This is a question I was given in the exam: show that if $\lambda, \kappa$ are uncountable cardinals then
$$(H\_\lambda,\in) \prec\_1 (H\_\kappa,\in)$$
where $H\_\lambda$ is the class of all sets $x$ such that $|TC(x)| < \lambda$. ($TC(x)$ refers to the transitive closure of $x$.)


Equivalently, given any $\Sigma\_1$ formula $\varphi(x\_1,x\_2,...,x\_n)$ with free variables shown and $a\_1,a\_2,...,a\_n \in H\_\lambda$ then $(H\_\lambda,\in) \models \varphi(a\_1,...,a\_n)$ if and only if $(H\_\kappa,\in) \models \varphi(a\_1,...,a\_n)$.


I have tried hard without success. Can anyone give me a hint?



$\Sigma_1$ elementary substructure
I assume that you mean to assume $\lambda\leq\kappa$. 


For any $\vec x\in H\_\lambda$, we may by the Lowenheim-Skolem theorem find an elementary substructure $X\subset H\_\kappa$ with $TC(\vec x)\subset X$ and $X$ having size less than $\lambda$. In particular, $X$ has a witness $z$ for your $\Sigma\_1$ statement $\exists z\,\psi(\vec x,z)$, so that $\psi(\vec x,z)$ for this particular $z$. Let $X\_0$ be the Mostowski collapse of $X$. This is a transitive set of size less than $\lambda$ and hence $X\_0\subset H\_\lambda$. Furthermore, $\vec x$ is not changed by the collapse. Thus, if $z\_0$ is the set $z$ collapses to, then we have $\psi(\vec x,z\_0)$ in $X\_0$ and hence also in $H\_\lambda$. So we've found a witness in $H\_\lambda$, and so $H\_\lambda$ is $\Sigma\_1$ elementary in $H\_\kappa$, as desired.


