Given a cardinal number $\aleph_\alpha$, is it known whether or not $\aleph_\alpha=\beth_\alpha$ is independent of ZFC?

One could define $\mathrm{CH}(\aleph_\alpha)$ as $\aleph_\alpha=\beth_\alpha$. For which cardinals is it true that $\mathrm{ZFC}\models\mathrm{CH(\kappa)}$?

Clearly, $\mathrm{ZFC}\models\mathrm{CH}(\aleph_0)$, and $\mathrm{CH}(\aleph_1)$ is equivalent to $\mathrm{CH}$ (and is thus independent of ZFC). Because $\mathrm{GCH}$ is indepenedent of ZFC, $\neg\mathrm{CH}(\kappa)$ cannot be proven for any cardinal $\kappa$.

If $\mathrm{ZFC}\models\mathrm{CH}(\aleph_{\alpha+1})$, then $\aleph_{\alpha+1}=\beth_{\alpha+1}$. Thus, if we assume $\aleph_\alpha<\beth_\alpha$, we get a contradiction, because $\aleph_\alpha<\beth_\alpha$ and then $\aleph_{\alpha+1}<\beth_{\alpha+1}$. So, if $\mathrm{CH}(\kappa^+)$ is provable, then $\mathrm{CH}(\kappa)$ is also provable. Thus, for cardinals $\kappa<\aleph_\omega$, $\mathrm{CH}(\kappa)$ is independent of ZFC.

It is known that $\alpha\leq\aleph_\alpha$, and thus if $\beth_\kappa=\kappa$ then $\aleph_\kappa=\kappa$ (i.e. all $\beth$-fixed points are also $\aleph$-fixed points). Therefore, $\beth_\kappa=\aleph_\kappa=\kappa$, and $\beth_\kappa=\aleph_\kappa$. So, for all $\beth$-fixed points $\kappa$, $\mathrm{CH}(\kappa)$ is provable from ZFC.

Are there any other known $\kappa$ with $\mathrm{CH}(\kappa)$ independent of ZFC? Is $\exists\kappa\neq\aleph_0(\mathrm{CH}(\kappa))$ independent of ZFC? (If so, $\beth$ fixed points are independent of ZFC as well.)

Edit: Of course $\beth$-fixed points are not independent of ZFC, they do exist by $\alpha\rightarrow\beth_\alpha$ being a normal function. So, ZFC actually proves the existence of cardinals larger than $\aleph_0$ which fulfill GCH.

Given a cardinal number $\aleph_\alpha$, is it known whether or not $\aleph_\alpha=\beth_\alpha$ is independent of ZFC?

One could define $\mathrm{CH}(\aleph_\alpha)$ as $\aleph_\alpha=\beth_\alpha$. For which cardinals is it true that $\mathrm{ZFC}\models\mathrm{CH(\kappa)}$?

Clearly, $\mathrm{ZFC}\models\mathrm{CH}(\aleph_0)$, and $\mathrm{CH}(\aleph_1)$ is equivalent to $\mathrm{CH}$ (and is thus independent of ZFC). Because $\mathrm{GCH}$ is indepenedent of ZFC, $\neg\mathrm{CH}(\kappa)$ cannot be proven for any cardinal $\kappa$.

If $\mathrm{ZFC}\models\mathrm{CH}(\aleph_{\alpha+1})$, then $\aleph_{\alpha+1}=\beth_{\alpha+1}$. Thus, if we assume $\aleph_\alpha<\beth_\alpha$, we get a contradiction, because $\aleph_\alpha<\beth_\alpha$ and then $\aleph_{\alpha+1}<\beth_{\alpha+1}$. So, if $\mathrm{CH}(\kappa^+)$ is provable, then $\mathrm{CH}(\kappa)$ is also provable. Thus, for cardinals $\kappa<\aleph_\omega$, $\mathrm{CH}(\kappa)$ is independent of ZFC.

It is known that $\alpha\leq\aleph_\alpha$, and thus if $\beth_\kappa=\kappa$ then $\aleph_\kappa=\kappa$ (i.e. all $\beth$-fixed points are also $\aleph$-fixed points). Therefore, $\beth_\kappa=\aleph_\kappa=\kappa$, and $\beth_\kappa=\aleph_\kappa$. So, for all $\beth$-fixed points $\kappa$, $\mathrm{CH}(\kappa)$ is provable from ZFC.

Are there any other known $\kappa$ with $\mathrm{CH}(\kappa)$ independent of ZFC? Is $\exists\kappa\neq\aleph_0(\mathrm{CH}(\kappa))$ independent of ZFC? (If so, $\beth$ fixed points are independent of ZFC as well.)

