Proving certain collections are sets using axioms

For these problems (among others, but obviously didn't want to put them all here) I have to determine whether or not the following are sets using the axioms of set theory and certain theorems (specifically, those that show that there is an empty set, that the intersection between two sets is also a set, and that there is no set that contains every set; we also assume that $A$ and $B$ are non empty sets):

a) {$X: X \neq \emptyset$}

b) {$X: X \not\subseteq B$}

c) {$A, B, \emptyset$}

I believe the main problem I might be having is a misunderstanding of the different axioms, so I would appreciate it if someone could help verify/revise my solutions for these three:

a) Since the collection of all sets is not empty, then it would be in the set. This implies that this collection is a set, which is a contradiction. Hence, this is not a set.

b) By the axiom of specification, the {$x \in X: x \notin B$} is a set. This is equivalent to {$X: X \not \subseteq B$}, hence it is a set.

c) By the axiom of pairing, we know that {$A, B$} is a set. Furthermore, we know $\emptyset$ is a set, so we may conclude by the axiom of union that the union of these two sets, {$A, B, \emptyset$}, is a set.

Thank you kindly for all and any help!

• Which axioms? More precisely, exactly what kind of axiom system are you talking about -- Zermelo-Fraenkel or Gödel-Bernays-von Neumann? – murray Feb 26 '17 at 21:10
• Axioms of Extension, Specifcation, Pairing, Union, Powers, and Regularity. I am uncertain of which system we are using unfortunately, but from looking at their basic backgrounds I'd guess it would be Zermelo-Fraenkel. – Gizmo Feb 26 '17 at 21:46

a) You are correct in saying that this is not a set, but your reasoning is a bit faulty. $\{X : X \neq \emptyset\}$ represents the collection of all sets which are non-empty. Since the collection (class) of all sets isn't a set then it can't be in the collection to begin with. Now suppose this collection, let's call it $S$ is in fact a set. Then $S \cup \{\emptyset\}$ is a set, but this is the collection of all sets (why?) a contradiction and so $S$ can't be a set.
b) You're wrong here. $\{x \in X : x \not \in B\}$ not equivalent to $\{X : X \not \subseteq B\}$ since the latter is the collection of all sets which aren't subsets of $B$ while the former is the set of elements in $X$ which aren't elements of $B$. If this collection is a set then you can in fact create the set of all sets (why? Try using the powerset and union axioms) and so it can't be a set.
c) This one's right, you could add that $\{\emptyset\}$ is a set because of the axiom of pairing and the fact that $\emptyset$ is a set.