How exactly does the oracle for a well-order of order type $omega_1^textCK$ operate?
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The concept of an oracle for Turing machines assumes that the oracle answers Yes/No to a particular question $Q$, assuming that $Q$ is formulated as a bitstring on the oracle tape (instead of answering Yes/No, the oracle can generate the answer as a bitstring on the oracle tape).
Let $O$ denote an oracle that âÂÂencodes a well-order of order type $omega_1^textCK$â or, in other words, âÂÂhas access to the well-order relation describing the well-ordering of $omega_1^textCK$ in terms of $mathbbN$âÂÂ. Can anyone explain what the set of questions for $O$ is and how a particular $Q$ (from the set of all possible questions) is converted to a bitstring?
logic ordinals turing-machines oracles
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up vote
1
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favorite
The concept of an oracle for Turing machines assumes that the oracle answers Yes/No to a particular question $Q$, assuming that $Q$ is formulated as a bitstring on the oracle tape (instead of answering Yes/No, the oracle can generate the answer as a bitstring on the oracle tape).
Let $O$ denote an oracle that âÂÂencodes a well-order of order type $omega_1^textCK$â or, in other words, âÂÂhas access to the well-order relation describing the well-ordering of $omega_1^textCK$ in terms of $mathbbN$âÂÂ. Can anyone explain what the set of questions for $O$ is and how a particular $Q$ (from the set of all possible questions) is converted to a bitstring?
logic ordinals turing-machines oracles
Sorry I think the phrase "well-ordering of $omega_1^textCK$ in terms of $mathbbN$" is off/incorrect. I know I used it in my recent answer/question (it might be a bit awkward to bump them for correction I suppose), but for the correct phrase see the beginning of my answer here.
â SSequence
Sep 8 at 15:57
OK I have corrected the misleading phrase from my previous (recent) answer/question too.
â SSequence
Sep 8 at 16:12
add a comment |Â
up vote
1
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up vote
1
down vote
favorite
The concept of an oracle for Turing machines assumes that the oracle answers Yes/No to a particular question $Q$, assuming that $Q$ is formulated as a bitstring on the oracle tape (instead of answering Yes/No, the oracle can generate the answer as a bitstring on the oracle tape).
Let $O$ denote an oracle that âÂÂencodes a well-order of order type $omega_1^textCK$â or, in other words, âÂÂhas access to the well-order relation describing the well-ordering of $omega_1^textCK$ in terms of $mathbbN$âÂÂ. Can anyone explain what the set of questions for $O$ is and how a particular $Q$ (from the set of all possible questions) is converted to a bitstring?
logic ordinals turing-machines oracles
The concept of an oracle for Turing machines assumes that the oracle answers Yes/No to a particular question $Q$, assuming that $Q$ is formulated as a bitstring on the oracle tape (instead of answering Yes/No, the oracle can generate the answer as a bitstring on the oracle tape).
Let $O$ denote an oracle that âÂÂencodes a well-order of order type $omega_1^textCK$â or, in other words, âÂÂhas access to the well-order relation describing the well-ordering of $omega_1^textCK$ in terms of $mathbbN$âÂÂ. Can anyone explain what the set of questions for $O$ is and how a particular $Q$ (from the set of all possible questions) is converted to a bitstring?
logic ordinals turing-machines oracles
logic ordinals turing-machines oracles
edited Sep 8 at 13:25
asked Sep 8 at 13:03
lyrically wicked
1677
1677
Sorry I think the phrase "well-ordering of $omega_1^textCK$ in terms of $mathbbN$" is off/incorrect. I know I used it in my recent answer/question (it might be a bit awkward to bump them for correction I suppose), but for the correct phrase see the beginning of my answer here.
â SSequence
Sep 8 at 15:57
OK I have corrected the misleading phrase from my previous (recent) answer/question too.
â SSequence
Sep 8 at 16:12
add a comment |Â
Sorry I think the phrase "well-ordering of $omega_1^textCK$ in terms of $mathbbN$" is off/incorrect. I know I used it in my recent answer/question (it might be a bit awkward to bump them for correction I suppose), but for the correct phrase see the beginning of my answer here.
â SSequence
Sep 8 at 15:57
OK I have corrected the misleading phrase from my previous (recent) answer/question too.
â SSequence
Sep 8 at 16:12
Sorry I think the phrase "well-ordering of $omega_1^textCK$ in terms of $mathbbN$" is off/incorrect. I know I used it in my recent answer/question (it might be a bit awkward to bump them for correction I suppose), but for the correct phrase see the beginning of my answer here.
