1),(-1,2),ldots,(0,0),(0,1),(0,2),ldots,(1,0),(1,1),(1,2),ldots.$$
In the structure, $(0,1]times mathbbZ_+$, the succesor of a given point $p=(r,n)$ will be obtained looking up, and it will be $(r,n+1)$. On the contrary, in the order $mathbbZ_+times (0,1]$, the successor of a point $p=(n,r)$ would be given by $(n,s)$ where $s$ is the next element after $r$ in the order $(0,1]$. Since $(0,1]$ is a dense linear order, such element $s$ cannot exists, and therefore $p=(n,r)$ does not have a successor in $mathbbZ_+times (0,1]$.
answered Jul 23 at 13:56
DarÃo G
3,942613
Dario G, okay, you cleared my confusion.
– Biswarup Saha
Jul 23 at 14:13
I think the last $(0,1]times Bbb Z_+$ should be $Bbb Z_+times (0,1]$ instead
– Alessandro Codenotti
Jul 23 at 16:39
Corrected! Thanks
– DarÃo G
Jul 23 at 16:42
add a comment |Â
up vote
0
down vote
accepted
When you want to visualize the dictionary order (or as it is probably best known, the lexicographic order) one good trick is to think that the order is obtained by first "looking up2 and then to the right. So, in the order $mathbbNtimes mathbbZ$, the elements greater than $(-1,0)$ will be $$(-1,1),(-1,2),ldots,(0,0),(0,1),(0,2),ldots,(1,0),(1,1),(1,2),ldots.$$
In the structure, $(0,1]times mathbbZ_+$, the succesor of a given point $p=(r,n)$ will be obtained looking up, and it will be $(r,n+1)$. On the contrary, in the order $mathbbZ_+times (0,1]$, the successor of a point $p=(n,r)$ would be given by $(n,s)$ where $s$ is the next element after $r$ in the order $(0,1]$. Since $(0,1]$ is a dense linear order, such element $s$ cannot exists, and therefore $p=(n,r)$ does not have a successor in $mathbbZ_+times (0,1]$.
answered Jul 23 at 13:56
DarÃo G
3,942613
Dario G, okay, you cleared my confusion.
– Biswarup Saha
Jul 23 at 14:13
I think the last $(0,1]times Bbb Z_+$ should be $Bbb Z_+times (0,1]$ instead
– Alessandro Codenotti
Jul 23 at 16:39
Corrected! Thanks
– DarÃo G
Jul 23 at 16:42
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
When you want to visualize the dictionary order (or as it is probably best known, the lexicographic order) one good trick is to think that the order is obtained by first "looking up2 and then to the right. So, in the order $mathbbNtimes mathbbZ$, the elements greater than $(-1,0)$ will be $$(-1,1),(-1,2),ldots,(0,0),(0,1),(0,2),ldots,(1,0),(1,1),(1,2),ldots.$$
In the structure, $(0,1]times mathbbZ_+$, the succesor of a given point $p=(r,n)$ will be obtained looking up, and it will be $(r,n+1)$. On the contrary, in the order $mathbbZ_+times (0,1]$, the successor of a point $p=(n,r)$ would be given by $(n,s)$ where $s$ is the next element after $r$ in the order $(0,1]$. Since $(0,1]$ is a dense linear order, such element $s$ cannot exists, and therefore $p=(n,r)$ does not have a successor in $mathbbZ_+times (0,1]$.
answered Jul 23 at 13:56
DarÃo G
3,942613
When you want to visualize the dictionary order (or as it is probably best known, the lexicographic order) one good trick is to think that the order is obtained by first "looking up2 and then to the right. So, in the order $mathbbNtimes mathbbZ$, the elements greater than $(-1,0)$ will be $$(-1,1),(-1,2),ldots,(0,0),(0,1),(0,2),ldots,(1,0),(1,1),(1,2),ldots.$$
In the structure, $(0,1]times mathbbZ_+$, the succesor of a given point $p=(r,n)$ will be obtained looking up, and it will be $(r,n+1)$. On the contrary, in the order $mathbbZ_+times (0,1]$, the successor of a point $p=(n,r)$ would be given by $(n,s)$ where $s$ is the next element after $r$ in the order $(0,1]$. Since $(0,1]$ is a dense linear order, such element $s$ cannot exists, and therefore $p=(n,r)$ does not have a successor in $mathbbZ_+times (0,1]$.
