Can we state the existence of infinite set without infinity axiom? [duplicate]












0












$begingroup$



This question already has an answer here:




  • Using the Definition of Dedekind-Infinite to Replace the Axiom of Infinity

    2 answers



  • How can I define $mathbb{N}$ if I postulate existence of a Dedekind-infinite set rather than existence of an inductive set?

    2 answers




I have a question about infinity axiom in ZF and maybe, it has nonsense. So I apologize in advance if it is the case.



In ZF, the infiny axiom can be state as $exists X(Xneqemptysetwedgeforall xin X(S(x)in X))$ and it says that exists an infinite set (moreover, it is equivalent to say that $omega$ exists) more exactly, it says that exists an inductive set. In this last point is where arise my question, is it possible to build an infinite set without infinite axiom?



Note that the question depends of that we means by "infinite". A usual way to say that a set $A$ is infinite is through of $preceq$-relation: $A$ is a infinite set iff $omegapreceq A$. And with this definition it is possible to show that ZF-Inf+$forall x(x text{is finite})$ is consistent.



Now, there exists another "definition" of infinite and it is "Dedekind-infinite": A set $A$ is Dedekind-infinite iff exists $Xsubsetneq A$ such that $Xapprox A.$



So, can we state the existence of a infinite set (with another "reasonable" definition of infinite) that it is not be infinite in the "usual way"?



For example, can we get that ZF+$negmathbf{Inf}$+$exists x(x text{is Dedekind-infinite})$ is consistent?






set-theory infinity axioms







share|cite|improve this question











$endgroup$



marked as duplicate by Asaf Karagila♦ set-theory
Users with the  set-theory badge can single-handedly close set-theory questions as duplicates and reopen them as needed.

StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;

$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');

$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
 Nov 28 '18 at 13:41


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • 3




    $begingroup$
    You might also want to read math.stackexchange.com/questions/1706661/…
    $endgroup$
    – Asaf Karagila♦
    Nov 28 '18 at 13:44
















0












$begingroup$



This question already has an answer here:




  • Using the Definition of Dedekind-Infinite to Replace the Axiom of Infinity

    2 answers



  • How can I define $mathbb{N}$ if I postulate existence of a Dedekind-infinite set rather than existence of an inductive set?

    2 answers




I have a question about infinity axiom in ZF and maybe, it has nonsense. So I apologize in advance if it is the case.



In ZF, the infiny axiom can be state as $exists X(Xneqemptysetwedgeforall xin X(S(x)in X))$ and it says that exists an infinite set (moreover, it is equivalent to say that $omega$ exists) more exactly, it says that exists an inductive set. In this last point is where arise my question, is it possible to build an infinite set without infinite axiom?



Note that the question depends of that we means by "infinite". A usual way to say that a set $A$ is infinite is through of $preceq$-relation: $A$ is a infinite set iff $omegapreceq A$. And with this definition it is possible to show that ZF-Inf+$forall x(x text{is finite})$ is consistent.



Now, there exists another "definition" of infinite and it is "Dedekind-infinite": A set $A$ is Dedekind-infinite iff exists $Xsubsetneq A$ such that $Xapprox A.$



So, can we state the existence of a infinite set (with another "reasonable" definition of infinite) that it is not be infinite in the "usual way"?



For example, can we get that ZF+$negmathbf{Inf}$+$exists x(x text{is Dedekind-infinite})$ is consistent?






set-theory infinity axioms







share|cite|improve this question











$endgroup$



marked as duplicate by Asaf Karagila♦ set-theory
Users with the  set-theory badge can single-handedly close set-theory questions as duplicates and reopen them as needed.

StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;

$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');

$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
 Nov 28 '18 at 13:41


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • 3




    $begingroup$
    You might also want to read math.stackexchange.com/questions/1706661/…
    $endgroup$
    – Asaf Karagila♦
    Nov 28 '18 at 13:44














0












0








0





$begingroup$



This question already has an answer here:




  • Using the Definition of Dedekind-Infinite to Replace the Axiom of Infinity

    2 answers



  • How can I define $mathbb{N}$ if I postulate existence of a Dedekind-infinite set rather than existence of an inductive set?

    2 answers




I have a question about infinity axiom in ZF and maybe, it has nonsense. So I apologize in advance if it is the case.



In ZF, the infiny axiom can be state as $exists X(Xneqemptysetwedgeforall xin X(S(x)in X))$ and it says that exists an infinite set (moreover, it is equivalent to say that $omega$ exists) more exactly, it says that exists an inductive set. In this last point is where arise my question, is it possible to build an infinite set without infinite axiom?



Note that the question depends of that we means by "infinite". A usual way to say that a set $A$ is infinite is through of $preceq$-relation: $A$ is a infinite set iff $omegapreceq A$. And with this definition it is possible to show that ZF-Inf+$forall x(x text{is finite})$ is consistent.



Now, there exists another "definition" of infinite and it is "Dedekind-infinite": A set $A$ is Dedekind-infinite iff exists $Xsubsetneq A$ such that $Xapprox A.$



So, can we state the existence of a infinite set (with another "reasonable" definition of infinite) that it is not be infinite in the "usual way"?



For example, can we get that ZF+$negmathbf{Inf}$+$exists x(x text{is Dedekind-infinite})$ is consistent?






set-theory infinity axioms







share|cite|improve this question











$endgroup$





This question already has an answer here:




  • Using the Definition of Dedekind-Infinite to Replace the Axiom of Infinity

    2 answers



  • How can I define $mathbb{N}$ if I postulate existence of a Dedekind-infinite set rather than existence of an inductive set?

