Mixing
Equivalent condition for a cardinal number to be of cofinality $aleph_0$
$begingroup$
Let cf$(alpha)$ denotes the co-final of the transfinite cardinal number $alpha.$
For every successor cardinal $alpha$ we have cf$(alpha)=alphaneq aleph_0.$
Thus cf$(alpha) = aleph_0,$ implies $alpha$ is a limit cardinal.
We can find some limit cardinal number $alpha,$ for example $alpha=aleph_{omega}$ with cf$(alpha)=aleph_0.$ Now, can we fined all (limit) cardinal numbers which satisfies
cf$(alpha)=aleph_0?$
set-theory
asked Jan 9 at 8:11
Ali BayatiAli Bayati
314
$endgroup$
add a comment |
$begingroup$
Let cf$(alpha)$ denotes the co-final of the transfinite cardinal number $alpha.$
For every successor cardinal $alpha$ we have cf$(alpha)=alphaneq aleph_0.$
Thus cf$(alpha) = aleph_0,$ implies $alpha$ is a limit cardinal.
We can find some limit cardinal number $alpha,$ for example $alpha=aleph_{omega}$ with cf$(alpha)=aleph_0.$ Now, can we fined all (limit) cardinal numbers which satisfies
cf$(alpha)=aleph_0?$
set-theory
asked Jan 9 at 8:11
Ali BayatiAli Bayati
314
$endgroup$
$begingroup$
$aleph_omega$ has cofinality $aleph_0$, but $aleph_{omega_1}$ has cofinality $aleph_1$. At least assuming the axiom of choice.
$endgroup$
– Asaf Karagila♦
Jan 9 at 8:12
$begingroup$
Prove that $operatorname{cf}(aleph_alpha)=operatorname{cf}(alpha)$ for limit ordinals (although cofinality of arbitrary ordinals is easier to measure with order types rather than cardinals, but for initial ordinals it is the same).
$endgroup$
– Asaf Karagila♦
Jan 9 at 8:14
$begingroup$
OK! I wrote wrong. But, is there an equivalent condition for cf$(alpha)=aleph_0?$
$endgroup$
– Ali Bayati
Jan 9 at 8:27
2
$begingroup$
I don't understand your last two comments. At all. My second comment provides an equivalent definition. It is not predicative, since you can have $aleph_alpha=alpha$ with cofinality $aleph_0$.
$endgroup$
– Asaf Karagila♦
Jan 9 at 8:48
add a comment |
$begingroup$
Let cf$(alpha)$ denotes the co-final of the transfinite cardinal number $alpha.$
For every successor cardinal $alpha$ we have cf$(alpha)=alphaneq aleph_0.$
Thus cf$(alpha) = aleph_0,$ implies $alpha$ is a limit cardinal.
We can find some limit cardinal number $alpha,$ for example $alpha=aleph_{omega}$ with cf$(alpha)=aleph_0.$ Now, can we fined all (limit) cardinal numbers which satisfies
cf$(alpha)=aleph_0?$
set-theory
asked Jan 9 at 8:11
Ali BayatiAli Bayati
314
$endgroup$
Let cf$(alpha)$ denotes the co-final of the transfinite cardinal number $alpha.$
For every successor cardinal $alpha$ we have cf$(alpha)=alphaneq aleph_0.$
Thus cf$(alpha) = aleph_0,$ implies $alpha$ is a limit cardinal.
We can find some limit cardinal number $alpha,$ for example $alpha=aleph_{omega}$ with cf$(alpha)=aleph_0.$ Now, can we fined all (limit) cardinal numbers which satisfies
cf$(alpha)=aleph_0?$
set-theory
set-theory
asked Jan 9 at 8:11
Ali BayatiAli Bayati
314
asked Jan 9 at 8:11
Ali BayatiAli Bayati
314
edited Jan 9 at 8:22
Ali Bayati
asked Jan 9 at 8:11
Ali BayatiAli Bayati
314
asked Jan 9 at 8:11
Ali BayatiAli Bayati
314
asked Jan 9 at 8:11
Ali BayatiAli Bayati
314
314
$begingroup$
$aleph_omega$ has cofinality $aleph_0$, but $aleph_{omega_1}$ has cofinality $aleph_1$. At least assuming the axiom of choice.