Edit: Of course $\beth$-fixed points are not independent of ZFC, they do exist by $\alpha\mapsto\beth_\alpha$ being a normal function. So, ZFC actually proves the existence of cardinals larger than $\aleph_0$ which fulfill GCH.

Given a cardinal number $\aleph_\alpha$, is it known whether or not $\aleph_\alpha=\beth_\alpha$ is independent of ZFC?

One could define $\mathrm{CH}(\aleph_\alpha)$ as $\aleph_\alpha=\beth_\alpha$. For which cardinals is it true that $\mathrm{ZFC}\models\mathrm{CH(\kappa)}$?

Clearly, $\mathrm{ZFC}\models\mathrm{CH}(\aleph_0)$, and $\mathrm{CH}(\aleph_1)$ is equivalent to $\mathrm{CH}$ (and is thus independent of ZFC). Because $\mathrm{GCH}$ is indepenedent of ZFC, $\neg\mathrm{CH}(\kappa)$ cannot be proven for any cardinal $\kappa$.

If $\mathrm{ZFC}\models\mathrm{CH}(\aleph_{\alpha+1})$, then $\aleph_{\alpha+1}=\beth_{\alpha+1}$. Thus, if we assume $\aleph_\alpha<\beth_\alpha$, we get a contradiction, because $\aleph_\alpha<\beth_\alpha$ and then $\aleph_{\alpha+1}<\beth_{\alpha+1}$. So, if $\mathrm{CH}(\kappa^+)$ is provable, then $\mathrm{CH}(\kappa)$ is also provable. Thus, for cardinals $\kappa<\aleph_\omega$, $\mathrm{CH}(\kappa)$ is independent of ZFC.

It is known that $\alpha\leq\aleph_\alpha$, and thus if $\beth_\kappa=\kappa$ then $\aleph_\kappa=\kappa$ (i.e. all $\beth$-fixed points are also $\aleph$-fixed points). Therefore, $\beth_\kappa=\aleph_\kappa=\kappa$, and $\beth_\kappa=\aleph_\kappa$. So, for all $\beth$-fixed points $\kappa$, $\mathrm{CH}(\kappa)$ is provable from ZFC.

Are there any other known $\kappa$ with $\mathrm{CH}(\kappa)$ independent of ZFC? Is $\exists\kappa(\mathrm{CH}(\kappa))$ independent of ZFC? (If so, $\beth$ fixed points are independent of ZFC as well.)

Given a cardinal number $\aleph_\alpha$, is it known whether or not $\aleph_\alpha=\beth_\alpha$ is independent of ZFC?

One could define $\mathrm{CH}(\aleph_\alpha)$ as $\aleph_\alpha=\beth_\alpha$. For which cardinals is it true that $\mathrm{ZFC}\models\mathrm{CH(\kappa)}$?

Clearly, $\mathrm{ZFC}\models\mathrm{CH}(\aleph_0)$, and $\mathrm{CH}(\aleph_1)$ is equivalent to $\mathrm{CH}$ (and is thus independent of ZFC). Because $\mathrm{GCH}$ is indepenedent of ZFC, $\neg\mathrm{CH}(\kappa)$ cannot be proven for any cardinal $\kappa$.

If $\mathrm{ZFC}\models\mathrm{CH}(\aleph_{\alpha+1})$, then $\aleph_{\alpha+1}=\beth_{\alpha+1}$. Thus, if we assume $\aleph_\alpha<\beth_\alpha$, we get a contradiction, because $\aleph_\alpha<\beth_\alpha$ and then $\aleph_{\alpha+1}<\beth_{\alpha+1}$. So, if $\mathrm{CH}(\kappa^+)$ is provable, then $\mathrm{CH}(\kappa)$ is also provable. Thus, for cardinals $\kappa<\aleph_\omega$, $\mathrm{CH}(\kappa)$ is independent of ZFC.

It is known that $\alpha\leq\aleph_\alpha$, and thus if $\beth_\kappa=\kappa$ then $\aleph_\kappa=\kappa$ (i.e. all $\beth$-fixed points are also $\aleph$-fixed points). Therefore, $\beth_\kappa=\aleph_\kappa=\kappa$, and $\beth_\kappa=\aleph_\kappa$. So, for all $\beth$-fixed points $\kappa$, $\mathrm{CH}(\kappa)$ is provable from ZFC.

Are there any other known $\kappa$ with $\mathrm{CH}(\kappa)$ independent of ZFC? Is $\exists\kappa\neq\aleph_0(\mathrm{CH}(\kappa))$ independent of ZFC? (If so, $\beth$ fixed points are independent of ZFC as well.)

Edit: Of course $\beth$-fixed points are not independent of ZFC, they do exist by $\alpha\rightarrow\beth_\alpha$ being a normal function. So, ZFC actually proves the existence of cardinals larger than $\aleph_0$ which fulfill GCH.