# Abstract It is proved that while $\Delta$ ([[Delta-extension]]) preserves the [[Numbers of logics|Hanf numbers]] of most logics, it fails, in a suitable Boolean extension, to preserve the Hanf-number of the logic with the [[Härtig quantifier]]. The preservation of some general [[Numbers of logics|Löwenheim-Skolem]] properties under $\Delta$ and variants of $\Delta$ is discussed. ## Projective classes >[!note] Notation >Suppose $K$ is a class of models of type $L$, $L^{\prime} \subseteq L$ and $\mathfrak{U}$ is a model of type $L^{\prime}$. > $ > E(\mathfrak{U}, K)=\left\{\mathfrak{B} \in K: \mathfrak{B} \upharpoonright L^{\prime}=\mathfrak{U}\right\} . > $ > The projection of $K$ along $L-L^{\prime}$ is defined as usual: > $ > \operatorname{Proj}_{L^{\prime}}(K)=\left\{\mathfrak{l l} \in \operatorname{Str}\left(L^{\prime}\right): E(\mathfrak{l l}, K) \neq \varnothing\right\}\left({ }^1\right) . > $ If $L^*$ is an abstract logic and $\varphi \in L^*$, we write $\operatorname{Proj}_L(\varphi)$ for $\operatorname{Proj}_L(\operatorname{Mod}(\varphi))$. >[!info] Definition > - $K$ is a *projective class* of $L^*$ if $K=\operatorname{Proj}_L(\varphi)$ for some $\varphi \in L^*$. > - The family of projective classes of $L^*$ is denoted $\Sigma\left(L^*\right)$. > - If $K$ and $\bar{K}$ are in $\Sigma\left(L^*\right)$ then $K$ is said to be *$\Delta$-definable in $L^\ast$*. > - The family of model classes which are $\Delta$-definable in $L^*$ is denoted $\Delta\left(L^*\right)$. > - The model class $K$ is a *simple projective class* of $L^*$, in symbols $K \in \Sigma_1^1\left(L^*\right)$, if $K=\operatorname{Proj}_{L^{\prime}}(\varphi)$ for some $\varphi \in L^*$ such that $L$ and $L^{\prime}$ have the same sorts. > - $\Delta_1^1\left(L^*\right)$ denotes the family of model classes $K$ such that $K$ and $\bar{K}$ are in $\Sigma_1^1\left(L^*\right)$. >[!note] Remark >Note the difference of terminology from [[Model classes]]. >[!note] Remark > Clearly $\Delta_1^1\left(L^*\right) \subseteq \Delta\left(L^*\right)$. Note that $\Sigma_1^1\left(L^{\mathrm{II}}\right)=L^{\mathrm{II}}$ and therefore $\Delta_1^1\left(L^{\mathrm{II}}\right)=L^{\mathrm{II}}$, but $L^{\mathrm{II}}$ $\neq \Delta\left(L^{\mathrm{II}}\right)$.