Isomorphism of $mathbb R$ and $mathbb{R} times mathbb R$
How to prove that the set $mathbb R times mathbb R$ with lexicographic order is not isomorphic with $mathbb R$ with linear order?
I know I have to show that there is no such functions $f:mathbb R rightarrowmathbb R times mathbb R$ which is an isomorphism but I dont't know how to write this proof.
order-theory
edited Nov 28 at 23:16
egreg
177k1484199
asked Nov 28 at 22:40
avan1235
1886
add a comment |
How to prove that the set $mathbb R times mathbb R$ with lexicographic order is not isomorphic with $mathbb R$ with linear order?
I know I have to show that there is no such functions $f:mathbb R rightarrowmathbb R times mathbb R$ which is an isomorphism but I dont't know how to write this proof.
order-theory
edited Nov 28 at 23:16
egreg
177k1484199
asked Nov 28 at 22:40
avan1235
1886
2
Perhaps show that there is a nonempty bounded subset of $mathbb R times mathbb R$ with no least upper bound.
– GEdgar
Nov 28 at 22:46
@GEdgar but how to start this proof and how is it connected with that there is no isomorphism function?
– avan1235
Nov 28 at 23:18
add a comment |
How to prove that the set $mathbb R times mathbb R$ with lexicographic order is not isomorphic with $mathbb R$ with linear order?
I know I have to show that there is no such functions $f:mathbb R rightarrowmathbb R times mathbb R$ which is an isomorphism but I dont't know how to write this proof.
order-theory
edited Nov 28 at 23:16
egreg
177k1484199
asked Nov 28 at 22:40
avan1235
1886
How to prove that the set $mathbb R times mathbb R$ with lexicographic order is not isomorphic with $mathbb R$ with linear order?
I know I have to show that there is no such functions $f:mathbb R rightarrowmathbb R times mathbb R$ which is an isomorphism but I dont't know how to write this proof.
order-theory
order-theory
edited Nov 28 at 23:16
egreg
177k1484199
asked Nov 28 at 22:40
avan1235
1886
edited Nov 28 at 23:16
egreg
177k1484199
asked Nov 28 at 22:40
avan1235
1886
edited Nov 28 at 23:16
egreg
177k1484199
edited Nov 28 at 23:16
egreg
177k1484199
edited Nov 28 at 23:16
egreg
177k1484199
177k1484199
asked Nov 28 at 22:40
avan1235
1886
asked Nov 28 at 22:40
avan1235
1886
asked Nov 28 at 22:40
avan1235
1886
1886
2
Perhaps show that there is a nonempty bounded subset of $mathbb R times mathbb R$ with no least upper bound.
– GEdgar
Nov 28 at 22:46
@GEdgar but how to start this proof and how is it connected with that there is no isomorphism function?
– avan1235
Nov 28 at 23:18
add a comment |
2
Perhaps show that there is a nonempty bounded subset of $mathbb R times mathbb R$ with no least upper bound.
– GEdgar
Nov 28 at 22:46
@GEdgar but how to start this proof and how is it connected with that there is no isomorphism function?
– avan1235
Nov 28 at 23:18
2
2
Perhaps show that there is a nonempty bounded subset of $mathbb R times mathbb R$ with no least upper bound.
– GEdgar
Nov 28 at 22:46
Perhaps show that there is a nonempty bounded subset of $mathbb R times mathbb R$ with no least upper bound.
– GEdgar
Nov 28 at 22:46
@GEdgar but how to start this proof and how is it connected with that there is no isomorphism function?
– avan1235
Nov 28 at 23:18
@GEdgar but how to start this proof and how is it connected with that there is no isomorphism function?
– avan1235
Nov 28 at 23:18
add a comment |
1 Answer
1
active
oldest
votes
Two isomorphic ordered sets share all properties that can be expressed in the language of ordered sets.
For instance, if $fcolon Xto Y$ is an order isomorphism and $A$ is a subset of $X$, then $A$ has a least upper bound if and only if $f(A)$ has a least upper bound.
Also, $A$ is upper bounded if and only if $f(A)$ is upper bounded.
Exercise: prove the two statements above.
