$omega^omega$ correspondence with $mathbb R$-irrationality
Here in the second comment I do not understand why $omega^omega$ corresponds to irrational numbers? :
In my experience one typically identifies $ω^ω$ with the irrational elements of R; and then we call them "reals" because they are equinumerous, and in particular "isomorphic up to a countable set".
QUESTION: What can we say about irrationality of $f=(2,2,2,2,...)$ goes to $(0,0,1,0,0,1,0,0,1,0,0,...)$
which is not irrational since it repeats the $0,0,1$ pattern forever,right? And yet, it doesn't end wiht eventually all $1$'s.
continuity real-numbers integers natural-numbers
asked 2 days ago
user122424
1,0921616
|
show 3 more comments
Here in the second comment I do not understand why $omega^omega$ corresponds to irrational numbers? :
In my experience one typically identifies $ω^ω$ with the irrational elements of R; and then we call them "reals" because they are equinumerous, and in particular "isomorphic up to a countable set".
QUESTION: What can we say about irrationality of $f=(2,2,2,2,...)$ goes to $(0,0,1,0,0,1,0,0,1,0,0,...)$
which is not irrational since it repeats the $0,0,1$ pattern forever,right? And yet, it doesn't end wiht eventually all $1$'s.
continuity real-numbers integers natural-numbers
asked 2 days ago
user122424
1,0921616
2
The usual correspondence maps elements of $omega^omega$ to infinite continued fractions, i.e., irrationals in $(0,1)$. It is not clear at all what the example you give is supposed to be.
– Andrés E. Caicedo
2 days ago
1
The mapping that Noah Schweber gave in his post is not the one I was talking about. He does mention the one I was talking about in the sentence right before: it is the mapping that takes $(a_1, a_2, a_3, ldots)$ to $frac 1{a_1 + frac1{a_2 + frac1{a_3 + cdots}}}$.
– Mees de Vries
2 days ago
@MeesdeVries That explains my question, at least partially. However, how do I prove that all such $f's$ from $omega$ to $omega$ get map to irrationals via this continued fractions? What if there is a tail of $0$'s? Do they also map to irrational numbers?
– user122424
2 days ago
1
In particular, $(2, 2, 2, 2, 2, 2, ldots)$ goes to an $x$ satisfying $x = frac1{2 + x}$, so $x = sqrt2 - 1$.
– Mees de Vries
2 days ago
OK. But what about $(2,0,0,0,0,0,0....)$ ?
– user122424
2 days ago
|
show 3 more comments
Here in the second comment I do not understand why $omega^omega$ corresponds to irrational numbers? :
In my experience one typically identifies $ω^ω$ with the irrational elements of R; and then we call them "reals" because they are equinumerous, and in particular "isomorphic up to a countable set".
QUESTION: What can we say about irrationality of $f=(2,2,2,2,...)$ goes to $(0,0,1,0,0,1,0,0,1,0,0,...)$
which is not irrational since it repeats the $0,0,1$ pattern forever,right? And yet, it doesn't end wiht eventually all $1$'s.
continuity real-numbers integers natural-numbers
asked 2 days ago
user122424
1,0921616
Here in the second comment I do not understand why $omega^omega$ corresponds to irrational numbers? :
In my experience one typically identifies $ω^ω$ with the irrational elements of R; and then we call them "reals" because they are equinumerous, and in particular "isomorphic up to a countable set".
QUESTION: What can we say about irrationality of $f=(2,2,2,2,...)$ goes to $(0,0,1,0,0,1,0,0,1,0,0,...)$
which is not irrational since it repeats the $0,0,1$ pattern forever,right? And yet, it doesn't end wiht eventually all $1$'s.
continuity real-numbers integers natural-numbers
continuity real-numbers integers natural-numbers
asked 2 days ago
user122424
1,0921616
asked 2 days ago
user122424
1,0921616
edited 2 days ago
asked 2 days ago
user122424
1,0921616
asked 2 days ago
user122424
1,0921616
asked 2 days ago
user122424
1,0921616
1,0921616
2
The usual correspondence maps elements of $omega^omega$ to infinite continued fractions, i.e., irrationals in $(0,1)$. It is not clear at all what the example you give is supposed to be.
– Andrés E. Caicedo
2 days ago
1
The mapping that Noah Schweber gave in his post is not the one I was talking about. He does mention the one I was talking about in the sentence right before: it is the mapping that takes $(a_1, a_2, a_3, ldots)$ to $frac 1{a_1 + frac1{a_2 + frac1{a_3 + cdots}}}$.