    2 answers




I have a question about infinity axiom in ZF and maybe, it has nonsense. So I apologize in advance if it is the case.



In ZF, the infiny axiom can be state as $exists X(Xneqemptysetwedgeforall xin X(S(x)in X))$ and it says that exists an infinite set (moreover, it is equivalent to say that $omega$ exists) more exactly, it says that exists an inductive set. In this last point is where arise my question, is it possible to build an infinite set without infinite axiom?



Note that the question depends of that we means by "infinite". A usual way to say that a set $A$ is infinite is through of $preceq$-relation: $A$ is a infinite set iff $omegapreceq A$. And with this definition it is possible to show that ZF-Inf+$forall x(x text{is finite})$ is consistent.



Now, there exists another "definition" of infinite and it is "Dedekind-infinite": A set $A$ is Dedekind-infinite iff exists $Xsubsetneq A$ such that $Xapprox A.$



So, can we state the existence of a infinite set (with another "reasonable" definition of infinite) that it is not be infinite in the "usual way"?



For example, can we get that ZF+$negmathbf{Inf}$+$exists x(x text{is Dedekind-infinite})$ is consistent?





This question already has an answer here:




  • Using the Definition of Dedekind-Infinite to Replace the Axiom of Infinity

    2 answers



  • How can I define $mathbb{N}$ if I postulate existence of a Dedekind-infinite set rather than existence of an inductive set?

    2 answers







set-theory infinity axioms




set-theory infinity axioms






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 28 '18 at 13:14







Gödel

















asked Nov 28 '18 at 13:09









GödelGödel

1,406319




1,406319




marked as duplicate by Asaf Karagila♦ set-theory
Users with the  set-theory badge can single-handedly close set-theory questions as duplicates and reopen them as needed.

StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;

$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');

$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
 Nov 28 '18 at 13:41


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.









marked as duplicate by Asaf Karagila♦ set-theory
Users with the  set-theory badge can single-handedly close set-theory questions as duplicates and reopen them as needed.

StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;

$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');

$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
 Nov 28 '18 at 13:41


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.










  • 3




    $begingroup$
    You might also want to read math.stackexchange.com/questions/1706661/…
    $endgroup$
    – Asaf Karagila♦
    Nov 28 '18 at 13:44














  • 3




    $begingroup$
    You might also want to read math.stackexchange.com/questions/1706661/…
    $endgroup$
    – Asaf Karagila♦
    Nov 28 '18 at 13:44








3




3




$begingroup$
You might also want to read math.stackexchange.com/questions/1706661/…
$endgroup$
– Asaf Karagila♦
Nov 28 '18 at 13:44




$begingroup$
You might also want to read math.stackexchange.com/questions/1706661/…
$endgroup$
– Asaf Karagila♦
Nov 28 '18 at 13:44










0






active

oldest

votes

















0






active

oldest

votes








0






active

oldest

votes









active

oldest

votes






active

oldest

votes

-

Popular posts from this blog

Biblatex bibliography style without URLs when DOI exists (in Overleaf with Zotero bibliography)

Image
2 2 I have a Zotero bibliography in Overleaf. I use biblatex . usepackage[backend=biber,citestyle=authoryear]{biblatex} addbibresource{zoteroALH.bib} Bibliography entries look like: @book{de_saint-gervais_uniformization_2016, title = {Uniformization of {Riemann} {Surfaces}}, isbn = {978-3-03719-145-3}, url = {https://www.ems-ph.org/books/book.php?proj_nr=198&srch=series%7Chem}, urldate = {2019-02-03}, author = {de Saint-Gervais, Paul Henri}, year = {2016}, doi = {10.4171/145} } In the bibliography down the paper the reference shows like this: There is the DOI, great, but also the URL which is unnecessary when the DOI is there. How can I remove the URL when the DOI is present? biblatex bibliographies overleaf zote
Read more

ComboBox Display Member on multiple fields

Image
2 I'm trying to set the .DisplayMember property of a ComboBox in C#, but I want to bind it to multiple columns in the .DataSouce . My SQL looks like this: SELECT PersNbr, PersFirstName, PersMiddleName, PersLastName FROM Pers WHERE PersNbr = :persNbr; I'm saving this query in a DataTable so each column selected has it's own column in the Datatable . I want to make the .DisplayMember a combination of PersFirstName + PersMiddleName + PersLastName so their full name appears like this: comboBox.DisplayMemeber = "PersFirstName" + "PersMiddleName" + "PersLastName" I know I can just to this on the query: SELECT PersNbr, (PersFirstName || PersMiddleName || PersLastName) PersName and then just do this: comboBox.DisplayMember = "PersName"; but I
Read more

Is it possible to collect Nectar points via Trainline?

Image
2 Since I am not eligible for any Railcard, the only available discount/bonus scheme is Nectar Card for me. GWR, SWR, LNER and Virgin Trains allow to collect Nectar points and I have already linked a card to my GWR account. A friend suggested me to use Trainline App, because it can search for bus tickets either. I had a look at the application, but could not see any option for Nectar. Is there a way to collect Nectar points when I buy tickets via Trainline? uk trains loyalty-programs trainline share | improve this question asked Dec 2 '18 at 15:50 ahmedus 3,224 5 19 49
Read more