Let $(G,<)$ be an archimedean left-ordered group, then $exists H in (mathbb{R},<)$ and $ (G,<) cong...
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Here is the proof in the book:
How do we show that $phi(g)$ converges?
My attempt:
Let $ phi_k(g)=frac{a_{2^k}}{2^k}$, then $|phi_{k+1}(g)-phi_k(g)|=|frac{a_{2^{k+1}}}{2^{k+1}}-frac{a_{2^k}}{2^k}|=|frac{a_{2^{k+1}}-2a_{2^k}}{2^{k+1}}|$
abstract-algebra group-theory
asked Nov 22 at 20:28
mathnoob
1,754322
add a comment |
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1
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Here is the proof in the book:
How do we show that $phi(g)$ converges?
My attempt:
Let $ phi_k(g)=frac{a_{2^k}}{2^k}$, then $|phi_{k+1}(g)-phi_k(g)|=|frac{a_{2^{k+1}}}{2^{k+1}}-frac{a_{2^k}}{2^k}|=|frac{a_{2^{k+1}}-2a_{2^k}}{2^{k+1}}|$
abstract-algebra group-theory
asked Nov 22 at 20:28
mathnoob
1,754322
Do you know that the ordering of an Archimedean left-ordered group is also right-invariant? Or that such groups are abelian?
– Fimpellizieri
Nov 27 at 2:14
yes I proved it before
– mathnoob
Nov 27 at 12:07
Actually I can show that the $phi_k$ converge, but I've not yet managed to show that $a_n/n$ also does.
– Fimpellizieri
Nov 27 at 16:28
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Here is the proof in the book:
How do we show that $phi(g)$ converges?
My attempt:
Let $ phi_k(g)=frac{a_{2^k}}{2^k}$, then $|phi_{k+1}(g)-phi_k(g)|=|frac{a_{2^{k+1}}}{2^{k+1}}-frac{a_{2^k}}{2^k}|=|frac{a_{2^{k+1}}-2a_{2^k}}{2^{k+1}}|$
abstract-algebra group-theory
asked Nov 22 at 20:28
mathnoob
1,754322
Here is the proof in the book:
How do we show that $phi(g)$ converges?
My attempt:
Let $ phi_k(g)=frac{a_{2^k}}{2^k}$, then $|phi_{k+1}(g)-phi_k(g)|=|frac{a_{2^{k+1}}}{2^{k+1}}-frac{a_{2^k}}{2^k}|=|frac{a_{2^{k+1}}-2a_{2^k}}{2^{k+1}}|$
abstract-algebra group-theory
abstract-algebra group-theory
asked Nov 22 at 20:28
mathnoob
1,754322
asked Nov 22 at 20:28
mathnoob
1,754322
edited Nov 23 at 4:03
asked Nov 22 at 20:28
mathnoob
1,754322
asked Nov 22 at 20:28
mathnoob
1,754322
asked Nov 22 at 20:28
mathnoob
1,754322
1,754322
Do you know that the ordering of an Archimedean left-ordered group is also right-invariant? Or that such groups are abelian?
– Fimpellizieri
Nov 27 at 2:14
yes I proved it before
– mathnoob
Nov 27 at 12:07
Actually I can show that the $phi_k$ converge, but I've not yet managed to show that $a_n/n$ also does.
– Fimpellizieri
Nov 27 at 16:28
add a comment |
Do you know that the ordering of an Archimedean left-ordered group is also right-invariant? Or that such groups are abelian?
– Fimpellizieri
Nov 27 at 2:14
yes I proved it before
– mathnoob
Nov 27 at 12:07
Actually I can show that the $phi_k$ converge, but I've not yet managed to show that $a_n/n$ also does.
– Fimpellizieri
Nov 27 at 16:28
Do you know that the ordering of an Archimedean left-ordered group is also right-invariant? Or that such groups are abelian?
– Fimpellizieri
Nov 27 at 2:14
Do you know that the ordering of an Archimedean left-ordered group is also right-invariant? Or that such groups are abelian?
– Fimpellizieri
Nov 27 at 2:14
yes I proved it before
– mathnoob
Nov 27 at 12:07
yes I proved it before
– mathnoob
Nov 27 at 12:07
Actually I can show that the $phi_k$ converge, but I've not yet managed to show that $a_n/n$ also does.
– Fimpellizieri
Nov 27 at 16:28
Actually I can show that the $phi_k$ converge, but I've not yet managed to show that $a_n/n$ also does.
– Fimpellizieri
Nov 27 at 16:28
add a comment |
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Do you know that the ordering of an Archimedean left-ordered group is also right-invariant? Or that such groups are abelian?
– Fimpellizieri
Nov 27 at 2:14
yes I proved it before
– mathnoob
Nov 27 at 12:07
Actually I can show that the $phi_k$ converge, but I've not yet managed to show that $a_n/n$ also does.
– Fimpellizieri
Nov 27 at 16:28