â SSequence
Sep 8 at 15:57
Sorry I think the phrase "well-ordering of $omega_1^textCK$ in terms of $mathbbN$" is off/incorrect. I know I used it in my recent answer/question (it might be a bit awkward to bump them for correction I suppose), but for the correct phrase see the beginning of my answer here.
â SSequence
Sep 8 at 15:57
OK I have corrected the misleading phrase from my previous (recent) answer/question too.
â SSequence
Sep 8 at 16:12
OK I have corrected the misleading phrase from my previous (recent) answer/question too.
â SSequence
Sep 8 at 16:12
add a comment |Â
2 Answers
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You're right that we can't really use a structure as an oracle, per se; instead, we look at copies of the structure, that is, ways of representing the structure as a set of natural numbers. Incidentally, everything I say below applies to general countable structures.
A relation $Esubseteqomegatimesomega$ is a copy of $omega_1^CK$ if it defines a structure isomorphic to $omega_1^CK$; that is, if $(omega;E)congomega_1^CK$. EDIT: If you prefer, you can think of a copy as a pair $(A; E)$ with $Asubseteqomega$ and $Esubseteqomega^2$ which is isomorphic to $omega_1^CK$; for our purposes, this won't make a difference - see the comments below. It might be easier to think about this in a simpler case: consider the relation given by $mEn$ iff
$m$ is odd and $n$ is even, or
$m,n$ have the same parity and $m<n$.
This is a copy of the ordinal $omega+omega$: it's describing the linear order $$1prec 3prec 5prec 7prec 9prec ...prec 0prec 2prec 4prec 6prec 8prec ...$$
Copies can be used as oracles in the obvious way, since a copy is just a set of pairs of natural numbers. Of course, there is no unique copy of $omega_1^CK$, although we may argue that one copy or another is particularly "natural." The types of questions we ask in this context range across all copies of a structure; for example, the statement "$omega_1^CK$ is not computable" is shorthand for "there is no computable copy of $omega_1^CK$."
Just to be sure about this (and perhaps for the benefit of other readers also), if we have $a,b in mathbbN$ but either one or both of the elements $a$ and $b$ are NOT "used" in the copy ......... then, based on the formal def. of copy, in that case do we have $(a,b) in E$ or not?
â SSequence
Sep 8 at 17:14
And I guess if all elements have to be "used" in a copy (by definition), then that renders the question in my previous comment redundant.
â SSequence
Sep 8 at 17:49
1
@SSequence It ultimately doesn't really make a difference. I think the actual right thing to do is to say that a copy is a pair $(A,E)$ where $Asubseteqomega$ is the domain and $Esubseteq A^2$; however, in this case it doesn't matter, since from $E$ you can already recover $A$ (fix some $iin A$; then $jin A$ iff $i=j$ or $(i,j)in E$ or $(j,i)in E$. So it's not a real issue. (Note that this trick doesn't work for all structures necessarily, so it can be important, but here it's not an issue.)
â Noah Schweber
Sep 8 at 18:06
Suppose that we have chosen some structure as a particular copy of $omega_1^textCK$ that the oracle has access to. Can I assume that, technically, all questions can be given for the oracle in the form of a pair of natural numbers $a$ and $b$, and the oracle answers Yes if $a$ precedes $b$ and No if $a$ does not precede $b$?
â lyrically wicked
Sep 9 at 4:05
@lyricallywicked Yes. Although I would use the word "structure" to refer to the "pre-copy thing" as opposed to the oracle, or rather one of the oracles, associated to it - that is, $omega_1^CK$ is the structure and the copies are the things we can use as oracles.
â Noah Schweber
Sep 9 at 4:18
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Since you used my phrase too, let me just mention first that a good way to say the same thing might be "well-ordering of/on $mathbbN$ with order-type $omega_1^CK$". To be fair, I can relate to your question very well because if you don't have a formal math background and have never read either an (i) elementary set theory book cover-to-cover (ii) an introductory formal proof book which also covers various types of "ordering" ..... then I think you might just have never encountered many of these terms. Up until I asked first few questions here, I didn't know how to express properly/formally (for which I could simply use that phrase) what I had been playing around with for few years. And it did cause a lot of difficulty for me in asking questions.
Anyway, here is my "informal/naive" take on this (I think an expert answer might be more authoritative place to look, but this might help you in getting some understanding of things). Informally, you can think of "well-ordering of/on $mathbbN$ with order-type $omega_1^CK$" simply as "inducing" a bijection between $omega_1^CK$ and $mathbbN$ (probably if one has to be more formal, one might word it a little differently?). More generally, you can replace $omega_1^CK$ with any (non-finite) $alpha$ that satisfies $alpha <omega_1$ ($omega_1$ being first uncountable).
