How does the universal set apply to the truth set? 
If the truth set is identical to it's universal set, then the statement is logically true statement and it's called a tautology.

Can someone please explain the universal set to me in this context? I thought the universal set has to be defined for each problem, so how is it defined, for, say,
$\neg (p\wedge q)\iff (\neg p) \vee (\neg q)$
I'm not taking a class, just interested in learning more about math. Thanks 
 A: The terms "truth set" and "universal set" aren't really common, but from context it sounds like they mean the following:

Fixing a propositional language $P$, the universal set associated to $P$ is the set of all valuations of propositional formulas built from $P$ (or essentially equivalently, the set of all maps $P\rightarrow\{True, False\}$), and for a formula $\varphi$ built from $P$ the truth set of $\varphi$ is the subset of the universal set consisting of all valuations making $\varphi$ true.

Now you are correct that $\varphi$ alone can't tell us what $P$ is - e.g. if $\varphi$ is $p\rightarrow q$, then $\{p,q\}$ and $\{p,q,r\}$ would each make sense. However, $\varphi$ does have a minimal language associated to it, namely the set of all propositional atoms which actually occur in $\varphi$, so we often default to that language.
Moreover, in this context - and indeed in many contexts - the choice of language doesn't matter:

Suppose $P_1,P_2$ are two languages each of which contains every propositional atom occurring in $\varphi$. Then $\varphi$ is a tautology in the sense of $P_1$ iff $\varphi$ is a tautology in the sense of $P_2$.

This is a good exercise: the key point is that whether or not a valuation $v$ makes a sentence $\varphi$ true is determined entirely by the restriction of $v$ to the propositional atoms actually occurring in $\varphi$. So we don't have to worry about the choice of language issue here.

That said, sometimes the choice of language does matter. For example, in the context of first-order logic it is important to specify the language when talking about the decidability or completeness of a theory: Presburger arithmetic, for example, is complete and decidable as a $\{+\}$-language but not as a $\{+,\cdot\}$-language.
