Is the set of the empty set a subset of a set that has listed the empty set and set of the empty set?

Answer 1

Yes. This, of course, is under only the assumption that the two relations of element and containment are well defined relations between any two objects that are presumed to exist. Such a question can not be postulated if the empty set is absent from existence, so it follows the existence of the empty set is presumed to exist.

When considering which objects the question depicts , the representation "a set that has listed the empty set and the set of the empty set" for such an object obtains ambiguous interpretations. However, the proposition is true under each interpretation, because the containment of the set containing the empty set is in reference to the object "a set" that has the proposed characteristics. Because one of those characteristics is that within the expression of this set, the empty set is listed, inclusion of the empty set in any set that is found to have these characteristics results in a set that also meets all these characteristics. Hence, for every interpretation of the question above, it follows that the answer is yes.

In the case that such ambiguity is accidental, or in the least unintentional to the problem, one may simply represent the sets that are depicted in the question using set notation:

{0} {0,{0}}

As one can see, the first set is a subset of the second set because there exists a collection of elements (in this case, the empty set 0) that are in the second set such that the set containing such a collection is identical to the first set.

In other words, the first set is equivalent to the second set intersected with itself. So it is clear that the answer is yes.