If you denote the corresponding (unique/unambiguous) bijective function as $f:alpha rightarrow mathbbN$ then the inverse of this $f^-1:mathbbN rightarrow alpha$ is also of course a bijective function. Now what is the well-order relation corresponding to this well-ordering? It can be simply defined by the function $lessequals:mathbbN^2 rightarrow 0,1$ such that:
$lessequals(x,y)=$ truth value of $f^-1(x) le f^-1(y)$
Note that $f^-1(x) le f^-1(y)$ is an ordinal comparison. You can also define the function $lessthan:mathbbN^2 rightarrow 0,1$ as:
$lessthan(x,y)=$ truth value of $f^-1(x) < f^-1(y)$
In the given context, it is the function $lessequals$ that can be called the well-order relation (well I am going by description in the wiki link here honestly). But, at any rate, one can observe that both these functions are computationally reducible to each other (somewhat trivially).
One can say that an ordinal $alpha$ is "computable/recursive" if there exists a well-ordering of $mathbbN$ with order-type $alpha$ (adding a minor exception for finite ordinals I suppose) such that the corresponding
well-order relation is computable. $omega^CK_1$ is defined as the smallest ordinal that is not computable/recursive.
There is one important fact here that, while somewhat trivial, is important to know. If $alpha$ is a recursive ordinal than so is any ordinal $beta$ (where $beta < alpha$).
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
You're right that we can't really use a structure as an oracle, per se; instead, we look at copies of the structure, that is, ways of representing the structure as a set of natural numbers. Incidentally, everything I say below applies to general countable structures.
A relation $Esubseteqomegatimesomega$ is a copy of $omega_1^CK$ if it defines a structure isomorphic to $omega_1^CK$; that is, if $(omega;E)congomega_1^CK$. EDIT: If you prefer, you can think of a copy as a pair $(A; E)$ with $Asubseteqomega$ and $Esubseteqomega^2$ which is isomorphic to $omega_1^CK$; for our purposes, this won't make a difference - see the comments below. It might be easier to think about this in a simpler case: consider the relation given by $mEn$ iff
$m$ is odd and $n$ is even, or
$m,n$ have the same parity and $m<n$.
This is a copy of the ordinal $omega+omega$: it's describing the linear order $$1prec 3prec 5prec 7prec 9prec ...prec 0prec 2prec 4prec 6prec 8prec ...$$
Copies can be used as oracles in the obvious way, since a copy is just a set of pairs of natural numbers. Of course, there is no unique copy of $omega_1^CK$, although we may argue that one copy or another is particularly "natural." The types of questions we ask in this context range across all copies of a structure; for example, the statement "$omega_1^CK$ is not computable" is shorthand for "there is no computable copy of $omega_1^CK$."
Just to be sure about this (and perhaps for the benefit of other readers also), if we have $a,b in mathbbN$ but either one or both of the elements $a$ and $b$ are NOT "used" in the copy ......... then, based on the formal def. of copy, in that case do we have $(a,b) in E$ or not?
â SSequence
Sep 8 at 17:14
And I guess if all elements have to be "used" in a copy (by definition), then that renders the question in my previous comment redundant.
â SSequence
Sep 8 at 17:49
1
@SSequence It ultimately doesn't really make a difference. I think the actual right thing to do is to say that a copy is a pair $(A,E)$ where $Asubseteqomega$ is the domain and $Esubseteq A^2$; however, in this case it doesn't matter, since from $E$ you can already recover $A$ (fix some $iin A$; then $jin A$ iff $i=j$ or $(i,j)in E$ or $(j,i)in E$. So it's not a real issue. (Note that this trick doesn't work for all structures necessarily, so it can be important, but here it's not an issue.)
â Noah Schweber
Sep 8 at 18:06
Suppose that we have chosen some structure as a particular copy of $omega_1^textCK$ that the oracle has access to. Can I assume that, technically, all questions can be given for the oracle in the form of a pair of natural numbers $a$ and $b$, and the oracle answers Yes if $a$ precedes $b$ and No if $a$ does not precede $b$?
â lyrically wicked
Sep 9 at 4:05
@lyricallywicked Yes. Although I would use the word "structure" to refer to the "pre-copy thing" as opposed to the oracle, or rather one of the oracles, associated to it - that is, $omega_1^CK$ is the structure and the copies are the things we can use as oracles.
â Noah Schweber
Sep 9 at 4:18
 |Â
show 1 more comment
up vote
2
down vote
accepted
You're right that we can't really use a structure as an oracle, per se; instead, we look at copies of the structure, that is, ways of representing the structure as a set of natural numbers. Incidentally, everything I say below applies to general countable structures.