answered Jul 23 at 13:56
DarÃo G
3,942613
edited Jul 23 at 16:41
answered Jul 23 at 13:56
DarÃo G
3,942613
answered Jul 23 at 13:56
DarÃo G
3,942613
answered Jul 23 at 13:56
DarÃo G
3,942613
3,942613
Dario G, okay, you cleared my confusion.
– Biswarup Saha
Jul 23 at 14:13
I think the last $(0,1]times Bbb Z_+$ should be $Bbb Z_+times (0,1]$ instead
– Alessandro Codenotti
Jul 23 at 16:39
Corrected! Thanks
– DarÃo G
Jul 23 at 16:42
add a comment |Â
Dario G, okay, you cleared my confusion.
– Biswarup Saha
Jul 23 at 14:13
I think the last $(0,1]times Bbb Z_+$ should be $Bbb Z_+times (0,1]$ instead
– Alessandro Codenotti
Jul 23 at 16:39
Corrected! Thanks
– DarÃo G
Jul 23 at 16:42
Dario G, okay, you cleared my confusion.
– Biswarup Saha
Jul 23 at 14:13
Dario G, okay, you cleared my confusion.
– Biswarup Saha
Jul 23 at 14:13
I think the last $(0,1]times Bbb Z_+$ should be $Bbb Z_+times (0,1]$ instead
– Alessandro Codenotti
Jul 23 at 16:39
I think the last $(0,1]times Bbb Z_+$ should be $Bbb Z_+times (0,1]$ instead
– Alessandro Codenotti
Jul 23 at 16:39
Corrected! Thanks
– DarÃo G
Jul 23 at 16:42
Corrected! Thanks
– DarÃo G
Jul 23 at 16:42
add a comment |Â
up vote
0
down vote
A simpler example may help your intuition regards dictionary
order. N is the positive integers.
(0,1,2) × N looks like 1,2,3,... 1,2,3,... 1,2,3,...
N × (0,1,2) looks like 0,1,2, 0,1,2, 0,1,2, ...
answered Jul 24 at 8:52
William Elliot
5,0942416
add a comment |Â
up vote
0
down vote
A simpler example may help your intuition regards dictionary
order. N is the positive integers.
(0,1,2) × N looks like 1,2,3,... 1,2,3,... 1,2,3,...
N × (0,1,2) looks like 0,1,2, 0,1,2, 0,1,2, ...
answered Jul 24 at 8:52
William Elliot
5,0942416
add a comment |Â
up vote
0
down vote
up vote
0
down vote
A simpler example may help your intuition regards dictionary
order. N is the positive integers.
(0,1,2) × N looks like 1,2,3,... 1,2,3,... 1,2,3,...
N × (0,1,2) looks like 0,1,2, 0,1,2, 0,1,2, ...
answered Jul 24 at 8:52
William Elliot
5,0942416
A simpler example may help your intuition regards dictionary
order. N is the positive integers.
(0,1,2) × N looks like 1,2,3,... 1,2,3,... 1,2,3,...
N × (0,1,2) looks like 0,1,2, 0,1,2, 0,1,2, ...
answered Jul 24 at 8:52
William Elliot
5,0942416
answered Jul 24 at 8:52
William Elliot
5,0942416
answered Jul 24 at 8:52
William Elliot
5,0942416
answered Jul 24 at 8:52
William Elliot
5,0942416
5,0942416
add a comment |Â
add a comment |Â
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1
Why do you need to visualize things? It's just a huge inconvenience having to visualize everything in mathematics. You have definitions. Use them.
– Asaf Karagila
Jul 23 at 13:54