– Mees de Vries
2 days ago
@MeesdeVries That explains my question, at least partially. However, how do I prove that all such $f's$ from $omega$ to $omega$ get map to irrationals via this continued fractions? What if there is a tail of $0$'s? Do they also map to irrational numbers?
– user122424
2 days ago
1
In particular, $(2, 2, 2, 2, 2, 2, ldots)$ goes to an $x$ satisfying $x = frac1{2 + x}$, so $x = sqrt2 - 1$.
– Mees de Vries
2 days ago
OK. But what about $(2,0,0,0,0,0,0....)$ ?
– user122424
2 days ago
|
show 3 more comments
2
The usual correspondence maps elements of $omega^omega$ to infinite continued fractions, i.e., irrationals in $(0,1)$. It is not clear at all what the example you give is supposed to be.
– Andrés E. Caicedo
2 days ago
1
The mapping that Noah Schweber gave in his post is not the one I was talking about. He does mention the one I was talking about in the sentence right before: it is the mapping that takes $(a_1, a_2, a_3, ldots)$ to $frac 1{a_1 + frac1{a_2 + frac1{a_3 + cdots}}}$.
– Mees de Vries
2 days ago
@MeesdeVries That explains my question, at least partially. However, how do I prove that all such $f's$ from $omega$ to $omega$ get map to irrationals via this continued fractions? What if there is a tail of $0$'s? Do they also map to irrational numbers?
– user122424
2 days ago
1
In particular, $(2, 2, 2, 2, 2, 2, ldots)$ goes to an $x$ satisfying $x = frac1{2 + x}$, so $x = sqrt2 - 1$.
– Mees de Vries
2 days ago
OK. But what about $(2,0,0,0,0,0,0....)$ ?
– user122424
2 days ago
2
2
The usual correspondence maps elements of $omega^omega$ to infinite continued fractions, i.e., irrationals in $(0,1)$. It is not clear at all what the example you give is supposed to be.
– Andrés E. Caicedo
2 days ago
The usual correspondence maps elements of $omega^omega$ to infinite continued fractions, i.e., irrationals in $(0,1)$. It is not clear at all what the example you give is supposed to be.
– Andrés E. Caicedo
2 days ago
1
1
The mapping that Noah Schweber gave in his post is not the one I was talking about. He does mention the one I was talking about in the sentence right before: it is the mapping that takes $(a_1, a_2, a_3, ldots)$ to $frac 1{a_1 + frac1{a_2 + frac1{a_3 + cdots}}}$.
– Mees de Vries
2 days ago
The mapping that Noah Schweber gave in his post is not the one I was talking about. He does mention the one I was talking about in the sentence right before: it is the mapping that takes $(a_1, a_2, a_3, ldots)$ to $frac 1{a_1 + frac1{a_2 + frac1{a_3 + cdots}}}$.
– Mees de Vries
2 days ago
@MeesdeVries That explains my question, at least partially. However, how do I prove that all such $f's$ from $omega$ to $omega$ get map to irrationals via this continued fractions? What if there is a tail of $0$'s? Do they also map to irrational numbers?
– user122424
2 days ago
@MeesdeVries That explains my question, at least partially. However, how do I prove that all such $f's$ from $omega$ to $omega$ get map to irrationals via this continued fractions? What if there is a tail of $0$'s? Do they also map to irrational numbers?
– user122424
2 days ago
1
1
In particular, $(2, 2, 2, 2, 2, 2, ldots)$ goes to an $x$ satisfying $x = frac1{2 + x}$, so $x = sqrt2 - 1$.
– Mees de Vries
2 days ago
In particular, $(2, 2, 2, 2, 2, 2, ldots)$ goes to an $x$ satisfying $x = frac1{2 + x}$, so $x = sqrt2 - 1$.
– Mees de Vries
2 days ago
OK. But what about $(2,0,0,0,0,0,0....)$ ?
– user122424
2 days ago
OK. But what about $(2,0,0,0,0,0,0....)$ ?
– user122424
2 days ago
|
show 3 more comments
1 Answer
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As noted above, you're conflating two different homeomorphisms between $omega^omega$ and $mathbb{R}setminusmathbb{Q}$. It seems you're most interested in the continued fraction map, so let me say a bit about that one.
