Prove that these orderings are isomorphic
0
$begingroup$
I want to prove that:
1) $mathbb{Z}+mathbb{Z}$ and $mathbb{Z} + mathbb{N}$
2) $mathbb{Q}$ and $mathbb{N}timesmathbb{Q}$ (lexicographical order in $mathbb{N}timesmathbb{Q}$)
are isomorphic or not.
I know the definition of isomorphic orderings but I don't know how to solve these kind of tasks. Can you give me a hint how to do it correct?
order-theory
asked Dec 9 '18 at 8:05
ErlGreyErlGrey
32
$endgroup$
|
show 1 more comment
0
$begingroup$
I want to prove that:
1) $mathbb{Z}+mathbb{Z}$ and $mathbb{Z} + mathbb{N}$
2) $mathbb{Q}$ and $mathbb{N}timesmathbb{Q}$ (lexicographical order in $mathbb{N}timesmathbb{Q}$)
are isomorphic or not.
I know the definition of isomorphic orderings but I don't know how to solve these kind of tasks. Can you give me a hint how to do it correct?
order-theory
asked Dec 9 '18 at 8:05
ErlGreyErlGrey
32
$endgroup$
$begingroup$
1 are not order isomophic. Find a feature in one that's not in the other. For example Z has no bottom element, N does.
$endgroup$
– William Elliot
Dec 10 '18 at 10:59
$begingroup$
@WilliamElliot, thank you, I succeed in solution of the first item and, yes, I have used this information in my solution. But I don't know how to prove the second item(I think it is true)
$endgroup$
– ErlGrey
Dec 10 '18 at 11:01
$begingroup$
What order does N×Q have?
$endgroup$
– William Elliot
Dec 10 '18 at 11:03
$begingroup$
This is a set of pairs $(x, y)$, where $xinmathbb{N}$ and $yinmathbb{Q}$, we use lexicographical order, so, there is no bootom element if you mean it
$endgroup$
– ErlGrey
Dec 10 '18 at 11:08
$begingroup$
There is a famous characterization of orderings isomorphic to $mathbb Q$: countable, no top element, no bottom element, between any two elements there is another. If this result is available, then you can easily see that $mathbb Q$ and $mathbb Ntimesmathbb Q$ are isomorphic.
$endgroup$
– bof
Dec 10 '18 at 11:45
|
show 1 more comment
0
0
0
$begingroup$
I want to prove that:
1) $mathbb{Z}+mathbb{Z}$ and $mathbb{Z} + mathbb{N}$
2) $mathbb{Q}$ and $mathbb{N}timesmathbb{Q}$ (lexicographical order in $mathbb{N}timesmathbb{Q}$)
are isomorphic or not.
I know the definition of isomorphic orderings but I don't know how to solve these kind of tasks. Can you give me a hint how to do it correct?
order-theory
asked Dec 9 '18 at 8:05
ErlGreyErlGrey
32
$endgroup$
I want to prove that:
1) $mathbb{Z}+mathbb{Z}$ and $mathbb{Z} + mathbb{N}$
2) $mathbb{Q}$ and $mathbb{N}timesmathbb{Q}$ (lexicographical order in $mathbb{N}timesmathbb{Q}$)
are isomorphic or not.
I know the definition of isomorphic orderings but I don't know how to solve these kind of tasks. Can you give me a hint how to do it correct?
order-theory
order-theory
asked Dec 9 '18 at 8:05
ErlGreyErlGrey
32
asked Dec 9 '18 at 8:05
ErlGreyErlGrey
32
edited Dec 10 '18 at 11:09
ErlGrey
asked Dec 9 '18 at 8:05
ErlGreyErlGrey
32
asked Dec 9 '18 at 8:05
ErlGreyErlGrey
32
asked Dec 9 '18 at 8:05
ErlGreyErlGrey
32
32
$begingroup$
1 are not order isomophic. Find a feature in one that's not in the other. For example Z has no bottom element, N does.
$endgroup$
– William Elliot
Dec 10 '18 at 10:59
$begingroup$
@WilliamElliot, thank you, I succeed in solution of the first item and, yes, I have used this information in my solution. But I don't know how to prove the second item(I think it is true)
$endgroup$
– ErlGrey
Dec 10 '18 at 11:01
$begingroup$
What order does N×Q have?
$endgroup$
– William Elliot
Dec 10 '18 at 11:03
$begingroup$
This is a set of pairs $(x, y)$, where $xinmathbb{N}$ and $yinmathbb{Q}$, we use lexicographical order, so, there is no bootom element if you mean it
$endgroup$
– ErlGrey
Dec 10 '18 at 11:08
$begingroup$
There is a famous characterization of orderings isomorphic to $mathbb Q$: countable, no top element, no bottom element, between any two elements there is another. If this result is available, then you can easily see that $mathbb Q$ and $mathbb Ntimesmathbb Q$ are isomorphic.
