The collection $\{x : \exists y(x\in y)\}$ is not a set. ("Set Theory: A First Course" by Daniel W. Cunningham) I am reading "Set Theory: A First Course" by Daniel W. Cunningham.
There is the following exercise in this book:

Exercise 2.1.33
The collection $\{x : \exists y(x\in y)\}$ is not a set.

My answer is here:
Asume that $\exists A\forall x(x\in A)$ holds.
Let $\forall x(x\in A_0)$ holds.
By Subset Axiom, $\exists S\forall x(x\in S\leftrightarrow(x\in A_0\land x\notin x))$ holds.
Let $\forall x(x\in B_0\leftrightarrow(x\in A_0\land x\notin x))$ holds.
$B_0\in B_0\lor B_0\notin B_0$ holds.
If $B_0\in B_0$ holds, then $B_0\in A_0\land B_0\notin B_0$ holds.
So, both $B_0\in B_0$ and $B_0\notin B_0$ hold.
This is a contradiction.
If $B_0\notin B_0$ holds, then $\neg (B_0\in A_0\land B_0\notin x)$ holds.
So, $B_0\notin A_0\lor B_0\in B_0$ holds.
Since $B_0\notin B_0$, $B_0\notin A_0$ holds.
This is a contradiction.
So, $\neg(\exists A\forall x(x\in A))$ holds.
Assume that $\exists C\forall x(x\in C\leftrightarrow\exists y(x\in y))$ holds.
Let $\forall x(x\in C_0\leftrightarrow\exists y(x\in y))$ holds.
By Pairing Axiom, $\forall u\forall v\exists D\forall x(x\in D\leftrightarrow x=u\lor x=v))$ holds.
So, $\forall x\exists D\forall z(z\in D\leftrightarrow z=x)$ holds.
So, $\forall x\exists D(x\in D\leftrightarrow x=x)$ holds.
So, $\forall x\exists D(x\in D)$ holds.
So, $\forall x\exists y(x\in y)$ holds.
So, $\exists C\forall x(x\in C)$ holds.
This is a contradiction.
So, $\neg(\exists C\forall x(x\in C\leftrightarrow\exists y(x\in y)))$ holds.

My question is here:
In this book, there doesn't exist a definition of a collection.
But the author wrote "The collection $\{x : \exists y(x\in y)\}$ is not a set.".
Is this OK?
I think we can write $\{x : \phi(x)\}$ if and only if $\exists S\forall x(x\in S\leftrightarrow\phi(x))$.
 A: A collection (see page 1) is an "informal" set, while sets in the precise definition of mathematical set theory are all and only those objects whose existence is asserted (proved) by the theory (page 23).
The "collecting" operation "is usually identified by enclosing its elements with braces (curly brackets)".
The empty set $\emptyset$ exists (Axiom 2) and the set $\{ x,y \}$ exists, provided sets $x,y$ exits (Axiom 4).
The set $\mathbb N$ of natural number does not exist ab initio, until in due course we will prove its existence.
Having said that, for every $x$ the set $\{ x \}$ exists, by Pairing (see Ex.10, page 27) and $x$ is its only element.
Thus, from formula $x \in \{ x \}$, by predicate logic we derive: $\exists y (x \in y)$, and this hold for every $x$.
Now we have to use the basic property of the "collection forming" operator $\{ x \mid P(x) \}$ (see page 3):

$x \in \{ z \mid P(z) \} \text {  iff  } P(x)$.

Using it with the formula above, we have that:

$x \in \{ z \mid ∃y(z ∈ y) \} \text {  iff  } ∃y(x ∈ y)$.

But we have proved that $\exists y (x \in y)$ holds, for every $x$, and thus $x \in \{ z \mid ∃y(z ∈ y) \}$, for every $x$.
But this amounts to saying that the collection $\{ z \mid ∃y(z ∈ y) \}$ is the "universal" one: the collection of which every set is a member.
But by Th.2.1.2 (page 30) we know that there is no universal set; thus, the collection $\{ z \mid ∃y(z ∈ y) \}$ is not a set.
