How do you define ordinal exponentiation without induction?
$begingroup$
I am writing some notes on introductory set theory, starting from the basic axioms all the way to cardinal arithmetic. Right now I am up to ordinal arithmetic. I have defined ordinal addition and ordinal multiplication but not in the most typical way. Usually, most authors define ordinal addition/multiplication by induction. They define ordinal addition as follows:
$$1. alpha+0=alpha.$$
$$2. alpha+beta ^{+}=(alpha+beta)^{+} $$
$$3. text{If λ is a limit ordinal, then } alpha+lambda=bigcup_{gammainlambda}left(alpha +gammaright).$$
Ordinal multiplication is defined analogously. The reason I am not defining ordinal addition and multiplication in this way is because these definitions involve using recursive class functions and I want to avoid classes and class functions. Instead, I define ordinal addition and multiplication synthetically. If $alpha$ and $beta$ are ordinals, then I define $alpha+beta$ to be the order type of the well-ordered set $left[alphatimesleft{ 0right} right]cupleft[betatimesleft{ 1right} right]$ (with the ordering inherited from $alpha$ and $beta$.) I define $alphabeta$ to be the order type of the well-ordered set $alphatimesbeta$ with the reverse lexicographic ordering. I then proved that the synthetic definitions of addition and multiplications agreed with the induction definitions. What I am having trouble with is defining ordinal exponentiation synthetically. There is way to do this, as shown here, but it involves using facts about finite sets. The problem is that I haven't defined finite sets and want to leave it till the next chapter on cardinal numbers. Is there another way to synthetically define ordinal exponentiation without using results about finite sets?
set-theory ordinals
asked Dec 13 '18 at 8:37
EigenfieldEigenfield
627518
$endgroup$
|
show 1 more comment
$begingroup$
I am writing some notes on introductory set theory, starting from the basic axioms all the way to cardinal arithmetic. Right now I am up to ordinal arithmetic. I have defined ordinal addition and ordinal multiplication but not in the most typical way. Usually, most authors define ordinal addition/multiplication by induction. They define ordinal addition as follows:
$$1. alpha+0=alpha.$$
$$2. alpha+beta ^{+}=(alpha+beta)^{+} $$
$$3. text{If λ is a limit ordinal, then } alpha+lambda=bigcup_{gammainlambda}left(alpha +gammaright).$$
Ordinal multiplication is defined analogously. The reason I am not defining ordinal addition and multiplication in this way is because these definitions involve using recursive class functions and I want to avoid classes and class functions. Instead, I define ordinal addition and multiplication synthetically. If $alpha$ and $beta$ are ordinals, then I define $alpha+beta$ to be the order type of the well-ordered set $left[alphatimesleft{ 0right} right]cupleft[betatimesleft{ 1right} right]$ (with the ordering inherited from $alpha$ and $beta$.) I define $alphabeta$ to be the order type of the well-ordered set $alphatimesbeta$ with the reverse lexicographic ordering. I then proved that the synthetic definitions of addition and multiplications agreed with the induction definitions. What I am having trouble with is defining ordinal exponentiation synthetically. There is way to do this, as shown here, but it involves using facts about finite sets. The problem is that I haven't defined finite sets and want to leave it till the next chapter on cardinal numbers. Is there another way to synthetically define ordinal exponentiation without using results about finite sets?
set-theory ordinals
asked Dec 13 '18 at 8:37
EigenfieldEigenfield
627518
$endgroup$
2
$begingroup$
You can talk about decreasing functions. In well-orders they must have finite ranges. Of course, you might have to quotient by some equivalence relation, but I think that if you are being careful enough, you can avoid that too.
$endgroup$
– Asaf Karagila♦
Dec 13 '18 at 10:16
2
$begingroup$
From my notes here: There is also a non-recursive definition using a last-differences ordering on the functions in $A^B$ that have finite support (functions $f:B rightarrow A$ such that $f(x)$ differs from $min A$ for finitely many $x$ in $B),$ where $A$ and $B$ are well-ordered sets with order types $a$ and $b.$ Ordinal exponentiation can also be defined directly from transfinite products, where all the factors are the same. Although this assumes that transfinite products have been dealt with, some texts use this approach.
$endgroup$
– Dave L. Renfro
Dec 13 '18 at 12:51
$begingroup$
Do you know any of the books which define ordinal exponentiation using transfinite products?
