$omega^omega$ correspondence with $mathbb R$-irrationality
$begingroup$
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 Jan 3 at 16:31
user122424user122424
1,1462716
$endgroup$
|
show 3 more comments
$begingroup$
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 Jan 3 at 16:31
user122424user122424
1,1462716
$endgroup$
2
$begingroup$
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.
$endgroup$
– Andrés E. Caicedo
Jan 3 at 16:40
1
$begingroup$
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 $frac 1{a_1 + frac1{a_2 + frac1{a_3 + cdots}}}$.
$endgroup$
– Mees de Vries
Jan 3 at 16:42
$begingroup$
@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?
$endgroup$
– user122424
Jan 3 at 16:48
1
$begingroup$
In particular, $(2, 2, 2, 2, 2, 2, ldots)$ goes to an $x$ satisfying $x = frac1{2 + x}$, so $x = sqrt2 - 1$.
$endgroup$
– Mees de Vries
Jan 3 at 16:49
1
$begingroup$
@zwim The OP is using the notation correctly: ordinals are sets, and in particular $omega$ is the set of all finite ordinals, and is generally used instead of "$mathbb{N}$" in logic.
$endgroup$
– Noah Schweber
Jan 3 at 17:45
|
show 3 more comments
$begingroup$
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 Jan 3 at 16:31
user122424user122424
1,1462716
$endgroup$
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 Jan 3 at 16:31
user122424user122424
1,1462716
asked Jan 3 at 16:31
user122424user122424
1,1462716
edited Jan 3 at 16:39
user122424
asked Jan 3 at 16:31
user122424user122424
1,1462716
asked Jan 3 at 16:31
user122424user122424
1,1462716
asked Jan 3 at 16:31
user122424user122424
1,1462716
1,1462716
2
$begingroup$
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.
$endgroup$
– Andrés E. Caicedo
Jan 3 at 16:40
1
$begingroup$
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 $frac 1{a_1 + frac1{a_2 + frac1{a_3 + cdots}}}$.
$endgroup$
– Mees de Vries
Jan 3 at 16:42
$begingroup$
@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?
$endgroup$
– user122424
Jan 3 at 16:48
1
$begingroup$
In particular, $(2, 2, 2, 2, 2, 2, ldots)$ goes to an $x$ satisfying $x = frac1{2 + x}$, so $x = sqrt2 - 1$.
$endgroup$
– Mees de Vries
Jan 3 at 16:49
1
$begingroup$
@zwim The OP is using the notation correctly: ordinals are sets, and in particular $omega$ is the set of all finite ordinals, and is generally used instead of "$mathbb{N}$" in logic.
$endgroup$
– Noah Schweber
Jan 3 at 17:45
|
show 3 more comments
2
$begingroup$
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.
$endgroup$
– Andrés E. Caicedo
Jan 3 at 16:40
1
$begingroup$
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 $frac 1{a_1 + frac1{a_2 + frac1{a_3 + cdots}}}$.
$endgroup$
– Mees de Vries
Jan 3 at 16:42
$begingroup$
@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?
$endgroup$
– user122424
Jan 3 at 16:48
1
$begingroup$
In particular, $(2, 2, 2, 2, 2, 2, ldots)$ goes to an $x$ satisfying $x = frac1{2 + x}$, so $x = sqrt2 - 1$.
$endgroup$
– Mees de Vries
Jan 3 at 16:49
1
$begingroup$
@zwim The OP is using the notation correctly: ordinals are sets, and in particular $omega$ is the set of all finite ordinals, and is generally used instead of "$mathbb{N}$" in logic.
$endgroup$
– Noah Schweber
Jan 3 at 17:45
2
2
$begingroup$
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.
$endgroup$
– Andrés E. Caicedo
Jan 3 at 16:40
$begingroup$
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.
$endgroup$
– Andrés E. Caicedo
Jan 3 at 16:40
1
1
$begingroup$
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 $frac 1{a_1 + frac1{a_2 + frac1{a_3 + cdots}}}$.
$endgroup$
– Mees de Vries
Jan 3 at 16:42
$begingroup$
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 $frac 1{a_1 + frac1{a_2 + frac1{a_3 + cdots}}}$.