Now, can you find a subset $A$ of $mathbb{R}timesmathbb{R}$ that's upper bounded but has no least upper bound? If $fcolonmathbb{R}timesmathbb{R}tomathbb{R}$ is an isomorphism, can you find a contradiction?
answered Nov 28 at 23:23
egreg
177k1484199
Is it good example to take a subset $A=lbrace{langle a,branglein mathbb{R}timesmathbb{R} | a=0, binmathbb{R} rbrace}$ and then it has for example upper bound equal to $langle1,0rangle$ but has no least upper bound (because we can always find a better upper bound)
– avan1235
Nov 28 at 23:31
@avan1235 That's a fine example.
– egreg
Nov 28 at 23:39
How can I try to show that two isomorphic ordered sets share all properties which can be expressed in the language of ordered sets?
– avan1235
Nov 28 at 23:44
@avan1235 That's not an easy problem to formulate (but then easy to prove); be content of proving it for specific properties like the ones I suggested.
– egreg
Nov 29 at 9:59
add a comment |
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1 Answer
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1 Answer
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Two isomorphic ordered sets share all properties that can be expressed in the language of ordered sets.
For instance, if $fcolon Xto Y$ is an order isomorphism and $A$ is a subset of $X$, then $A$ has a least upper bound if and only if $f(A)$ has a least upper bound.
Also, $A$ is upper bounded if and only if $f(A)$ is upper bounded.
Exercise: prove the two statements above.
Now, can you find a subset $A$ of $mathbb{R}timesmathbb{R}$ that's upper bounded but has no least upper bound? If $fcolonmathbb{R}timesmathbb{R}tomathbb{R}$ is an isomorphism, can you find a contradiction?
answered Nov 28 at 23:23
egreg
177k1484199
Is it good example to take a subset $A=lbrace{langle a,branglein mathbb{R}timesmathbb{R} | a=0, binmathbb{R} rbrace}$ and then it has for example upper bound equal to $langle1,0rangle$ but has no least upper bound (because we can always find a better upper bound)
– avan1235
Nov 28 at 23:31
@avan1235 That's a fine example.
– egreg
Nov 28 at 23:39
How can I try to show that two isomorphic ordered sets share all properties which can be expressed in the language of ordered sets?
– avan1235
Nov 28 at 23:44
@avan1235 That's not an easy problem to formulate (but then easy to prove); be content of proving it for specific properties like the ones I suggested.
– egreg
Nov 29 at 9:59
add a comment |
Two isomorphic ordered sets share all properties that can be expressed in the language of ordered sets.
For instance, if $fcolon Xto Y$ is an order isomorphism and $A$ is a subset of $X$, then $A$ has a least upper bound if and only if $f(A)$ has a least upper bound.
Also, $A$ is upper bounded if and only if $f(A)$ is upper bounded.
Exercise: prove the two statements above.
Now, can you find a subset $A$ of $mathbb{R}timesmathbb{R}$ that's upper bounded but has no least upper bound? If $fcolonmathbb{R}timesmathbb{R}tomathbb{R}$ is an isomorphism, can you find a contradiction?
answered Nov 28 at 23:23
egreg
177k1484199
Is it good example to take a subset $A=lbrace{langle a,branglein mathbb{R}timesmathbb{R} | a=0, binmathbb{R} rbrace}$ and then it has for example upper bound equal to $langle1,0rangle$ but has no least upper bound (because we can always find a better upper bound)
– avan1235
Nov 28 at 23:31
@avan1235 That's a fine example.
– egreg
Nov 28 at 23:39
How can I try to show that two isomorphic ordered sets share all properties which can be expressed in the language of ordered sets?
– avan1235
Nov 28 at 23:44
@avan1235 That's not an easy problem to formulate (but then easy to prove); be content of proving it for specific properties like the ones I suggested.
– egreg
Nov 29 at 9:59
add a comment |
Two isomorphic ordered sets share all properties that can be expressed in the language of ordered sets.
For instance, if $fcolon Xto Y$ is an order isomorphism and $A$ is a subset of $X$, then $A$ has a least upper bound if and only if $f(A)$ has a least upper bound.
Also, $A$ is upper bounded if and only if $f(A)$ is upper bounded.
Exercise: prove the two statements above.
