Skip to main content
Well-foundedness
Source Link
Joel David Hamkins
  • 236.2k
  • 44
  • 777
  • 1.4k

This is an interesting idea, but I don't believe that your argument succeeds.

Notice first that if it were correct, then it would also refute the existence of $I_1$ rank-into-rank cardinals, that is, $j:V_{\lambda+1}\to V_{\lambda+1}$, since $\lambda=\kappa_\omega$ and the singularity of this cardinal is revealed in $V_{\lambda+1}$.

Some people might object that you haven't said what you mean by iterating $j$, but this is not actually a problem, for there is a natural way to do this. Any class embedding $j:V\to M$ can be iterated, since the definition of $j$ can be extended to any proper class $A$ by defining $j(A)=\bigcup_\alpha j(A\cap V_\alpha)$. This way, we get $j(j)$, which will be an elementary embedding from $M$ to $j(M)$. Thus, if we start with $j:V\to V$, we may indeed iterate $j$ to form a system $V\to V\to V\to\cdots$ as you describe.

The problem is that you claimed that the direct limit of this system is $V$, but I don't see why this should be true; this is a gap in your proof. It may seem reasonable to suppose that the direct limit of a system of structures, all of which are the same, is again that same structure, but in fact there are counterexamples to this general principle. (For a quick example, embed the discrete order $\mathbb{Z}$ into itself by stretching by a factor of two; the direct limit of the iteration of this process is the dense order $\mathbb{Q}$.)

Indeed, let me argue directly that the direct limit is definitely not $V$. The reason is that the direct limit model has only $\lambda$ many subsets of $\lambda$, where $\lambda=\kappa_\omega$. This is simply because every subset of $\lambda$ in the direct limit is born at some stage $M_n$ as a subset of $\kappa_n$, having the form $j_{n,\omega}(A)$ for some $A\subset\kappa_n$. But there are only $\lambda$ many $A$ like that, and so the limit model is missing most subsets of $\lambda$.

Note that the limit of the iteration is well-founded, by essentially the usual Kunen argument. Namely, if it were ill-founded, then we may consider the least $\gamma$ with $j_{0,\omega}(\gamma)$ in the ill-founded part of the limit. By pushing $\gamma$ into $M_n$, we may conclude on the one hand that $j_{0,n}(\gamma)$ must be least sent into the ill-founded part, but on the other hand we may also drop down to smaller ordinal born at that stage still in the ill-founded part, making a contradiction. (To formalize this precisely, one may avoid the complications caused by $j$ being a proper class simply by chopping off $j$ to the set $j\upharpoonright V_\alpha$, such that the iteration whose iteration is ill-founded.) This argument relies fundamentally on the fact that the iteration of $j$ in $M_0$ produces the same limit as the iteration of $j(j)$ in $M_1$.

Finally, if on the other hand you had formed your iteration by using the same embedding $j_{n,n+1}=j$ at each stage, then this is clearly ill-founded, since the threads starting with $\kappa$ at each stage form a descending sequence of ordinals in the limit.

This is an interesting idea, but I don't believe that your argument succeeds.

Notice first that if it were correct, then it would also refute the existence of $I_1$ rank-into-rank cardinals, that is, $j:V_{\lambda+1}\to V_{\lambda+1}$, since $\lambda=\kappa_\omega$ and the singularity of this cardinal is revealed in $V_{\lambda+1}$.

Some people might object that you haven't said what you mean by iterating $j$, but this is not actually a problem, for there is a natural way to do this. Any class embedding $j:V\to M$ can be iterated, since the definition of $j$ can be extended to any proper class $A$ by defining $j(A)=\bigcup_\alpha j(A\cap V_\alpha)$. This way, we get $j(j)$, which will be an elementary embedding from $M$ to $j(M)$. Thus, if we start with $j:V\to V$, we may indeed iterate $j$ to form a system $V\to V\to V\to\cdots$ as you describe.

The problem is that you claimed that the direct limit of this system is $V$, but I don't see why this should be true; this is a gap in your proof. It may seem reasonable to suppose that the direct limit of a system of structures, all of which are the same, is again that same structure, but in fact there are counterexamples to this general principle. (For a quick example, embed the discrete order $\mathbb{Z}$ into itself by stretching by a factor of two; the direct limit of the iteration of this process is the dense order $\mathbb{Q}$.)