$endgroup$
– Mees de Vries
Jan 3 at 16:42
$begingroup$
@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?
$endgroup$
– user122424
Jan 3 at 16:48
$begingroup$
@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?
$endgroup$
– user122424
Jan 3 at 16:48
1
1
$begingroup$
In particular, $(2, 2, 2, 2, 2, 2, ldots)$ goes to an $x$ satisfying $x = frac1{2 + x}$, so $x = sqrt2 - 1$.
$endgroup$
– Mees de Vries
Jan 3 at 16:49
$begingroup$
In particular, $(2, 2, 2, 2, 2, 2, ldots)$ goes to an $x$ satisfying $x = frac1{2 + x}$, so $x = sqrt2 - 1$.
$endgroup$
– Mees de Vries
Jan 3 at 16:49
1
1
$begingroup$
@zwim The OP is using the notation correctly: ordinals are sets, and in particular $omega$ is the set of all finite ordinals, and is generally used instead of "$mathbb{N}$" in logic.
$endgroup$
– Noah Schweber
Jan 3 at 17:45
$begingroup$
@zwim The OP is using the notation correctly: ordinals are sets, and in particular $omega$ is the set of all finite ordinals, and is generally used instead of "$mathbb{N}$" in logic.
$endgroup$
– Noah Schweber
Jan 3 at 17:45
|
show 3 more comments
1 Answer
1
active
oldest
votes
$begingroup$
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 Jan 3 at 17:49
Noah SchweberNoah Schweber
126k10150289
$endgroup$
1
$begingroup$
@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.
$endgroup$
– Noah Schweber
Jan 3 at 20:29
$begingroup$
Is there a continuous bijection between $omega^omega$ and entire $mathbb R$?
$endgroup$
– user122424
Jan 19 at 19:06
$begingroup$
@user122424 Yes - indeed, if $mathcal{X}$ is any "interesting" (= nonempty and perfect) Polish space, then there is a continuous bijection from $omega^omega$ to $mathcal{X}$. I believe this was first proved by Sierpinski; it should be in Kechris' descriptive set theory book. Of course, order matters here - as I noted above, there is not even any nonconstant continuous map from $mathbb{R}$ to $omega^omega$.
$endgroup$
– Noah Schweber
Jan 19 at 19:23
add a comment |
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$begingroup$
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 Jan 3 at 17:49
Noah SchweberNoah Schweber
126k10150289
$endgroup$
1
$begingroup$
@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.
$endgroup$
– Noah Schweber
Jan 3 at 20:29
$begingroup$
Is there a continuous bijection between $omega^omega$ and entire $mathbb R$?
$endgroup$
– user122424
Jan 19 at 19:06
$begingroup$
@user122424 Yes - indeed, if $mathcal{X}$ is any "interesting" (= nonempty and perfect) Polish space, then there is a continuous bijection from $omega^omega$ to $mathcal{X}$. I believe this was first proved by Sierpinski; it should be in Kechris' descriptive set theory book. Of course, order matters here - as I noted above, there is not even any nonconstant continuous map from $mathbb{R}$ to $omega^omega$.
$endgroup$
– Noah Schweber
Jan 19 at 19:23
add a comment |
$begingroup$
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 Jan 3 at 17:49
Noah SchweberNoah Schweber
126k10150289
$endgroup$
1
$begingroup$
@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.
$endgroup$
– Noah Schweber
Jan 3 at 20:29
$begingroup$
Is there a continuous bijection between $omega^omega$ and entire $mathbb R$?
$endgroup$
– user122424
Jan 19 at 19:06
$begingroup$
@user122424 Yes - indeed, if $mathcal{X}$ is any "interesting" (= nonempty and perfect) Polish space, then there is a continuous bijection from $omega^omega$ to $mathcal{X}$. I believe this was first proved by Sierpinski; it should be in Kechris' descriptive set theory book. Of course, order matters here - as I noted above, there is not even any nonconstant continuous map from $mathbb{R}$ to $omega^omega$.