$endgroup$
– Eigenfield
Dec 13 '18 at 18:37
1
$begingroup$
One such book is Kamke's Theory of Sets --- Chapter IV, Article 9 (Powers of Ordinal Numbers), p. 104: The power ${mu}^{alpha}$ is now once more defined as a product of equal factors. By specializing the general definition of the product, we obtain the following definition of the power: Put $ldots$" Incidentally, the use of "is now once more" does not mean that ordinal exponentiation is being defined again in this manner, but rather it is a reference to the analogous way that Kamke defined exponentiation of cardinal numbers back on p. 41.
$endgroup$
– Dave L. Renfro
Dec 13 '18 at 21:44
1
$begingroup$
Before defining exponentiation, you surely defined $omega$, right? Then you can describe finite ordinals without using cardinality: It's just the ordinals that are less than $omega$.
$endgroup$
– celtschk
Dec 17 '18 at 8:38
|
show 1 more comment
$begingroup$
I am writing some notes on introductory set theory, starting from the basic axioms all the way to cardinal arithmetic. Right now I am up to ordinal arithmetic. I have defined ordinal addition and ordinal multiplication but not in the most typical way. Usually, most authors define ordinal addition/multiplication by induction. They define ordinal addition as follows:
$$1. alpha+0=alpha.$$
$$2. alpha+beta ^{+}=(alpha+beta)^{+} $$
$$3. text{If λ is a limit ordinal, then } alpha+lambda=bigcup_{gammainlambda}left(alpha +gammaright).$$
Ordinal multiplication is defined analogously. The reason I am not defining ordinal addition and multiplication in this way is because these definitions involve using recursive class functions and I want to avoid classes and class functions. Instead, I define ordinal addition and multiplication synthetically. If $alpha$ and $beta$ are ordinals, then I define $alpha+beta$ to be the order type of the well-ordered set $left[alphatimesleft{ 0right} right]cupleft[betatimesleft{ 1right} right]$ (with the ordering inherited from $alpha$ and $beta$.) I define $alphabeta$ to be the order type of the well-ordered set $alphatimesbeta$ with the reverse lexicographic ordering. I then proved that the synthetic definitions of addition and multiplications agreed with the induction definitions. What I am having trouble with is defining ordinal exponentiation synthetically. There is way to do this, as shown here, but it involves using facts about finite sets. The problem is that I haven't defined finite sets and want to leave it till the next chapter on cardinal numbers. Is there another way to synthetically define ordinal exponentiation without using results about finite sets?
set-theory ordinals
asked Dec 13 '18 at 8:37
EigenfieldEigenfield
627518
$endgroup$
I am writing some notes on introductory set theory, starting from the basic axioms all the way to cardinal arithmetic. Right now I am up to ordinal arithmetic. I have defined ordinal addition and ordinal multiplication but not in the most typical way. Usually, most authors define ordinal addition/multiplication by induction. They define ordinal addition as follows:
$$1. alpha+0=alpha.$$
$$2. alpha+beta ^{+}=(alpha+beta)^{+} $$
$$3. text{If λ is a limit ordinal, then } alpha+lambda=bigcup_{gammainlambda}left(alpha +gammaright).$$
Ordinal multiplication is defined analogously. The reason I am not defining ordinal addition and multiplication in this way is because these definitions involve using recursive class functions and I want to avoid classes and class functions. Instead, I define ordinal addition and multiplication synthetically. If $alpha$ and $beta$ are ordinals, then I define $alpha+beta$ to be the order type of the well-ordered set $left[alphatimesleft{ 0right} right]cupleft[betatimesleft{ 1right} right]$ (with the ordering inherited from $alpha$ and $beta$.) I define $alphabeta$ to be the order type of the well-ordered set $alphatimesbeta$ with the reverse lexicographic ordering. I then proved that the synthetic definitions of addition and multiplications agreed with the induction definitions. What I am having trouble with is defining ordinal exponentiation synthetically. There is way to do this, as shown here, but it involves using facts about finite sets. The problem is that I haven't defined finite sets and want to leave it till the next chapter on cardinal numbers. Is there another way to synthetically define ordinal exponentiation without using results about finite sets?
set-theory ordinals
set-theory ordinals
asked Dec 13 '18 at 8:37
EigenfieldEigenfield
627518
asked Dec 13 '18 at 8:37
EigenfieldEigenfield
627518
asked Dec 13 '18 at 8:37
EigenfieldEigenfield
627518
asked Dec 13 '18 at 8:37
EigenfieldEigenfield
627518
asked Dec 13 '18 at 8:37
EigenfieldEigenfield
627518
627518
2
$begingroup$
You can talk about decreasing functions. In well-orders they must have finite ranges. Of course, you might have to quotient by some equivalence relation, but I think that if you are being careful enough, you can avoid that too.