A relation $Esubseteqomegatimesomega$ is a copy of $omega_1^CK$ if it defines a structure isomorphic to $omega_1^CK$; that is, if $(omega;E)congomega_1^CK$. EDIT: If you prefer, you can think of a copy as a pair $(A; E)$ with $Asubseteqomega$ and $Esubseteqomega^2$ which is isomorphic to $omega_1^CK$; for our purposes, this won't make a difference - see the comments below. It might be easier to think about this in a simpler case: consider the relation given by $mEn$ iff
$m$ is odd and $n$ is even, or
$m,n$ have the same parity and $m<n$.
This is a copy of the ordinal $omega+omega$: it's describing the linear order $$1prec 3prec 5prec 7prec 9prec ...prec 0prec 2prec 4prec 6prec 8prec ...$$
Copies can be used as oracles in the obvious way, since a copy is just a set of pairs of natural numbers. Of course, there is no unique copy of $omega_1^CK$, although we may argue that one copy or another is particularly "natural." The types of questions we ask in this context range across all copies of a structure; for example, the statement "$omega_1^CK$ is not computable" is shorthand for "there is no computable copy of $omega_1^CK$."
Just to be sure about this (and perhaps for the benefit of other readers also), if we have $a,b in mathbbN$ but either one or both of the elements $a$ and $b$ are NOT "used" in the copy ......... then, based on the formal def. of copy, in that case do we have $(a,b) in E$ or not?
â SSequence
Sep 8 at 17:14
And I guess if all elements have to be "used" in a copy (by definition), then that renders the question in my previous comment redundant.
â SSequence
Sep 8 at 17:49
1
@SSequence It ultimately doesn't really make a difference. I think the actual right thing to do is to say that a copy is a pair $(A,E)$ where $Asubseteqomega$ is the domain and $Esubseteq A^2$; however, in this case it doesn't matter, since from $E$ you can already recover $A$ (fix some $iin A$; then $jin A$ iff $i=j$ or $(i,j)in E$ or $(j,i)in E$. So it's not a real issue. (Note that this trick doesn't work for all structures necessarily, so it can be important, but here it's not an issue.)
â Noah Schweber
Sep 8 at 18:06
Suppose that we have chosen some structure as a particular copy of $omega_1^textCK$ that the oracle has access to. Can I assume that, technically, all questions can be given for the oracle in the form of a pair of natural numbers $a$ and $b$, and the oracle answers Yes if $a$ precedes $b$ and No if $a$ does not precede $b$?
â lyrically wicked
Sep 9 at 4:05
@lyricallywicked Yes. Although I would use the word "structure" to refer to the "pre-copy thing" as opposed to the oracle, or rather one of the oracles, associated to it - that is, $omega_1^CK$ is the structure and the copies are the things we can use as oracles.
â Noah Schweber
Sep 9 at 4:18
 |Â
show 1 more comment
up vote
2
down vote
accepted
up vote
2
down vote
accepted
You're right that we can't really use a structure as an oracle, per se; instead, we look at copies of the structure, that is, ways of representing the structure as a set of natural numbers. Incidentally, everything I say below applies to general countable structures.
A relation $Esubseteqomegatimesomega$ is a copy of $omega_1^CK$ if it defines a structure isomorphic to $omega_1^CK$; that is, if $(omega;E)congomega_1^CK$. EDIT: If you prefer, you can think of a copy as a pair $(A; E)$ with $Asubseteqomega$ and $Esubseteqomega^2$ which is isomorphic to $omega_1^CK$; for our purposes, this won't make a difference - see the comments below. It might be easier to think about this in a simpler case: consider the relation given by $mEn$ iff
$m$ is odd and $n$ is even, or
$m,n$ have the same parity and $m<n$.
This is a copy of the ordinal $omega+omega$: it's describing the linear order $$1prec 3prec 5prec 7prec 9prec ...prec 0prec 2prec 4prec 6prec 8prec ...$$
Copies can be used as oracles in the obvious way, since a copy is just a set of pairs of natural numbers. Of course, there is no unique copy of $omega_1^CK$, although we may argue that one copy or another is particularly "natural." The types of questions we ask in this context range across all copies of a structure; for example, the statement "$omega_1^CK$ is not computable" is shorthand for "there is no computable copy of $omega_1^CK$."
You're right that we can't really use a structure as an oracle, per se; instead, we look at copies of the structure, that is, ways of representing the structure as a set of natural numbers. Incidentally, everything I say below applies to general countable structures.