Given an infinite sequence $A=(a_i)_{iinomega}$ of natural numbers - which in logic include $0$, so you'll see some "$1+$"s here which you won't in other contexts - the corresponding continued fraction is $$F_A=(1+a_0)+{1over (1+a_1)+{1over (1+a_2)+{1over ...}}}.$$ The following is a basic fact about continued fractions (and not a result from logic):
$F_A$ is always defined and irrational and every irrational $alpha$ is equal to $F_A$ for exactly one $A$.
See here for a proof. This implies immediately that the map $Amapsto F_A$ is a bijection from $omega^omega$ to $mathbb{R}setminus mathbb{Q}$, and with a bit of thought we can also see that it's continuous in both directions.
Note that the proof of the basic fact above is nontrivial. This is the main virtue of the bijection mentioned in my answer to your previous question: it's less natural but much easier to see that it is indeed a homeomorphism.
answered 2 days ago
Noah Schweber
122k10148284
1
@zwim I don't think you read my answer carefully - I used $1+a_i$ instead of $a_i$ to deal with precisely this issue. E.g. taking $a_0=a_1=a_2=...=0$, we get $F_A=varphi$. I believe my answer is correct as written.
– Noah Schweber
2 days ago
add a comment |
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As noted above, you're conflating two different homeomorphisms between $omega^omega$ and $mathbb{R}setminusmathbb{Q}$. It seems you're most interested in the continued fraction map, so let me say a bit about that one.
Given an infinite sequence $A=(a_i)_{iinomega}$ of natural numbers - which in logic include $0$, so you'll see some "$1+$"s here which you won't in other contexts - the corresponding continued fraction is $$F_A=(1+a_0)+{1over (1+a_1)+{1over (1+a_2)+{1over ...}}}.$$ The following is a basic fact about continued fractions (and not a result from logic):
$F_A$ is always defined and irrational and every irrational $alpha$ is equal to $F_A$ for exactly one $A$.
See here for a proof. This implies immediately that the map $Amapsto F_A$ is a bijection from $omega^omega$ to $mathbb{R}setminus mathbb{Q}$, and with a bit of thought we can also see that it's continuous in both directions.
Note that the proof of the basic fact above is nontrivial. This is the main virtue of the bijection mentioned in my answer to your previous question: it's less natural but much easier to see that it is indeed a homeomorphism.
answered 2 days ago
Noah Schweber
122k10148284
1
@zwim I don't think you read my answer carefully - I used $1+a_i$ instead of $a_i$ to deal with precisely this issue. E.g. taking $a_0=a_1=a_2=...=0$, we get $F_A=varphi$. I believe my answer is correct as written.
– Noah Schweber
2 days ago
add a comment |
As noted above, you're conflating two different homeomorphisms between $omega^omega$ and $mathbb{R}setminusmathbb{Q}$. It seems you're most interested in the continued fraction map, so let me say a bit about that one.
Given an infinite sequence $A=(a_i)_{iinomega}$ of natural numbers - which in logic include $0$, so you'll see some "$1+$"s here which you won't in other contexts - the corresponding continued fraction is $$F_A=(1+a_0)+{1over (1+a_1)+{1over (1+a_2)+{1over ...}}}.$$ The following is a basic fact about continued fractions (and not a result from logic):
$F_A$ is always defined and irrational and every irrational $alpha$ is equal to $F_A$ for exactly one $A$.
See here for a proof. This implies immediately that the map $Amapsto F_A$ is a bijection from $omega^omega$ to $mathbb{R}setminus mathbb{Q}$, and with a bit of thought we can also see that it's continuous in both directions.
Note that the proof of the basic fact above is nontrivial. This is the main virtue of the bijection mentioned in my answer to your previous question: it's less natural but much easier to see that it is indeed a homeomorphism.
answered 2 days ago
Noah Schweber
122k10148284
1
@zwim I don't think you read my answer carefully - I used $1+a_i$ instead of $a_i$ to deal with precisely this issue. E.g. taking $a_0=a_1=a_2=...=0$, we get $F_A=varphi$. I believe my answer is correct as written.
– Noah Schweber
2 days ago
add a comment |
As noted above, you're conflating two different homeomorphisms between $omega^omega$ and $mathbb{R}setminusmathbb{Q}$. It seems you're most interested in the continued fraction map, so let me say a bit about that one.