$endgroup$
– Noah Schweber
Jan 19 at 19:23
add a comment |
$begingroup$
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 Jan 3 at 17:49
Noah SchweberNoah Schweber
126k10150289
$endgroup$
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 Jan 3 at 17:49
Noah SchweberNoah Schweber
126k10150289
edited Jan 3 at 17:57
answered Jan 3 at 17:49
Noah SchweberNoah Schweber
126k10150289
answered Jan 3 at 17:49
Noah SchweberNoah Schweber
126k10150289
answered Jan 3 at 17:49
Noah SchweberNoah Schweber
126k10150289
126k10150289
1
$begingroup$
@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.
$endgroup$
– Noah Schweber
Jan 3 at 20:29
$begingroup$
Is there a continuous bijection between $omega^omega$ and entire $mathbb R$?
$endgroup$
– user122424
Jan 19 at 19:06
$begingroup$
@user122424 Yes - indeed, if $mathcal{X}$ is any "interesting" (= nonempty and perfect) Polish space, then there is a continuous bijection from $omega^omega$ to $mathcal{X}$. I believe this was first proved by Sierpinski; it should be in Kechris' descriptive set theory book. Of course, order matters here - as I noted above, there is not even any nonconstant continuous map from $mathbb{R}$ to $omega^omega$.
$endgroup$
– Noah Schweber
Jan 19 at 19:23
add a comment |
1
$begingroup$
@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.
$endgroup$
– Noah Schweber
Jan 3 at 20:29
$begingroup$
Is there a continuous bijection between $omega^omega$ and entire $mathbb R$?
$endgroup$
– user122424
Jan 19 at 19:06
$begingroup$
@user122424 Yes - indeed, if $mathcal{X}$ is any "interesting" (= nonempty and perfect) Polish space, then there is a continuous bijection from $omega^omega$ to $mathcal{X}$. I believe this was first proved by Sierpinski; it should be in Kechris' descriptive set theory book. Of course, order matters here - as I noted above, there is not even any nonconstant continuous map from $mathbb{R}$ to $omega^omega$.
$endgroup$
– Noah Schweber
Jan 19 at 19:23
1
1
$begingroup$
@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.
$endgroup$
– Noah Schweber
Jan 3 at 20:29
$begingroup$
@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.
$endgroup$
– Noah Schweber
Jan 3 at 20:29
$begingroup$
Is there a continuous bijection between $omega^omega$ and entire $mathbb R$?
$endgroup$
– user122424
Jan 19 at 19:06
$begingroup$
Is there a continuous bijection between $omega^omega$ and entire $mathbb R$?
$endgroup$
– user122424
Jan 19 at 19:06
$begingroup$
@user122424 Yes - indeed, if $mathcal{X}$ is any "interesting" (= nonempty and perfect) Polish space, then there is a continuous bijection from $omega^omega$ to $mathcal{X}$. I believe this was first proved by Sierpinski; it should be in Kechris' descriptive set theory book. Of course, order matters here - as I noted above, there is not even any nonconstant continuous map from $mathbb{R}$ to $omega^omega$.
$endgroup$
– Noah Schweber
Jan 19 at 19:23
$begingroup$
@user122424 Yes - indeed, if $mathcal{X}$ is any "interesting" (= nonempty and perfect) Polish space, then there is a continuous bijection from $omega^omega$ to $mathcal{X}$. I believe this was first proved by Sierpinski; it should be in Kechris' descriptive set theory book. Of course, order matters here - as I noted above, there is not even any nonconstant continuous map from $mathbb{R}$ to $omega^omega$.
$endgroup$
– Noah Schweber
Jan 19 at 19:23
add a comment |
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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.
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– Andrés E. Caicedo
Jan 3 at 16:40
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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 $frac 1{a_1 + frac1{a_2 + frac1{a_3 + cdots}}}$.
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– Mees de Vries
Jan 3 at 16:42
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@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?
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– user122424
Jan 3 at 16:48
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In particular, $(2, 2, 2, 2, 2, 2, ldots)$ goes to an $x$ satisfying $x = frac1{2 + x}$, so $x = sqrt2 - 1$.
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– Mees de Vries
Jan 3 at 16:49
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@zwim The OP is using the notation correctly: ordinals are sets, and in particular $omega$ is the set of all finite ordinals, and is generally used instead of "$mathbb{N}$" in logic.
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– Noah Schweber
Jan 3 at 17:45