A relation $Esubseteqomegatimesomega$ is a copy of $omega_1^CK$ if it defines a structure isomorphic to $omega_1^CK$; that is, if $(omega;E)congomega_1^CK$. EDIT: If you prefer, you can think of a copy as a pair $(A; E)$ with $Asubseteqomega$ and $Esubseteqomega^2$ which is isomorphic to $omega_1^CK$; for our purposes, this won't make a difference - see the comments below. It might be easier to think about this in a simpler case: consider the relation given by $mEn$ iff
$m$ is odd and $n$ is even, or
$m,n$ have the same parity and $m<n$.
This is a copy of the ordinal $omega+omega$: it's describing the linear order $$1prec 3prec 5prec 7prec 9prec ...prec 0prec 2prec 4prec 6prec 8prec ...$$
Copies can be used as oracles in the obvious way, since a copy is just a set of pairs of natural numbers. Of course, there is no unique copy of $omega_1^CK$, although we may argue that one copy or another is particularly "natural." The types of questions we ask in this context range across all copies of a structure; for example, the statement "$omega_1^CK$ is not computable" is shorthand for "there is no computable copy of $omega_1^CK$."
edited Sep 9 at 4:39
answered Sep 8 at 16:36
Noah Schweber
113k9143267
113k9143267
Just to be sure about this (and perhaps for the benefit of other readers also), if we have $a,b in mathbbN$ but either one or both of the elements $a$ and $b$ are NOT "used" in the copy ......... then, based on the formal def. of copy, in that case do we have $(a,b) in E$ or not?
â SSequence
Sep 8 at 17:14
And I guess if all elements have to be "used" in a copy (by definition), then that renders the question in my previous comment redundant.
â SSequence
Sep 8 at 17:49
1
@SSequence It ultimately doesn't really make a difference. I think the actual right thing to do is to say that a copy is a pair $(A,E)$ where $Asubseteqomega$ is the domain and $Esubseteq A^2$; however, in this case it doesn't matter, since from $E$ you can already recover $A$ (fix some $iin A$; then $jin A$ iff $i=j$ or $(i,j)in E$ or $(j,i)in E$. So it's not a real issue. (Note that this trick doesn't work for all structures necessarily, so it can be important, but here it's not an issue.)
â Noah Schweber
Sep 8 at 18:06
Suppose that we have chosen some structure as a particular copy of $omega_1^textCK$ that the oracle has access to. Can I assume that, technically, all questions can be given for the oracle in the form of a pair of natural numbers $a$ and $b$, and the oracle answers Yes if $a$ precedes $b$ and No if $a$ does not precede $b$?
â lyrically wicked
Sep 9 at 4:05
@lyricallywicked Yes. Although I would use the word "structure" to refer to the "pre-copy thing" as opposed to the oracle, or rather one of the oracles, associated to it - that is, $omega_1^CK$ is the structure and the copies are the things we can use as oracles.
â Noah Schweber
Sep 9 at 4:18
 |Â
show 1 more comment
Just to be sure about this (and perhaps for the benefit of other readers also), if we have $a,b in mathbbN$ but either one or both of the elements $a$ and $b$ are NOT "used" in the copy ......... then, based on the formal def. of copy, in that case do we have $(a,b) in E$ or not?
â SSequence
Sep 8 at 17:14
And I guess if all elements have to be "used" in a copy (by definition), then that renders the question in my previous comment redundant.
â SSequence
Sep 8 at 17:49
1
@SSequence It ultimately doesn't really make a difference. I think the actual right thing to do is to say that a copy is a pair $(A,E)$ where $Asubseteqomega$ is the domain and $Esubseteq A^2$; however, in this case it doesn't matter, since from $E$ you can already recover $A$ (fix some $iin A$; then $jin A$ iff $i=j$ or $(i,j)in E$ or $(j,i)in E$. So it's not a real issue. (Note that this trick doesn't work for all structures necessarily, so it can be important, but here it's not an issue.)
â Noah Schweber
Sep 8 at 18:06
Suppose that we have chosen some structure as a particular copy of $omega_1^textCK$ that the oracle has access to. Can I assume that, technically, all questions can be given for the oracle in the form of a pair of natural numbers $a$ and $b$, and the oracle answers Yes if $a$ precedes $b$ and No if $a$ does not precede $b$?
â lyrically wicked
Sep 9 at 4:05
@lyricallywicked Yes. Although I would use the word "structure" to refer to the "pre-copy thing" as opposed to the oracle, or rather one of the oracles, associated to it - that is, $omega_1^CK$ is the structure and the copies are the things we can use as oracles.