$endgroup$
– Asaf Karagila♦
Jan 9 at 8:12
$begingroup$
Prove that $operatorname{cf}(aleph_alpha)=operatorname{cf}(alpha)$ for limit ordinals (although cofinality of arbitrary ordinals is easier to measure with order types rather than cardinals, but for initial ordinals it is the same).
$endgroup$
– Asaf Karagila♦
Jan 9 at 8:14
$begingroup$
OK! I wrote wrong. But, is there an equivalent condition for cf$(alpha)=aleph_0?$
$endgroup$
– Ali Bayati
Jan 9 at 8:27
2
$begingroup$
I don't understand your last two comments. At all. My second comment provides an equivalent definition. It is not predicative, since you can have $aleph_alpha=alpha$ with cofinality $aleph_0$.
$endgroup$
– Asaf Karagila♦
Jan 9 at 8:48
add a comment |
$begingroup$
$aleph_omega$ has cofinality $aleph_0$, but $aleph_{omega_1}$ has cofinality $aleph_1$. At least assuming the axiom of choice.
$endgroup$
– Asaf Karagila♦
Jan 9 at 8:12
$begingroup$
Prove that $operatorname{cf}(aleph_alpha)=operatorname{cf}(alpha)$ for limit ordinals (although cofinality of arbitrary ordinals is easier to measure with order types rather than cardinals, but for initial ordinals it is the same).
$endgroup$
– Asaf Karagila♦
Jan 9 at 8:14
$begingroup$
OK! I wrote wrong. But, is there an equivalent condition for cf$(alpha)=aleph_0?$
$endgroup$
– Ali Bayati
Jan 9 at 8:27
2
$begingroup$
I don't understand your last two comments. At all. My second comment provides an equivalent definition. It is not predicative, since you can have $aleph_alpha=alpha$ with cofinality $aleph_0$.
$endgroup$
– Asaf Karagila♦
Jan 9 at 8:48
$begingroup$
$aleph_omega$ has cofinality $aleph_0$, but $aleph_{omega_1}$ has cofinality $aleph_1$. At least assuming the axiom of choice.
$endgroup$
– Asaf Karagila♦
Jan 9 at 8:12
$begingroup$
$aleph_omega$ has cofinality $aleph_0$, but $aleph_{omega_1}$ has cofinality $aleph_1$. At least assuming the axiom of choice.
$endgroup$
– Asaf Karagila♦
Jan 9 at 8:12
$begingroup$
Prove that $operatorname{cf}(aleph_alpha)=operatorname{cf}(alpha)$ for limit ordinals (although cofinality of arbitrary ordinals is easier to measure with order types rather than cardinals, but for initial ordinals it is the same).
$endgroup$
– Asaf Karagila♦
Jan 9 at 8:14
$begingroup$
Prove that $operatorname{cf}(aleph_alpha)=operatorname{cf}(alpha)$ for limit ordinals (although cofinality of arbitrary ordinals is easier to measure with order types rather than cardinals, but for initial ordinals it is the same).
$endgroup$
– Asaf Karagila♦
Jan 9 at 8:14
$begingroup$
OK! I wrote wrong. But, is there an equivalent condition for cf$(alpha)=aleph_0?$
$endgroup$
– Ali Bayati
Jan 9 at 8:27
$begingroup$
OK! I wrote wrong. But, is there an equivalent condition for cf$(alpha)=aleph_0?$
$endgroup$
– Ali Bayati
Jan 9 at 8:27
2
2
$begingroup$
I don't understand your last two comments. At all. My second comment provides an equivalent definition. It is not predicative, since you can have $aleph_alpha=alpha$ with cofinality $aleph_0$.
