Definable sets in set theory A formula $\phi$ with a single free variable in the language of set theory with a single free-variable defines a set $x$ if $\phi(x)$ holds but $\phi(y)$ does not hold if $y \neq x$.
Is every set definable by a  $\Sigma_1$ formula countable?
Is every set definable by a $\Pi_1$ formula countable?
 A: The answer to the first question is yes, every set definable by a $\Sigma_1$-formula is countable. This is an immediate consequence of the Levy reflection theorem, which states that $H_{\omega_1}\prec_{\Sigma_1} V$. So the assertion that there is a set satisfying that definition is also $\Sigma_1$ and therefore must be in $H_{\omega_1}$ and hence it is countable. 
(The Levy reflection theorem itself is not difficult to prove: fix a witness for any $\Sigma_1$ property, and then take a countable elementary substructure inside a bigger $H_\theta$, and collapse it. The collapse of the witness will also witness the property, since any transitive set is correct about $\Delta_0$ assertions.) 
For the second question, the answer is no, since $x=P(\omega)$ is defined by the $\Pi_1$ formula asserting first that every element of $x$ is a subset of $\omega$ (that part is $\Delta_0$) and secondly that every $z$ that is a subset of $\omega$ is an element of $x$. We don't need $\omega$ as a parameter, since it is definable as the unique limit ordinal not containing any limit ordinals as elements, and that is $\Delta_0$, so it doesn't increase the complexity of the definition of $P(\omega)$. 