â Noah Schweber
Sep 9 at 4:18
Just to be sure about this (and perhaps for the benefit of other readers also), if we have $a,b in mathbbN$ but either one or both of the elements $a$ and $b$ are NOT "used" in the copy ......... then, based on the formal def. of copy, in that case do we have $(a,b) in E$ or not?
â SSequence
Sep 8 at 17:14
Just to be sure about this (and perhaps for the benefit of other readers also), if we have $a,b in mathbbN$ but either one or both of the elements $a$ and $b$ are NOT "used" in the copy ......... then, based on the formal def. of copy, in that case do we have $(a,b) in E$ or not?
â SSequence
Sep 8 at 17:14
And I guess if all elements have to be "used" in a copy (by definition), then that renders the question in my previous comment redundant.
â SSequence
Sep 8 at 17:49
And I guess if all elements have to be "used" in a copy (by definition), then that renders the question in my previous comment redundant.
â SSequence
Sep 8 at 17:49
1
1
@SSequence It ultimately doesn't really make a difference. I think the actual right thing to do is to say that a copy is a pair $(A,E)$ where $Asubseteqomega$ is the domain and $Esubseteq A^2$; however, in this case it doesn't matter, since from $E$ you can already recover $A$ (fix some $iin A$; then $jin A$ iff $i=j$ or $(i,j)in E$ or $(j,i)in E$. So it's not a real issue. (Note that this trick doesn't work for all structures necessarily, so it can be important, but here it's not an issue.)
â Noah Schweber
Sep 8 at 18:06
@SSequence It ultimately doesn't really make a difference. I think the actual right thing to do is to say that a copy is a pair $(A,E)$ where $Asubseteqomega$ is the domain and $Esubseteq A^2$; however, in this case it doesn't matter, since from $E$ you can already recover $A$ (fix some $iin A$; then $jin A$ iff $i=j$ or $(i,j)in E$ or $(j,i)in E$. So it's not a real issue. (Note that this trick doesn't work for all structures necessarily, so it can be important, but here it's not an issue.)
â Noah Schweber
Sep 8 at 18:06
Suppose that we have chosen some structure as a particular copy of $omega_1^textCK$ that the oracle has access to. Can I assume that, technically, all questions can be given for the oracle in the form of a pair of natural numbers $a$ and $b$, and the oracle answers Yes if $a$ precedes $b$ and No if $a$ does not precede $b$?
â lyrically wicked
Sep 9 at 4:05
Suppose that we have chosen some structure as a particular copy of $omega_1^textCK$ that the oracle has access to. Can I assume that, technically, all questions can be given for the oracle in the form of a pair of natural numbers $a$ and $b$, and the oracle answers Yes if $a$ precedes $b$ and No if $a$ does not precede $b$?
â lyrically wicked
Sep 9 at 4:05
@lyricallywicked Yes. Although I would use the word "structure" to refer to the "pre-copy thing" as opposed to the oracle, or rather one of the oracles, associated to it - that is, $omega_1^CK$ is the structure and the copies are the things we can use as oracles.
â Noah Schweber
Sep 9 at 4:18
@lyricallywicked Yes. Although I would use the word "structure" to refer to the "pre-copy thing" as opposed to the oracle, or rather one of the oracles, associated to it - that is, $omega_1^CK$ is the structure and the copies are the things we can use as oracles.
â Noah Schweber
Sep 9 at 4:18
 |Â
show 1 more comment
up vote
0
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Since you used my phrase too, let me just mention first that a good way to say the same thing might be "well-ordering of/on $mathbbN$ with order-type $omega_1^CK$". To be fair, I can relate to your question very well because if you don't have a formal math background and have never read either an (i) elementary set theory book cover-to-cover (ii) an introductory formal proof book which also covers various types of "ordering" ..... then I think you might just have never encountered many of these terms. Up until I asked first few questions here, I didn't know how to express properly/formally (for which I could simply use that phrase) what I had been playing around with for few years. And it did cause a lot of difficulty for me in asking questions.
Anyway, here is my "informal/naive" take on this (I think an expert answer might be more authoritative place to look, but this might help you in getting some understanding of things). Informally, you can think of "well-ordering of/on $mathbbN$ with order-type $omega_1^CK$" simply as "inducing" a bijection between $omega_1^CK$ and $mathbbN$ (probably if one has to be more formal, one might word it a little differently?). More generally, you can replace $omega_1^CK$ with any (non-finite) $alpha$ that satisfies $alpha <omega_1$ ($omega_1$ being first uncountable).