$endgroup$
– Asaf Karagila♦
Jan 9 at 8:48
$begingroup$
I don't understand your last two comments. At all. My second comment provides an equivalent definition. It is not predicative, since you can have $aleph_alpha=alpha$ with cofinality $aleph_0$.
$endgroup$
– Asaf Karagila♦
Jan 9 at 8:48
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Asaf Karagila has essentially answered this question in comments (and if he posts an answer I'll delete this one for precedence reasons), but let me flesh it out a bit.
Remember that cofinality makes sense for arbitrary infinite limit ordinals, not just infinite cardinals. The characterization is:
$aleph_alpha$ has cofinality $omega$ iff $alpha=0$ or $alpha$ (is a limit ordinal and) has cofinality $omega$.
As an aside, remember that $aleph_alpha$ and $omega_alpha$ are literally the same thing, the difference between the notation is merely that the former indicates that we're thinking about only cardinals while the second indicates that we're thinking about general ordinals. I'm going to use both notations since in this particular question I think it will be clearer, but really I think the ideal approach is to ditch the $aleph$-notation completely except for situations when we're using cardinal arithmetic - that is cardinal addition, multiplication, and exponentiation (which we're not doing here).
So, for example, the cardinals $aleph_omega, aleph_{omega^3+omega^2cdot 13}$, and $aleph_{omega_{omega_omega}}$ each have cofinality $omega$, while $aleph_{37}$ and $aleph_{omega_1}$ do not since $cf(37)=1$ and $cf(omega_1)=omega_1$.
Proving the fact above is a good exercise; here are a couple hints:
In the right-to-left direction, suppose $cf(alpha)=omega$. Let $(beta_i)_{iinomega}$ be cofinal in $alpha$; what can you say about the sequence $(aleph_{beta_i})_{iinomega}$?
In the left-to-right direction, we do the same thing but in the other direction. Suppose $aleph_alpha$ has cofinality $omega$ and $alphanot=0$. Let $(aleph_{beta_i})_{iinomega}$ be cofinal in $aleph_alpha$: what can you say about the sequence $(beta_i)_{iinomega}$, and what does that tell us about $alpha$?
answered Jan 20 at 14:09
Noah SchweberNoah Schweber
123k10150286
$endgroup$
add a comment |
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%2f3067205%2fequivalent-condition-for-a-cardinal-number-to-be-of-cofinality-aleph-0%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Asaf Karagila has essentially answered this question in comments (and if he posts an answer I'll delete this one for precedence reasons), but let me flesh it out a bit.
Remember that cofinality makes sense for arbitrary infinite limit ordinals, not just infinite cardinals. The characterization is:
$aleph_alpha$ has cofinality $omega$ iff $alpha=0$ or $alpha$ (is a limit ordinal and) has cofinality $omega$.
As an aside, remember that $aleph_alpha$ and $omega_alpha$ are literally the same thing, the difference between the notation is merely that the former indicates that we're thinking about only cardinals while the second indicates that we're thinking about general ordinals. I'm going to use both notations since in this particular question I think it will be clearer, but really I think the ideal approach is to ditch the $aleph$-notation completely except for situations when we're using cardinal arithmetic - that is cardinal addition, multiplication, and exponentiation (which we're not doing here).
So, for example, the cardinals $aleph_omega, aleph_{omega^3+omega^2cdot 13}$, and $aleph_{omega_{omega_omega}}$ each have cofinality $omega$, while $aleph_{37}$ and $aleph_{omega_1}$ do not since $cf(37)=1$ and $cf(omega_1)=omega_1$.
Proving the fact above is a good exercise; here are a couple hints:
In the right-to-left direction, suppose $cf(alpha)=omega$. Let $(beta_i)_{iinomega}$ be cofinal in $alpha$; what can you say about the sequence $(aleph_{beta_i})_{iinomega}$?