Given an infinite sequence $A=(a_i)_{iinomega}$ of natural numbers - which in logic include $0$, so you'll see some "$1+$"s here which you won't in other contexts - the corresponding continued fraction is $$F_A=(1+a_0)+{1over (1+a_1)+{1over (1+a_2)+{1over ...}}}.$$ The following is a basic fact about continued fractions (and not a result from logic):
$F_A$ is always defined and irrational and every irrational $alpha$ is equal to $F_A$ for exactly one $A$.
See here for a proof. This implies immediately that the map $Amapsto F_A$ is a bijection from $omega^omega$ to $mathbb{R}setminus mathbb{Q}$, and with a bit of thought we can also see that it's continuous in both directions.
Note that the proof of the basic fact above is nontrivial. This is the main virtue of the bijection mentioned in my answer to your previous question: it's less natural but much easier to see that it is indeed a homeomorphism.
answered 2 days ago
Noah Schweber
122k10148284
As noted above, you're conflating two different homeomorphisms between $omega^omega$ and $mathbb{R}setminusmathbb{Q}$. It seems you're most interested in the continued fraction map, so let me say a bit about that one.
Given an infinite sequence $A=(a_i)_{iinomega}$ of natural numbers - which in logic include $0$, so you'll see some "$1+$"s here which you won't in other contexts - the corresponding continued fraction is $$F_A=(1+a_0)+{1over (1+a_1)+{1over (1+a_2)+{1over ...}}}.$$ The following is a basic fact about continued fractions (and not a result from logic):
$F_A$ is always defined and irrational and every irrational $alpha$ is equal to $F_A$ for exactly one $A$.
See here for a proof. This implies immediately that the map $Amapsto F_A$ is a bijection from $omega^omega$ to $mathbb{R}setminus mathbb{Q}$, and with a bit of thought we can also see that it's continuous in both directions.
Note that the proof of the basic fact above is nontrivial. This is the main virtue of the bijection mentioned in my answer to your previous question: it's less natural but much easier to see that it is indeed a homeomorphism.
answered 2 days ago
Noah Schweber
122k10148284
edited 2 days ago
answered 2 days ago
Noah Schweber
122k10148284
answered 2 days ago
Noah Schweber
122k10148284
answered 2 days ago
Noah Schweber
122k10148284
122k10148284
1
@zwim I don't think you read my answer carefully - I used $1+a_i$ instead of $a_i$ to deal with precisely this issue. E.g. taking $a_0=a_1=a_2=...=0$, we get $F_A=varphi$. I believe my answer is correct as written.
– Noah Schweber
2 days ago
add a comment |
1
@zwim I don't think you read my answer carefully - I used $1+a_i$ instead of $a_i$ to deal with precisely this issue. E.g. taking $a_0=a_1=a_2=...=0$, we get $F_A=varphi$. I believe my answer is correct as written.
– Noah Schweber
2 days ago
1
1
@zwim I don't think you read my answer carefully - I used $1+a_i$ instead of $a_i$ to deal with precisely this issue. E.g. taking $a_0=a_1=a_2=...=0$, we get $F_A=varphi$. I believe my answer is correct as written.
– Noah Schweber
2 days ago
@zwim I don't think you read my answer carefully - I used $1+a_i$ instead of $a_i$ to deal with precisely this issue. E.g. taking $a_0=a_1=a_2=...=0$, we get $F_A=varphi$. I believe my answer is correct as written.
– Noah Schweber
2 days ago
add a comment |
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2
The usual correspondence maps elements of $omega^omega$ to infinite continued fractions, i.e., irrationals in $(0,1)$. It is not clear at all what the example you give is supposed to be.
– Andrés E. Caicedo
2 days ago
1
The mapping that Noah Schweber gave in his post is not the one I was talking about. He does mention the one I was talking about in the sentence right before: it is the mapping that takes $(a_1, a_2, a_3, ldots)$ to $frac 1{a_1 + frac1{a_2 + frac1{a_3 + cdots}}}$.
– Mees de Vries
2 days ago
@MeesdeVries That explains my question, at least partially. However, how do I prove that all such $f's$ from $omega$ to $omega$ get map to irrationals via this continued fractions? What if there is a tail of $0$'s? Do they also map to irrational numbers?
– user122424
2 days ago
1
In particular, $(2, 2, 2, 2, 2, 2, ldots)$ goes to an $x$ satisfying $x = frac1{2 + x}$, so $x = sqrt2 - 1$.
– Mees de Vries
2 days ago
OK. But what about $(2,0,0,0,0,0,0....)$ ?
– user122424
2 days ago