If you denote the corresponding (unique/unambiguous) bijective function as $f:alpha rightarrow mathbbN$ then the inverse of this $f^-1:mathbbN rightarrow alpha$ is also of course a bijective function. Now what is the well-order relation corresponding to this well-ordering? It can be simply defined by the function $lessequals:mathbbN^2 rightarrow 0,1$ such that:
$lessequals(x,y)=$ truth value of $f^-1(x) le f^-1(y)$
Note that $f^-1(x) le f^-1(y)$ is an ordinal comparison. You can also define the function $lessthan:mathbbN^2 rightarrow 0,1$ as:
$lessthan(x,y)=$ truth value of $f^-1(x) < f^-1(y)$
In the given context, it is the function $lessequals$ that can be called the well-order relation (well I am going by description in the wiki link here honestly). But, at any rate, one can observe that both these functions are computationally reducible to each other (somewhat trivially).
One can say that an ordinal $alpha$ is "computable/recursive" if there exists a well-ordering of $mathbbN$ with order-type $alpha$ (adding a minor exception for finite ordinals I suppose) such that the corresponding
well-order relation is computable. $omega^CK_1$ is defined as the smallest ordinal that is not computable/recursive.
There is one important fact here that, while somewhat trivial, is important to know. If $alpha$ is a recursive ordinal than so is any ordinal $beta$ (where $beta < alpha$).
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Since you used my phrase too, let me just mention first that a good way to say the same thing might be "well-ordering of/on $mathbbN$ with order-type $omega_1^CK$". To be fair, I can relate to your question very well because if you don't have a formal math background and have never read either an (i) elementary set theory book cover-to-cover (ii) an introductory formal proof book which also covers various types of "ordering" ..... then I think you might just have never encountered many of these terms. Up until I asked first few questions here, I didn't know how to express properly/formally (for which I could simply use that phrase) what I had been playing around with for few years. And it did cause a lot of difficulty for me in asking questions.
Anyway, here is my "informal/naive" take on this (I think an expert answer might be more authoritative place to look, but this might help you in getting some understanding of things). Informally, you can think of "well-ordering of/on $mathbbN$ with order-type $omega_1^CK$" simply as "inducing" a bijection between $omega_1^CK$ and $mathbbN$ (probably if one has to be more formal, one might word it a little differently?). More generally, you can replace $omega_1^CK$ with any (non-finite) $alpha$ that satisfies $alpha <omega_1$ ($omega_1$ being first uncountable).
If you denote the corresponding (unique/unambiguous) bijective function as $f:alpha rightarrow mathbbN$ then the inverse of this $f^-1:mathbbN rightarrow alpha$ is also of course a bijective function. Now what is the well-order relation corresponding to this well-ordering? It can be simply defined by the function $lessequals:mathbbN^2 rightarrow 0,1$ such that:
$lessequals(x,y)=$ truth value of $f^-1(x) le f^-1(y)$
Note that $f^-1(x) le f^-1(y)$ is an ordinal comparison. You can also define the function $lessthan:mathbbN^2 rightarrow 0,1$ as:
$lessthan(x,y)=$ truth value of $f^-1(x) < f^-1(y)$
In the given context, it is the function $lessequals$ that can be called the well-order relation (well I am going by description in the wiki link here honestly). But, at any rate, one can observe that both these functions are computationally reducible to each other (somewhat trivially).
One can say that an ordinal $alpha$ is "computable/recursive" if there exists a well-ordering of $mathbbN$ with order-type $alpha$ (adding a minor exception for finite ordinals I suppose) such that the corresponding
well-order relation is computable. $omega^CK_1$ is defined as the smallest ordinal that is not computable/recursive.
There is one important fact here that, while somewhat trivial, is important to know. If $alpha$ is a recursive ordinal than so is any ordinal $beta$ (where $beta < alpha$).
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Since you used my phrase too, let me just mention first that a good way to say the same thing might be "well-ordering of/on $mathbbN$ with order-type $omega_1^CK$". To be fair, I can relate to your question very well because if you don't have a formal math background and have never read either an (i) elementary set theory book cover-to-cover (ii) an introductory formal proof book which also covers various types of "ordering" ..... then I think you might just have never encountered many of these terms. Up until I asked first few questions here, I didn't know how to express properly/formally (for which I could simply use that phrase) what I had been playing around with for few years. And it did cause a lot of difficulty for me in asking questions.