In the left-to-right direction, we do the same thing but in the other direction. Suppose $aleph_alpha$ has cofinality $omega$ and $alphanot=0$. Let $(aleph_{beta_i})_{iinomega}$ be cofinal in $aleph_alpha$: what can you say about the sequence $(beta_i)_{iinomega}$, and what does that tell us about $alpha$?
answered Jan 20 at 14:09
Noah SchweberNoah Schweber
123k10150286
$endgroup$
add a comment |
$begingroup$
Asaf Karagila has essentially answered this question in comments (and if he posts an answer I'll delete this one for precedence reasons), but let me flesh it out a bit.
Remember that cofinality makes sense for arbitrary infinite limit ordinals, not just infinite cardinals. The characterization is:
$aleph_alpha$ has cofinality $omega$ iff $alpha=0$ or $alpha$ (is a limit ordinal and) has cofinality $omega$.
As an aside, remember that $aleph_alpha$ and $omega_alpha$ are literally the same thing, the difference between the notation is merely that the former indicates that we're thinking about only cardinals while the second indicates that we're thinking about general ordinals. I'm going to use both notations since in this particular question I think it will be clearer, but really I think the ideal approach is to ditch the $aleph$-notation completely except for situations when we're using cardinal arithmetic - that is cardinal addition, multiplication, and exponentiation (which we're not doing here).
So, for example, the cardinals $aleph_omega, aleph_{omega^3+omega^2cdot 13}$, and $aleph_{omega_{omega_omega}}$ each have cofinality $omega$, while $aleph_{37}$ and $aleph_{omega_1}$ do not since $cf(37)=1$ and $cf(omega_1)=omega_1$.
Proving the fact above is a good exercise; here are a couple hints:
In the right-to-left direction, suppose $cf(alpha)=omega$. Let $(beta_i)_{iinomega}$ be cofinal in $alpha$; what can you say about the sequence $(aleph_{beta_i})_{iinomega}$?
In the left-to-right direction, we do the same thing but in the other direction. Suppose $aleph_alpha$ has cofinality $omega$ and $alphanot=0$. Let $(aleph_{beta_i})_{iinomega}$ be cofinal in $aleph_alpha$: what can you say about the sequence $(beta_i)_{iinomega}$, and what does that tell us about $alpha$?
answered Jan 20 at 14:09
Noah SchweberNoah Schweber
123k10150286
$endgroup$
add a comment |
$begingroup$
Asaf Karagila has essentially answered this question in comments (and if he posts an answer I'll delete this one for precedence reasons), but let me flesh it out a bit.
Remember that cofinality makes sense for arbitrary infinite limit ordinals, not just infinite cardinals. The characterization is:
$aleph_alpha$ has cofinality $omega$ iff $alpha=0$ or $alpha$ (is a limit ordinal and) has cofinality $omega$.
As an aside, remember that $aleph_alpha$ and $omega_alpha$ are literally the same thing, the difference between the notation is merely that the former indicates that we're thinking about only cardinals while the second indicates that we're thinking about general ordinals. I'm going to use both notations since in this particular question I think it will be clearer, but really I think the ideal approach is to ditch the $aleph$-notation completely except for situations when we're using cardinal arithmetic - that is cardinal addition, multiplication, and exponentiation (which we're not doing here).
So, for example, the cardinals $aleph_omega, aleph_{omega^3+omega^2cdot 13}$, and $aleph_{omega_{omega_omega}}$ each have cofinality $omega$, while $aleph_{37}$ and $aleph_{omega_1}$ do not since $cf(37)=1$ and $cf(omega_1)=omega_1$.
Proving the fact above is a good exercise; here are a couple hints:
In the right-to-left direction, suppose $cf(alpha)=omega$. Let $(beta_i)_{iinomega}$ be cofinal in $alpha$; what can you say about the sequence $(aleph_{beta_i})_{iinomega}$?
