This is going to take some build up to completely describe what is a very strange question I seem to have walked into by accident:

For a model M and object $\theta$, we may define the two predicates

$$ \mathcal{C}(M, \theta ) \iff M \vDash \forall \alpha \in \dot{\theta}\ \forall f: \alpha \to \dot{\theta}\ \exists \gamma \in \dot{\theta}\ \forall \xi \in \alpha\ (f(\xi) \neq \gamma)$$

(there is nothing that can be brought into one-to-one correspondence with $\theta$ with in $M$)

$$ \mathbb{C}(M, \theta) \iff M \vDash \forall f\in C^{\infty}(\mathbb{R})\ \exists O \subset \mathbb{R} ( O \triangle f[\theta]\ \text{ is meager}) $$

(the image of $\theta$ under every continuous function with in M, has meager difference with an open set.)

Now, here comes the question: Can these two statements every be compared?

• It turns out the answer depends strongly on the particular context in which the statement or question was asked. Formally these statements have no meaning without an interpretation which witnesses the consistency of the principles which govern the interpretation. As statements of language they are just symbols and words arranged so that once we give them context they have meaning.

• In this particular situation there is no simple correct answer, because just like the emotions of the people that will read this particular paragraph and the apology this mathematician is humbly attempting to convey to another mathematician, it is impossible to separate regularity from cardinality without first asserting that for some rational $\delta$ and ordinal $\omega$, we have $\delta < \omega$ or $\omega < \delta$.

  In much the same way an apology is different from the emotions it illicits, the word emotion is hollow without an instance of an apology to interpret it. As such these two statements and types of objects must always remain incompatible; because otherwise they will inevitably lead to a contradiction.

-- MB.

This is going to take some build up to completely describe what is a very strange question I seem to have walked into by accident:

For every partial order $\mathbb{P}$ and regular cardinal $\lambda > \omega$ we can define the following two statements

$$ \mathcal{C}(\mathbb{P}, \lambda) \iff 1 \Vdash_{\mathbb{P}} \forall \alpha \in \check{\lambda}\ \forall f: \alpha \to \check{\lambda}\ \exists \gamma \in \check{\lambda}\ \forall \xi \in \alpha\ (f(\xi) \neq \gamma)$$

(this is the formalized version of the statement "$\mathbb{P}$ preserves $\lambda$ is a cardinal" in the forcing language, this statement is normally certified by reasoning which does not involve the forcing relation and depends on the structure of $\mathbb{P}$-names)

and

$$ Cof(\mathbb{P}, \lambda) \iff 1 \Vdash_{\mathbb{P}} \forall \alpha \in \check{\lambda}\ \forall f:\alpha \to \check{\lambda}\ \exists \gamma \in \check{\lambda} \ (\sup(ran(f)) \le \gamma) $$

(Again a forcing language version of the statement $\mathbb{P}$ preserves $\forall \alpha \in \lambda \ (cf(\alpha) < cf(\lambda))$: we had to be careful here because we need to be able to distinguish between the two (If this is not the correct way to formalize this please let me know.))

Now, here comes the question: Does the following conjunction:

$\exists \lambda > \omega\ \exists\ \mathbb{P}$ such that

• $\lambda$ is a Regular cardinal.
• $\vert \mathbb{P} \vert = \lambda^{+}$
• $\forall \mu \ (\mu$ is a cardinal $\implies \mathcal{C}(\mu,\mathbb{P}))$
• $\neg Cof(\lambda, \mathbb{P})$

Imply there is an inner model with a measurable cardinal? (changed based on the answers.)

(Namba for $\omega_2$ and threading a generic square collapse cardinals; moreover if $ 0^\sharp $ exists then $\aleph_\omega^{V}$ is regular in $L$ producing a model in some sense)

Edit:

(It was not my intention to scare a lot of nice mice)

(also, mice need to be more damn direct and stop subtly hinting things.... didn't realize what was going on until just now....)

This is going to take some build up to completely describe what is a very strange question I seem to have walked into by accident:

For every partial order $\mathbb{P}$ and regular cardinal $\lambda > \omega$ we can define the following two statements

$$ \mathcal{C}(\mathbb{P}, \lambda) \iff 1 \Vdash_{\mathbb{P}} \forall \alpha \in \check{\lambda}\ \forall f: \alpha \to \check{\lambda}\ \exists \gamma \in \check{\lambda}\ \forall \xi \in \alpha\ (f(\xi) \neq \gamma)$$

(this is the formalized version of the statement "$\mathbb{P}$ preserves $\lambda$ is a cardinal" in the forcing language, this statement is normally certified by reasoning which does not involve the forcing relation and depends on the structure of $\mathbb{P}$-names)

and

$$ Cof(\mathbb{P}, \lambda) \iff 1 \Vdash_{\mathbb{P}} \forall \alpha \in \check{\lambda}\ \forall f:\alpha \to \check{\lambda}\ \exists \gamma \in \check{\lambda} \ (\sup(ran(f)) \le \gamma) $$

(Again a forcing language version of the statement $\mathbb{P}$ preserves $\forall \alpha \in \lambda \ (cf(\alpha) < cf(\lambda))$: we had to be careful here because we need to be able to distinguish between the two (If this is not the correct way to formalize this please let me know.))