Indeed, let me argue directly that the direct limit is definitely not $V$. The reason is that the direct limit model has only $\lambda$ many subsets of $\lambda$, where $\lambda=\kappa_\omega$. This is simply because every subset of $\lambda$ in the direct limit is born at some stage $M_n$ as a subset of $\kappa_n$, having the form $j_{n,\omega}(A)$ for some $A\subset\kappa_n$. But there are only $\lambda$ many $A$ like that, and so the limit model is missing most subsets of $\lambda$.

This is an interesting idea, but I don't believe that your argument succeeds.

Notice first that if it were correct, then it would also refute the existence of $I_1$ rank-into-rank cardinals, that is, $j:V_{\lambda+1}\to V_{\lambda+1}$, since $\lambda=\kappa_\omega$ and the singularity of this cardinal is revealed in $V_{\lambda+1}$.

Some people might object that you haven't said what you mean by iterating $j$, but this is not actually a problem, for there is a natural way to do this. Any class embedding $j:V\to M$ can be iterated, since the definition of $j$ can be extended to any proper class $A$ by defining $j(A)=\bigcup_\alpha j(A\cap V_\alpha)$. This way, we get $j(j)$, which will be an elementary embedding from $M$ to $j(M)$. Thus, if we start with $j:V\to V$, we may indeed iterate $j$ to form a system $V\to V\to V\to\cdots$ as you describe.

The problem is that you claimed that the direct limit of this system is $V$, but I don't see why this should be true; this is a gap in your proof. It may seem reasonable to suppose that the direct limit of a system of structures, all of which are the same, is again that same structure, but in fact there are counterexamples to this general principle. (For a quick example, embed the discrete order $\mathbb{Z}$ into itself by stretching by a factor of two; the direct limit of the iteration of this process is the dense order $\mathbb{Q}$.)

Indeed, let me argue directly that the direct limit is definitely not $V$. The reason is that the direct limit model has only $\lambda$ many subsets of $\lambda$, where $\lambda=\kappa_\omega$. This is simply because every subset of $\lambda$ in the direct limit is born at some stage $M_n$ as a subset of $\kappa_n$, having the form $j_{n,\omega}(A)$ for some $A\subset\kappa_n$. But there are only $\lambda$ many $A$ like that, and so the limit model is missing most subsets of $\lambda$.

Note that the limit of the iteration is well-founded, by essentially the usual Kunen argument. Namely, if it were ill-founded, then we may consider the least $\gamma$ with $j_{0,\omega}(\gamma)$ in the ill-founded part of the limit. By pushing $\gamma$ into $M_n$, we may conclude on the one hand that $j_{0,n}(\gamma)$ must be least sent into the ill-founded part, but on the other hand we may also drop down to smaller ordinal born at that stage still in the ill-founded part, making a contradiction. (To formalize this precisely, one may avoid the complications caused by $j$ being a proper class simply by chopping off $j$ to the set $j\upharpoonright V_\alpha$, such that the iteration whose iteration is ill-founded.) This argument relies fundamentally on the fact that the iteration of $j$ in $M_0$ produces the same limit as the iteration of $j(j)$ in $M_1$.

Finally, if on the other hand you had formed your iteration by using the same embedding $j_{n,n+1}=j$ at each stage, then this is clearly ill-founded, since the threads starting with $\kappa$ at each stage form a descending sequence of ordinals in the limit.

Direct limit is not V
Source Link
Joel David Hamkins
  • 236.2k
  • 44
  • 777
  • 1.4k

This is an interesting idea, but I don't believe that your argument succeeds.

Notice first that if it were correct, then it would also refute the existence of $I_1$ rank-into-rank cardinals, that is, $j:V_{\lambda+1}\to V_{\lambda+1}$, since $\lambda=\kappa_\omega$ and the singularity of this cardinal is revealed in $V_{\lambda+1}$.

Some people might object that you haven't said what you mean by iterating $j$, but this is not actually a problem, for there is a natural way to do this. Any class embedding $j:V\to M$ can be iterated, since the definition of $j$ can be extended to any proper class $A$ by defining $j(A)=\bigcup_\alpha j(A\cap V_\alpha)$. This way, we get $j(j)$, which will be an elementary embedding from $M$ to $j(M)$. Thus, if we start with $j:V\to V$, we may indeed iterate $j$ to form a system $V\to V\to V\to\cdots$ as you describe.

