# Zero sharp

$0^{\#}$ is a $\Sigma_3^1$ real number which cannot be proven to exist in $\text{ZFC}$. It's existence contradicts the Axiom of constructibility, $V=L$. In fact, it's existence is somewhat equivalent to $L$ being completely different from $V$.

## Definition

$0^{\#}$ is defined as the set of all Gödel numberings of first-order formula $\varphi$ such that $L\models\varphi(\aleph_0,\aleph_1...\aleph_n)$ for some $n$. Because of the stability of $\aleph_\omega$, $0^{\#}$ is equivalent to the set of all Gödel numberings of first-order formula $\varphi$ such that $L_{\aleph_{\omega}}\models\varphi(\aleph_0,\aleph_1...\aleph_n)$. This definition implies the existence of Silver Indiscernables. Moreover, it implies:

- Given any set $X\in L$ which is first-order definable in $L$, $X\in L_{\omega_1}$. This of course implies that $\aleph_1$ is not first-order definable in $L$, because $\aleph_1\not\in L_{\omega_1}$. This is already a disproof of $V=L$ (because $\aleph_1$ is first-order definable).
- For every $\alpha\in\omega_1^L$, every uncountable cardinal is $\alpha$-iterable, $\geq$ an $\alpha$-Erdős, and totally ineffable in $L$.
- There are $\mathfrak{c}$ many reals which are not constructible (that is, $x\not\in L$).

The existence of $0^\#$ is implied by:

- Chang's Conjecture.
- The existence of an $\omega_1$-iterable cardinal.
- The negation of the singular cardinal hypothesis ($\text{SCH}$).
- The axiom of determinacy ($\text{AD}$).

## $0^{\#}$ cardinal

$0^{\#}$ exists iff there is a nontrivial elementary embedding $j:L\rightarrow L$ (by a theorem of Kunen). The critical point of such an embedding is sometimes called a $0^{\#}$ cardinal, and sometimes called a $j:L\rightarrow L$ cardinal. These cardinals do not coincide with measurable cardinals for a long time. While the least measurable cardinal is $\Sigma_1^2$-describable, each of these cardinals is totally indescribable. Furthermore, the least measurable cardinal $\kappa$ such that $V_\kappa$ satisfies the existence of a measurable cardinal is not a $j:L\rightarrow L$ cardinal, and the least measurable cardinal $\kappa$ such that $V_\kappa$ satisfies the existence of such a cardinal is not a $j:L\rightarrow L$ cardinal, and so on.

However, the existence of a measurable suffices to prove the existence and consistency of a $j:L\rightarrow L$ cardinal.

*More information to be added here.*

## References

- Jech, Thomas J. Set Theory (The 3rd Millennium Ed.). Springer, 2003.