Now, here comes the question: Does the following conjunction:

$\exists \lambda > \omega\ \exists\ \mathbb{P}$ such that

• $\lambda$ is a Regular cardinal.
• $\vert \mathbb{P} \vert = \lambda^{+}$
• $\forall \mu \ (\mu$ is a cardinal $\implies \mathcal{C}(\mu,\mathbb{P}))$
• $\neg Cof(\lambda, \mathbb{P})$

Imply there is an inner model with a measurable cardinal? (changed based on the answers.)

(Namba for $\omega_2$ and threading a generic square collapse cardinals; moreover if $ 0^\sharp $ exists then $\aleph_\omega^{V}$ is regular in $L$ producing a model in some sense)

Edit:

(It was not my intention to scare a lot of nice mice)

(also, mice need to be more damn direct and stop subtly hinting things.... didn't realize what was going on until just now....)

This is going to take some build up to completely describe what is a very strange question I seem to have walked into by accident:

For a model M and object $\theta$, we may define the two predicates

$$ \mathcal{C}(M, \theta ) \iff M \vDash \forall \alpha \in \dot{\theta}\ \forall f: \alpha \to \dot{\theta}\ \exists \gamma \in \dot{\theta}\ \forall \xi \in \alpha\ (f(\xi) \neq \gamma)$$

(there is nothing that can be brought into one-to-one correspondence with $\theta$ with in $M$)

$$ \mathbb{C}(M, \theta) \iff M \vDash \forall f\in C^{\infty}(\mathbb{R})\ \exists O \subset \mathbb{R} ( O \triangle f[\theta]\ \text{ is meager}) $$

(the image of $\theta$ under every continuous function with in M, has meager difference with an open set.)

Now, here comes the question: Can these two statements every be compared?

• It turns out the answer depends strongly on the particular context in which the statement or question was asked. Formally these statements have no meaning without an interpretation which witnesses the consistency of the principles which govern the interpretation. As statements of language they are just symbols and words arranged so that once we give them context they have meaning.

• In this particular situation there is no simple correct answer, because just like the emotions of the people that will read this particular paragraph and the apology this mathematician is humbly attempting to convey to another mathematician, it is impossible to separate regularity from cardinality without first asserting that for some rational $\delta$ and ordinal $\omega$, we have $\delta < \omega$ or $\omega < \delta$.

  In much the same way an apology is different from the emotions it illicits, the word emotion is hollow without an instance of an apology to interpret it. As such these two statements and types of objects must always remain incompatible; because otherwise they will inevitably lead to a contradiction.

-- MB.

No longer relevant, please remove the question.

This is going to take some build up to completely describe what is a very strange question I seem to have walked into by accident:

For every partial order $\mathbb{P}$ and regular cardinal $\lambda > \omega$ we can define the following two statements

$$ \mathcal{C}(\mathbb{P}, \lambda) \iff 1 \Vdash_{\mathbb{P}} \forall \alpha \in \check{\lambda}\ \forall f: \alpha \to \check{\lambda}\ \exists \gamma \in \check{\lambda}\ \forall \xi \in \alpha\ (f(\xi) \neq \gamma)$$

(this is the formalized version of the statement "$\mathbb{P}$ preserves $\lambda$ is a cardinal" in the forcing language, this statement is normally certified by reasoning which does not involve the forcing relation and depends on the structure of $\mathbb{P}$-names)

and

$$ Cof(\mathbb{P}, \lambda) \iff 1 \Vdash_{\mathbb{P}} \forall \alpha \in \check{\lambda}\ \forall f:\alpha \to \check{\lambda}\ \exists \gamma \in \check{\lambda} \ (\sup(ran(f)) \le \gamma) $$

(Again a forcing language version of the statement $\mathbb{P}$ preserves $\forall \alpha \in \lambda \ (cf(\alpha) < cf(\lambda))$: we had to be careful here because we need to be able to distinguish between the two (If this is not the correct way to formalize this please let me know.))

Now, here comes the question: Does the following conjunction:

$\exists \lambda > \omega\ \exists\ \mathbb{P}$ such that

• $\lambda$ is a Regular cardinal.
• $\vert \mathbb{P} \vert = \lambda^{+}$
• $\forall \mu \ (\mu$ is a cardinal $\implies \mathcal{C}(\mu,\mathbb{P}))$
• $\neg Cof(\lambda, \mathbb{P})$

Imply there is an inner model with a measurable cardinal? (changed based on the answers.)

(Namba for $\omega_2$ and threading a generic square collapse cardinals; moreover if $ 0^\sharp $ exists then $\aleph_\omega^{V}$ is regular in $L$ producing a model in some sense)

Edit:

(It was not my intention to scare a lot of nice mice)

(also, mice need to be more damn direct and stop subtly hinting things.... didn't realize what was going on until just now....)