The problem is that you claimed that the direct limit of this system is $V$, but I don't see why this should be true; this seems to beis a gap in your proof. It may seem reasonable to suppose that the direct limit of a system of structures, all of which are the same, is again that same structure, but in fact there are counterexamples to this general principle. (For a quick example, embed the discrete order $\mathbb{Z}$ into itself by stretching by a factor of two; the direct limit of the iteration of this process is the dense order $\mathbb{Q}$.)

Indeed, let me argue directly that the direct limit is definitely not $V$. The reason is that the direct limit model has only $\lambda$ many subsets of $\lambda$, where $\lambda=\kappa_\omega$. This is simply because every subset of $\lambda$ in the direct limit is born at some stage $M_n$ as a subset of $\kappa_n$, having the form $j_{n,\omega}(A)$ for some $A\subset\kappa_n$. But there are only $\lambda$ many $A$ like that, and so the limit model is missing most subsets of $\lambda$.

This is an interesting idea, but I don't believe that your argument succeeds.

Notice first that if it were correct, then it would also refute the existence of $I_1$ rank-into-rank cardinals, that is, $j:V_{\lambda+1}\to V_{\lambda+1}$, since $\lambda=\kappa_\omega$ and the singularity of this cardinal is revealed in $V_{\lambda+1}$.

Some people might object that you haven't said what you mean by iterating $j$, but this is not actually a problem, for there is a natural way to do this. Any class embedding $j:V\to M$ can be iterated, since the definition of $j$ can be extended to any proper class $A$ by defining $j(A)=\bigcup_\alpha j(A\cap V_\alpha)$. This way, we get $j(j)$, which will be an elementary embedding from $M$ to $j(M)$. Thus, if we start with $j:V\to V$, we may indeed iterate $j$ to form a system $V\to V\to V\to\cdots$ as you describe.

The problem is that you claimed that the direct limit of this system is $V$, but I don't see why this should be true; this seems to be a gap in your proof. It may seem reasonable to suppose that the direct limit of a system of structures, all of which are the same, is again that same structure, but in fact there are counterexamples to this general principle. (For a quick example, embed the discrete order $\mathbb{Z}$ into itself by stretching by a factor of two; the direct limit of the iteration of this process is the dense order $\mathbb{Q}$.)

This is an interesting idea, but I don't believe that your argument succeeds.

Notice first that if it were correct, then it would also refute the existence of $I_1$ rank-into-rank cardinals, that is, $j:V_{\lambda+1}\to V_{\lambda+1}$, since $\lambda=\kappa_\omega$ and the singularity of this cardinal is revealed in $V_{\lambda+1}$.

Some people might object that you haven't said what you mean by iterating $j$, but this is not actually a problem, for there is a natural way to do this. Any class embedding $j:V\to M$ can be iterated, since the definition of $j$ can be extended to any proper class $A$ by defining $j(A)=\bigcup_\alpha j(A\cap V_\alpha)$. This way, we get $j(j)$, which will be an elementary embedding from $M$ to $j(M)$. Thus, if we start with $j:V\to V$, we may indeed iterate $j$ to form a system $V\to V\to V\to\cdots$ as you describe.

The problem is that you claimed that the direct limit of this system is $V$, but I don't see why this should be true; this is a gap in your proof. It may seem reasonable to suppose that the direct limit of a system of structures, all of which are the same, is again that same structure, but in fact there are counterexamples to this general principle. (For a quick example, embed the discrete order $\mathbb{Z}$ into itself by stretching by a factor of two; the direct limit of the iteration of this process is the dense order $\mathbb{Q}$.)

Indeed, let me argue directly that the direct limit is definitely not $V$. The reason is that the direct limit model has only $\lambda$ many subsets of $\lambda$, where $\lambda=\kappa_\omega$. This is simply because every subset of $\lambda$ in the direct limit is born at some stage $M_n$ as a subset of $\kappa_n$, having the form $j_{n,\omega}(A)$ for some $A\subset\kappa_n$. But there are only $\lambda$ many $A$ like that, and so the limit model is missing most subsets of $\lambda$.

Well-foundedness may not be so problematic
Source Link
Joel David Hamkins
  • 236.2k
  • 44
  • 777
  • 1.4k

This is an interesting idea, but I don't believe that your argument succeeds.

Notice first that if it were correct, then it would also refute the existence of $I_1$ rank-into-rank cardinals, that is, $j:V_{\lambda+1}\to V_{\lambda+1}$, since $\lambda=\kappa_\omega$ and the singularity of this cardinal is revealed in $V_{\lambda+1}$.