$endgroup$
– Asaf Karagila♦
Dec 13 '18 at 10:16
2
$begingroup$
From my notes here: There is also a non-recursive definition using a last-differences ordering on the functions in $A^B$ that have finite support (functions $f:B rightarrow A$ such that $f(x)$ differs from $min A$ for finitely many $x$ in $B),$ where $A$ and $B$ are well-ordered sets with order types $a$ and $b.$ Ordinal exponentiation can also be defined directly from transfinite products, where all the factors are the same. Although this assumes that transfinite products have been dealt with, some texts use this approach.
$endgroup$
– Dave L. Renfro
Dec 13 '18 at 12:51
$begingroup$
Do you know any of the books which define ordinal exponentiation using transfinite products?
$endgroup$
– Eigenfield
Dec 13 '18 at 18:37
1
$begingroup$
One such book is Kamke's Theory of Sets --- Chapter IV, Article 9 (Powers of Ordinal Numbers), p. 104: The power ${mu}^{alpha}$ is now once more defined as a product of equal factors. By specializing the general definition of the product, we obtain the following definition of the power: Put $ldots$" Incidentally, the use of "is now once more" does not mean that ordinal exponentiation is being defined again in this manner, but rather it is a reference to the analogous way that Kamke defined exponentiation of cardinal numbers back on p. 41.
$endgroup$
– Dave L. Renfro
Dec 13 '18 at 21:44
1
$begingroup$
Before defining exponentiation, you surely defined $omega$, right? Then you can describe finite ordinals without using cardinality: It's just the ordinals that are less than $omega$.
$endgroup$
– celtschk
Dec 17 '18 at 8:38
|
show 1 more comment
2
$begingroup$
You can talk about decreasing functions. In well-orders they must have finite ranges. Of course, you might have to quotient by some equivalence relation, but I think that if you are being careful enough, you can avoid that too.
$endgroup$
– Asaf Karagila♦
Dec 13 '18 at 10:16
2
$begingroup$
From my notes here: There is also a non-recursive definition using a last-differences ordering on the functions in $A^B$ that have finite support (functions $f:B rightarrow A$ such that $f(x)$ differs from $min A$ for finitely many $x$ in $B),$ where $A$ and $B$ are well-ordered sets with order types $a$ and $b.$ Ordinal exponentiation can also be defined directly from transfinite products, where all the factors are the same. Although this assumes that transfinite products have been dealt with, some texts use this approach.
$endgroup$
– Dave L. Renfro
Dec 13 '18 at 12:51
$begingroup$
Do you know any of the books which define ordinal exponentiation using transfinite products?
$endgroup$
– Eigenfield
Dec 13 '18 at 18:37
1
$begingroup$
One such book is Kamke's Theory of Sets --- Chapter IV, Article 9 (Powers of Ordinal Numbers), p. 104: The power ${mu}^{alpha}$ is now once more defined as a product of equal factors. By specializing the general definition of the product, we obtain the following definition of the power: Put $ldots$" Incidentally, the use of "is now once more" does not mean that ordinal exponentiation is being defined again in this manner, but rather it is a reference to the analogous way that Kamke defined exponentiation of cardinal numbers back on p. 41.
$endgroup$
– Dave L. Renfro
Dec 13 '18 at 21:44
1
$begingroup$
Before defining exponentiation, you surely defined $omega$, right? Then you can describe finite ordinals without using cardinality: It's just the ordinals that are less than $omega$.
$endgroup$
– celtschk
Dec 17 '18 at 8:38
2
2
$begingroup$
You can talk about decreasing functions. In well-orders they must have finite ranges. Of course, you might have to quotient by some equivalence relation, but I think that if you are being careful enough, you can avoid that too.
$endgroup$
– Asaf Karagila♦
Dec 13 '18 at 10:16
$begingroup$
You can talk about decreasing functions. In well-orders they must have finite ranges. Of course, you might have to quotient by some equivalence relation, but I think that if you are being careful enough, you can avoid that too.
$endgroup$
– Asaf Karagila♦
Dec 13 '18 at 10:16
2
2
$begingroup$
From my notes here: There is also a non-recursive definition using a last-differences ordering on the functions in $A^B$ that have finite support (functions $f:B rightarrow A$ such that $f(x)$ differs from $min A$ for finitely many $x$ in $B),$ where $A$ and $B$ are well-ordered sets with order types $a$ and $b.$ Ordinal exponentiation can also be defined directly from transfinite products, where all the factors are the same. Although this assumes that transfinite products have been dealt with, some texts use this approach.