Now, can you find a subset $A$ of $mathbb{R}timesmathbb{R}$ that's upper bounded but has no least upper bound? If $fcolonmathbb{R}timesmathbb{R}tomathbb{R}$ is an isomorphism, can you find a contradiction?
answered Nov 28 at 23:23
egreg
177k1484199
Two isomorphic ordered sets share all properties that can be expressed in the language of ordered sets.
For instance, if $fcolon Xto Y$ is an order isomorphism and $A$ is a subset of $X$, then $A$ has a least upper bound if and only if $f(A)$ has a least upper bound.
Also, $A$ is upper bounded if and only if $f(A)$ is upper bounded.
Exercise: prove the two statements above.
Now, can you find a subset $A$ of $mathbb{R}timesmathbb{R}$ that's upper bounded but has no least upper bound? If $fcolonmathbb{R}timesmathbb{R}tomathbb{R}$ is an isomorphism, can you find a contradiction?
answered Nov 28 at 23:23
egreg
177k1484199
answered Nov 28 at 23:23
egreg
177k1484199
answered Nov 28 at 23:23
egreg
177k1484199
answered Nov 28 at 23:23
egreg
177k1484199
177k1484199
Is it good example to take a subset $A=lbrace{langle a,branglein mathbb{R}timesmathbb{R} | a=0, binmathbb{R} rbrace}$ and then it has for example upper bound equal to $langle1,0rangle$ but has no least upper bound (because we can always find a better upper bound)
– avan1235
Nov 28 at 23:31
@avan1235 That's a fine example.
– egreg
Nov 28 at 23:39
How can I try to show that two isomorphic ordered sets share all properties which can be expressed in the language of ordered sets?
– avan1235
Nov 28 at 23:44
@avan1235 That's not an easy problem to formulate (but then easy to prove); be content of proving it for specific properties like the ones I suggested.
– egreg
Nov 29 at 9:59
add a comment |
Is it good example to take a subset $A=lbrace{langle a,branglein mathbb{R}timesmathbb{R} | a=0, binmathbb{R} rbrace}$ and then it has for example upper bound equal to $langle1,0rangle$ but has no least upper bound (because we can always find a better upper bound)
– avan1235
Nov 28 at 23:31
@avan1235 That's a fine example.
– egreg
Nov 28 at 23:39
How can I try to show that two isomorphic ordered sets share all properties which can be expressed in the language of ordered sets?
– avan1235
Nov 28 at 23:44
@avan1235 That's not an easy problem to formulate (but then easy to prove); be content of proving it for specific properties like the ones I suggested.
– egreg
Nov 29 at 9:59
Is it good example to take a subset $A=lbrace{langle a,branglein mathbb{R}timesmathbb{R} | a=0, binmathbb{R} rbrace}$ and then it has for example upper bound equal to $langle1,0rangle$ but has no least upper bound (because we can always find a better upper bound)
– avan1235
Nov 28 at 23:31
Is it good example to take a subset $A=lbrace{langle a,branglein mathbb{R}timesmathbb{R} | a=0, binmathbb{R} rbrace}$ and then it has for example upper bound equal to $langle1,0rangle$ but has no least upper bound (because we can always find a better upper bound)
– avan1235
Nov 28 at 23:31
@avan1235 That's a fine example.
– egreg
Nov 28 at 23:39
@avan1235 That's a fine example.
– egreg
Nov 28 at 23:39
How can I try to show that two isomorphic ordered sets share all properties which can be expressed in the language of ordered sets?
– avan1235
Nov 28 at 23:44
How can I try to show that two isomorphic ordered sets share all properties which can be expressed in the language of ordered sets?
– avan1235
Nov 28 at 23:44
@avan1235 That's not an easy problem to formulate (but then easy to prove); be content of proving it for specific properties like the ones I suggested.
– egreg
Nov 29 at 9:59
@avan1235 That's not an easy problem to formulate (but then easy to prove); be content of proving it for specific properties like the ones I suggested.
– egreg
Nov 29 at 9:59
add a comment |
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2
Perhaps show that there is a nonempty bounded subset of $mathbb R times mathbb R$ with no least upper bound.
– GEdgar
Nov 28 at 22:46
@GEdgar but how to start this proof and how is it connected with that there is no isomorphism function?
– avan1235
Nov 28 at 23:18