Is $L_{\omega+1}$ uncountable? I'm trying to understand the constructible universe, but I'm having trouble understanding what $\operatorname{Def}(X)$ means (especially since I don't really understand the $\models$ symbol).
Is $L_{\omega+1}$ an uncountable set? Why or why not? If not, what's the smallest $\alpha$ for which $L_\alpha$ is uncountable?
Also, if anyone can explain the $\models$ notation used in the article (or provide a link to an explanation), that would be very helpful.
 A: The set $\mathrm{Def}(X)$ consists of all subsets of $X$ first-order definable in the structure $(X,\in)$ from parameters. What matters here is that there are only countably many formulas, and only $|X^{<\omega}|$ possible tuples of parameters from $X$, so there are only $\aleph_0|X^{<\omega}|$ possible elements in $\mathrm{Def}(X)$. If $X$ is countable, this means that $\mathrm{Def}(X)$ is also countable.
If you are not comfortable with the description of $\mathrm{Def}$ in terms of first-order definability, note that this set can also be realized by applying to $X$ a countable number of (finitary) operations, explicitly listed for instance in Chapter V of the first edition of Kunen's set theory book. These operations correspond to Boolean operators (underlying the fact that if $\phi,\psi$ are formulas, so are $\lnot\phi$ and $\phi\land\psi$) and projections (underlying the fact that if $\phi$ is a formula, so is $\exists x\,\phi$). Again, this implies, by the same counting as in the previous paragraph, that $\mathrm{Def}(X)$ is countable if $X$ is. 
An easy inductive argument now gives us that if $\alpha$ is a countable ordinal, then $L_\alpha$ is countable as well. On the other hand, another (slightly more involved) inductive argument shows that $L_\alpha\subsetneq L_{\alpha+1}$ for all $\alpha$, which implies that $L_{\omega_1}$ is uncountable.  