Some people might object that you haven't said what you mean by iterating $j$, but this is not actually a problem, for there is a natural way to do this. Any class embedding $j:V\to M$ can be iterated, since the definition of $j$ can be extended to any proper class $A$ by defining $j(A)=\bigcup_\alpha j(A\cap V_\alpha)$. This way, we get $j(j)$, which will be an elementary embedding from $M$ to $j(M)$. Thus, if we start with $j:V\to V$, we may indeed iterate $j$ to form a system $V\to V\to V\to\cdots$ as you describe.

The problem is that you claimed that the direct limit of this system is $V$, but I don't see why this should be true; this seems to be a gap in your proof. It may seem reasonable to suppose that the direct limit of a system of structures, all of which are the same, to beis again that same structure, but in fact there are counterexamples to this general principle. (For a quick example, embed the discrete order $\mathbb{Z}$ into itself by stretching by a factor of two; the direct limit of the iteration of this process is the dense order $\mathbb{Q}$.)

Indeed, since for the Kunen inconsistency we do not want to assume that the embedding $j$ is definable, but rather merely that it is a class in the sense of Gödel-Bernays set theory, I don't see even that the usual arguments show that the direct limit is well-founded.

This is an interesting idea, but I don't believe that your argument succeeds.

Notice first that if it were correct, then it would also refute the existence of $I_1$ rank-into-rank cardinals, that is, $j:V_{\lambda+1}\to V_{\lambda+1}$, since $\lambda=\kappa_\omega$ and the singularity of this cardinal is revealed in $V_{\lambda+1}$.

Some people might object that you haven't said what you mean by iterating $j$, but this is not actually a problem, for there is a natural way to do this. Any class embedding $j:V\to M$ can be iterated, since the definition of $j$ can be extended to any proper class $A$ by defining $j(A)=\bigcup_\alpha j(A\cap V_\alpha)$. This way, we get $j(j)$, which will be an elementary embedding from $M$ to $j(M)$. Thus, if we start with $j:V\to V$, we may indeed iterate $j$ to form a system $V\to V\to V\to\cdots$ as you describe.

The problem is that you claimed that the direct limit of this system is $V$, but I don't see why this should be true; this seems to be a gap in your proof. It may seem reasonable to suppose that the direct limit of a system of structures, all of which are the same, to be again that same structure, but in fact there are counterexamples to this general principle. (For a quick example, embed the discrete order $\mathbb{Z}$ into itself by stretching by a factor of two; the direct limit of the iteration of this process is the dense order $\mathbb{Q}$.)

Indeed, since for the Kunen inconsistency we do not want to assume that the embedding $j$ is definable, but rather merely that it is a class in the sense of Gödel-Bernays set theory, I don't see even that the usual arguments show that the direct limit is well-founded.

This is an interesting idea, but I don't believe that your argument succeeds.

Notice first that if it were correct, then it would also refute the existence of $I_1$ rank-into-rank cardinals, that is, $j:V_{\lambda+1}\to V_{\lambda+1}$, since $\lambda=\kappa_\omega$ and the singularity of this cardinal is revealed in $V_{\lambda+1}$.

Some people might object that you haven't said what you mean by iterating $j$, but this is not actually a problem, for there is a natural way to do this. Any class embedding $j:V\to M$ can be iterated, since the definition of $j$ can be extended to any proper class $A$ by defining $j(A)=\bigcup_\alpha j(A\cap V_\alpha)$. This way, we get $j(j)$, which will be an elementary embedding from $M$ to $j(M)$. Thus, if we start with $j:V\to V$, we may indeed iterate $j$ to form a system $V\to V\to V\to\cdots$ as you describe.

The problem is that you claimed that the direct limit of this system is $V$, but I don't see why this should be true; this seems to be a gap in your proof. It may seem reasonable to suppose that the direct limit of a system of structures, all of which are the same, is again that same structure, but in fact there are counterexamples to this general principle. (For a quick example, embed the discrete order $\mathbb{Z}$ into itself by stretching by a factor of two; the direct limit of the iteration of this process is the dense order $\mathbb{Q}$.)

added 15 characters in body; deleted 2 characters in body; added 8 characters in body
Source Link
Joel David Hamkins
  • 236.2k
  • 44
  • 777
  • 1.4k
Loading
Source Link
Joel David Hamkins
  • 236.2k
  • 44
  • 777
  • 1.4k
Loading