$endgroup$
– bof
Dec 10 '18 at 11:45
|
show 1 more comment
$begingroup$
1 are not order isomophic. Find a feature in one that's not in the other. For example Z has no bottom element, N does.
$endgroup$
– William Elliot
Dec 10 '18 at 10:59
$begingroup$
@WilliamElliot, thank you, I succeed in solution of the first item and, yes, I have used this information in my solution. But I don't know how to prove the second item(I think it is true)
$endgroup$
– ErlGrey
Dec 10 '18 at 11:01
$begingroup$
What order does N×Q have?
$endgroup$
– William Elliot
Dec 10 '18 at 11:03
$begingroup$
This is a set of pairs $(x, y)$, where $xinmathbb{N}$ and $yinmathbb{Q}$, we use lexicographical order, so, there is no bootom element if you mean it
$endgroup$
– ErlGrey
Dec 10 '18 at 11:08
$begingroup$
There is a famous characterization of orderings isomorphic to $mathbb Q$: countable, no top element, no bottom element, between any two elements there is another. If this result is available, then you can easily see that $mathbb Q$ and $mathbb Ntimesmathbb Q$ are isomorphic.
$endgroup$
– bof
Dec 10 '18 at 11:45
$begingroup$
1 are not order isomophic. Find a feature in one that's not in the other. For example Z has no bottom element, N does.
$endgroup$
– William Elliot
Dec 10 '18 at 10:59
$begingroup$
1 are not order isomophic. Find a feature in one that's not in the other. For example Z has no bottom element, N does.
$endgroup$
– William Elliot
Dec 10 '18 at 10:59
$begingroup$
@WilliamElliot, thank you, I succeed in solution of the first item and, yes, I have used this information in my solution. But I don't know how to prove the second item(I think it is true)
$endgroup$
– ErlGrey
Dec 10 '18 at 11:01
$begingroup$
@WilliamElliot, thank you, I succeed in solution of the first item and, yes, I have used this information in my solution. But I don't know how to prove the second item(I think it is true)
$endgroup$
– ErlGrey
Dec 10 '18 at 11:01
$begingroup$
What order does N×Q have?
$endgroup$
– William Elliot
Dec 10 '18 at 11:03
$begingroup$
What order does N×Q have?
$endgroup$
– William Elliot
Dec 10 '18 at 11:03
$begingroup$
This is a set of pairs $(x, y)$, where $xinmathbb{N}$ and $yinmathbb{Q}$, we use lexicographical order, so, there is no bootom element if you mean it
$endgroup$
– ErlGrey
Dec 10 '18 at 11:08
$begingroup$
This is a set of pairs $(x, y)$, where $xinmathbb{N}$ and $yinmathbb{Q}$, we use lexicographical order, so, there is no bootom element if you mean it
$endgroup$
– ErlGrey
Dec 10 '18 at 11:08
$begingroup$
There is a famous characterization of orderings isomorphic to $mathbb Q$: countable, no top element, no bottom element, between any two elements there is another. If this result is available, then you can easily see that $mathbb Q$ and $mathbb Ntimesmathbb Q$ are isomorphic.
$endgroup$
– bof
Dec 10 '18 at 11:45
$begingroup$
There is a famous characterization of orderings isomorphic to $mathbb Q$: countable, no top element, no bottom element, between any two elements there is another. If this result is available, then you can easily see that $mathbb Q$ and $mathbb Ntimesmathbb Q$ are isomorphic.
$endgroup$
– bof
Dec 10 '18 at 11:45
|
show 1 more comment
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$begingroup$
1 are not order isomophic. Find a feature in one that's not in the other. For example Z has no bottom element, N does.
$endgroup$
– William Elliot
Dec 10 '18 at 10:59
$begingroup$
@WilliamElliot, thank you, I succeed in solution of the first item and, yes, I have used this information in my solution. But I don't know how to prove the second item(I think it is true)
$endgroup$
– ErlGrey
Dec 10 '18 at 11:01
$begingroup$
What order does N×Q have?
$endgroup$
– William Elliot
Dec 10 '18 at 11:03
$begingroup$
This is a set of pairs $(x, y)$, where $xinmathbb{N}$ and $yinmathbb{Q}$, we use lexicographical order, so, there is no bootom element if you mean it
$endgroup$
– ErlGrey
Dec 10 '18 at 11:08
$begingroup$
There is a famous characterization of orderings isomorphic to $mathbb Q$: countable, no top element, no bottom element, between any two elements there is another. If this result is available, then you can easily see that $mathbb Q$ and $mathbb Ntimesmathbb Q$ are isomorphic.
$endgroup$
– bof
Dec 10 '18 at 11:45