$endgroup$
– Dave L. Renfro
Dec 13 '18 at 12:51
$begingroup$
From my notes here: There is also a non-recursive definition using a last-differences ordering on the functions in $A^B$ that have finite support (functions $f:B rightarrow A$ such that $f(x)$ differs from $min A$ for finitely many $x$ in $B),$ where $A$ and $B$ are well-ordered sets with order types $a$ and $b.$ Ordinal exponentiation can also be defined directly from transfinite products, where all the factors are the same. Although this assumes that transfinite products have been dealt with, some texts use this approach.
$endgroup$
– Dave L. Renfro
Dec 13 '18 at 12:51
$begingroup$
Do you know any of the books which define ordinal exponentiation using transfinite products?
$endgroup$
– Eigenfield
Dec 13 '18 at 18:37
$begingroup$
Do you know any of the books which define ordinal exponentiation using transfinite products?
$endgroup$
– Eigenfield
Dec 13 '18 at 18:37
1
1
$begingroup$
One such book is Kamke's Theory of Sets --- Chapter IV, Article 9 (Powers of Ordinal Numbers), p. 104: The power ${mu}^{alpha}$ is now once more defined as a product of equal factors. By specializing the general definition of the product, we obtain the following definition of the power: Put $ldots$" Incidentally, the use of "is now once more" does not mean that ordinal exponentiation is being defined again in this manner, but rather it is a reference to the analogous way that Kamke defined exponentiation of cardinal numbers back on p. 41.
$endgroup$
– Dave L. Renfro
Dec 13 '18 at 21:44
$begingroup$
One such book is Kamke's Theory of Sets --- Chapter IV, Article 9 (Powers of Ordinal Numbers), p. 104: The power ${mu}^{alpha}$ is now once more defined as a product of equal factors. By specializing the general definition of the product, we obtain the following definition of the power: Put $ldots$" Incidentally, the use of "is now once more" does not mean that ordinal exponentiation is being defined again in this manner, but rather it is a reference to the analogous way that Kamke defined exponentiation of cardinal numbers back on p. 41.
$endgroup$
– Dave L. Renfro
Dec 13 '18 at 21:44
1
1
$begingroup$
Before defining exponentiation, you surely defined $omega$, right? Then you can describe finite ordinals without using cardinality: It's just the ordinals that are less than $omega$.
$endgroup$
– celtschk
Dec 17 '18 at 8:38
$begingroup$
Before defining exponentiation, you surely defined $omega$, right? Then you can describe finite ordinals without using cardinality: It's just the ordinals that are less than $omega$.
$endgroup$
– celtschk
Dec 17 '18 at 8:38
|
show 1 more comment
0
active
oldest
votes
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3037759%2fhow-do-you-define-ordinal-exponentiation-without-induction%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3037759%2fhow-do-you-define-ordinal-exponentiation-without-induction%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
2
$begingroup$
You can talk about decreasing functions. In well-orders they must have finite ranges. Of course, you might have to quotient by some equivalence relation, but I think that if you are being careful enough, you can avoid that too.
$endgroup$
– Asaf Karagila♦
Dec 13 '18 at 10:16
2
$begingroup$
From my notes here: There is also a non-recursive definition using a last-differences ordering on the functions in $A^B$ that have finite support (functions $f:B rightarrow A$ such that $f(x)$ differs from $min A$ for finitely many $x$ in $B),$ where $A$ and $B$ are well-ordered sets with order types $a$ and $b.$ Ordinal exponentiation can also be defined directly from transfinite products, where all the factors are the same. Although this assumes that transfinite products have been dealt with, some texts use this approach.
$endgroup$
– Dave L. Renfro
Dec 13 '18 at 12:51
$begingroup$
Do you know any of the books which define ordinal exponentiation using transfinite products?
$endgroup$
– Eigenfield
Dec 13 '18 at 18:37
1
$begingroup$
One such book is Kamke's Theory of Sets --- Chapter IV, Article 9 (Powers of Ordinal Numbers), p. 104: The power ${mu}^{alpha}$ is now once more defined as a product of equal factors. By specializing the general definition of the product, we obtain the following definition of the power: Put $ldots$" Incidentally, the use of "is now once more" does not mean that ordinal exponentiation is being defined again in this manner, but rather it is a reference to the analogous way that Kamke defined exponentiation of cardinal numbers back on p. 41.
$endgroup$
– Dave L. Renfro
Dec 13 '18 at 21:44
1
$begingroup$
Before defining exponentiation, you surely defined $omega$, right? Then you can describe finite ordinals without using cardinality: It's just the ordinals that are less than $omega$.
$endgroup$
– celtschk
Dec 17 '18 at 8:38