Anyway, here is my "informal/naive" take on this (I think an expert answer might be more authoritative place to look, but this might help you in getting some understanding of things). Informally, you can think of "well-ordering of/on $mathbbN$ with order-type $omega_1^CK$" simply as "inducing" a bijection between $omega_1^CK$ and $mathbbN$ (probably if one has to be more formal, one might word it a little differently?). More generally, you can replace $omega_1^CK$ with any (non-finite) $alpha$ that satisfies $alpha <omega_1$ ($omega_1$ being first uncountable).
If you denote the corresponding (unique/unambiguous) bijective function as $f:alpha rightarrow mathbbN$ then the inverse of this $f^-1:mathbbN rightarrow alpha$ is also of course a bijective function. Now what is the well-order relation corresponding to this well-ordering? It can be simply defined by the function $lessequals:mathbbN^2 rightarrow 0,1$ such that:
$lessequals(x,y)=$ truth value of $f^-1(x) le f^-1(y)$
Note that $f^-1(x) le f^-1(y)$ is an ordinal comparison. You can also define the function $lessthan:mathbbN^2 rightarrow 0,1$ as:
$lessthan(x,y)=$ truth value of $f^-1(x) < f^-1(y)$
In the given context, it is the function $lessequals$ that can be called the well-order relation (well I am going by description in the wiki link here honestly). But, at any rate, one can observe that both these functions are computationally reducible to each other (somewhat trivially).
One can say that an ordinal $alpha$ is "computable/recursive" if there exists a well-ordering of $mathbbN$ with order-type $alpha$ (adding a minor exception for finite ordinals I suppose) such that the corresponding
well-order relation is computable. $omega^CK_1$ is defined as the smallest ordinal that is not computable/recursive.
There is one important fact here that, while somewhat trivial, is important to know. If $alpha$ is a recursive ordinal than so is any ordinal $beta$ (where $beta < alpha$).
Since you used my phrase too, let me just mention first that a good way to say the same thing might be "well-ordering of/on $mathbbN$ with order-type $omega_1^CK$". To be fair, I can relate to your question very well because if you don't have a formal math background and have never read either an (i) elementary set theory book cover-to-cover (ii) an introductory formal proof book which also covers various types of "ordering" ..... then I think you might just have never encountered many of these terms. Up until I asked first few questions here, I didn't know how to express properly/formally (for which I could simply use that phrase) what I had been playing around with for few years. And it did cause a lot of difficulty for me in asking questions.
Anyway, here is my "informal/naive" take on this (I think an expert answer might be more authoritative place to look, but this might help you in getting some understanding of things). Informally, you can think of "well-ordering of/on $mathbbN$ with order-type $omega_1^CK$" simply as "inducing" a bijection between $omega_1^CK$ and $mathbbN$ (probably if one has to be more formal, one might word it a little differently?). More generally, you can replace $omega_1^CK$ with any (non-finite) $alpha$ that satisfies $alpha <omega_1$ ($omega_1$ being first uncountable).
If you denote the corresponding (unique/unambiguous) bijective function as $f:alpha rightarrow mathbbN$ then the inverse of this $f^-1:mathbbN rightarrow alpha$ is also of course a bijective function. Now what is the well-order relation corresponding to this well-ordering? It can be simply defined by the function $lessequals:mathbbN^2 rightarrow 0,1$ such that:
$lessequals(x,y)=$ truth value of $f^-1(x) le f^-1(y)$
Note that $f^-1(x) le f^-1(y)$ is an ordinal comparison. You can also define the function $lessthan:mathbbN^2 rightarrow 0,1$ as:
$lessthan(x,y)=$ truth value of $f^-1(x) < f^-1(y)$
In the given context, it is the function $lessequals$ that can be called the well-order relation (well I am going by description in the wiki link here honestly). But, at any rate, one can observe that both these functions are computationally reducible to each other (somewhat trivially).
One can say that an ordinal $alpha$ is "computable/recursive" if there exists a well-ordering of $mathbbN$ with order-type $alpha$ (adding a minor exception for finite ordinals I suppose) such that the corresponding
well-order relation is computable. $omega^CK_1$ is defined as the smallest ordinal that is not computable/recursive.
There is one important fact here that, while somewhat trivial, is important to know. If $alpha$ is a recursive ordinal than so is any ordinal $beta$ (where $beta < alpha$).
edited Sep 8 at 16:16
answered Sep 8 at 15:25
SSequence
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Sorry I think the phrase "well-ordering of $omega_1^textCK$ in terms of $mathbbN$" is off/incorrect. I know I used it in my recent answer/question (it might be a bit awkward to bump them for correction I suppose), but for the correct phrase see the beginning of my answer here.
â SSequence
Sep 8 at 15:57
OK I have corrected the misleading phrase from my previous (recent) answer/question too.
â SSequence
Sep 8 at 16:12