Ultrafilter condition: If A is a subset of X, then either A or X \ A is an element of U My confusion concerns ultrafilters on sets that are themselves power sets. 
If $X=\{\emptyset,\ \{1\},\ \{2\},\ \{3\},\ \{4\},\ \{1,2\},\ \{1,3\},\ \{1,4\},\ \{2,3\},\ \{2,4\},\ \{3,4\},\ \{1,2,3\},\ \{1,2,4\},\ \{1,3,4\},\ \{2,3,4\},\ \{1,2,3,4\}\ \}$ and the upset $\{1\}=U=\{\ \{1\},\ \{1,2\},\ \{1,3\},\ \{1,4\},\ \{1,2,3\},\ \{1,2,4\},\ \{1,3,4\},\ \{1,2,3,4\}\ \}$ is supposedly a principal ultrafilter (for visual delineation see https://en.wikipedia.org/wiki/Filter_%28mathematics%29), then how do I satisfy the criteria that "If $A$ is a subset of $X$, then either $A$ or $X\setminus A$ is an element of $U$"?
For example, could I let $A=\{\emptyset,\{2\}\}$ such that $A\notin U$, but $X\setminus A\notin U$ because $\{2,3,4\}\in X\setminus A,\{2,3,4\}\notin U$?
My understanding of the distinction between an element and a subset is unrefined, particularly with regard to power sets. 

It is fine to have filters on a power set of another set. But then the filter is a subset of the power set of the power set of that another set; rather than our original set which is the power set of another set.
But in the example that you gave, note that $W\in U$ if and only if $\{1\}\in W$. Indeed $\{1\}\notin A$, but therefore it is in its complement and so $X\setminus A\in U$.
If things like that confuse you, just replace the sets with other elements, $1,\ldots,16$ for example, and consider the ultrafilter on that set. Then simply translate it back to your original $X$.