In the left-to-right direction, we do the same thing but in the other direction. Suppose $aleph_alpha$ has cofinality $omega$ and $alphanot=0$. Let $(aleph_{beta_i})_{iinomega}$ be cofinal in $aleph_alpha$: what can you say about the sequence $(beta_i)_{iinomega}$, and what does that tell us about $alpha$?
answered Jan 20 at 14:09
Noah SchweberNoah Schweber
123k10150286
$endgroup$
Asaf Karagila has essentially answered this question in comments (and if he posts an answer I'll delete this one for precedence reasons), but let me flesh it out a bit.
Remember that cofinality makes sense for arbitrary infinite limit ordinals, not just infinite cardinals. The characterization is:
$aleph_alpha$ has cofinality $omega$ iff $alpha=0$ or $alpha$ (is a limit ordinal and) has cofinality $omega$.
As an aside, remember that $aleph_alpha$ and $omega_alpha$ are literally the same thing, the difference between the notation is merely that the former indicates that we're thinking about only cardinals while the second indicates that we're thinking about general ordinals. I'm going to use both notations since in this particular question I think it will be clearer, but really I think the ideal approach is to ditch the $aleph$-notation completely except for situations when we're using cardinal arithmetic - that is cardinal addition, multiplication, and exponentiation (which we're not doing here).
So, for example, the cardinals $aleph_omega, aleph_{omega^3+omega^2cdot 13}$, and $aleph_{omega_{omega_omega}}$ each have cofinality $omega$, while $aleph_{37}$ and $aleph_{omega_1}$ do not since $cf(37)=1$ and $cf(omega_1)=omega_1$.
Proving the fact above is a good exercise; here are a couple hints:
In the right-to-left direction, suppose $cf(alpha)=omega$. Let $(beta_i)_{iinomega}$ be cofinal in $alpha$; what can you say about the sequence $(aleph_{beta_i})_{iinomega}$?
In the left-to-right direction, we do the same thing but in the other direction. Suppose $aleph_alpha$ has cofinality $omega$ and $alphanot=0$. Let $(aleph_{beta_i})_{iinomega}$ be cofinal in $aleph_alpha$: what can you say about the sequence $(beta_i)_{iinomega}$, and what does that tell us about $alpha$?
answered Jan 20 at 14:09
Noah SchweberNoah Schweber
123k10150286
answered Jan 20 at 14:09
Noah SchweberNoah Schweber
123k10150286
answered Jan 20 at 14:09
Noah SchweberNoah Schweber
123k10150286
answered Jan 20 at 14:09
Noah SchweberNoah Schweber
123k10150286
123k10150286
add a comment |
add a comment |
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%2f3067205%2fequivalent-condition-for-a-cardinal-number-to-be-of-cofinality-aleph-0%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
$begingroup$
$aleph_omega$ has cofinality $aleph_0$, but $aleph_{omega_1}$ has cofinality $aleph_1$. At least assuming the axiom of choice.
$endgroup$
– Asaf Karagila♦
Jan 9 at 8:12
$begingroup$
Prove that $operatorname{cf}(aleph_alpha)=operatorname{cf}(alpha)$ for limit ordinals (although cofinality of arbitrary ordinals is easier to measure with order types rather than cardinals, but for initial ordinals it is the same).
$endgroup$
– Asaf Karagila♦
Jan 9 at 8:14
$begingroup$
OK! I wrote wrong. But, is there an equivalent condition for cf$(alpha)=aleph_0?$
$endgroup$
– Ali Bayati
Jan 9 at 8:27
2
$begingroup$
I don't understand your last two comments. At all. My second comment provides an equivalent definition. It is not predicative, since you can have $aleph_alpha=alpha$ with cofinality $aleph_0$.
$endgroup$
– Asaf Karagila♦
Jan 9 at 8:48