Suppose [math]F := \{x:S(x)\}[/math], where [math]S(x)[/math] is the statement, "[math]x=x[/math]". Since [math]S(x)[/math] is TRUE for every possible object [math]x[/math], [math]F[/math] should contain every possible object, [math]x[/math]. In other words, [math]F[/math] is the set of of everything! But one can prove that there is no set that contains everything.

For this, suppose there exists a set of everything; call it [math]E[/math].

Claim 1: Every set is a subset of [math]E[/math].

Proof:
Let [math]A[/math] be a set.
Consider [math]a \in A[/math].
But since [math]E[/math] is the set containing everything, [math]a \in E[/math].

This proves [math]A \subseteq E[/math].

Now, let [math]R[/math] be a set defined as [math]R:=\{x:x \notin x\}[/math], that is, the set of all sets that don't contain themselves.

Claim 2: [math]R \notin E[/math].

Proof:
Suppose [math]R \in E[/math].
Since [math]R[/math] is a set, from Claim 1, we have [math]R \subseteq E[/math].
Now, since [math]R \in E[/math], and [math]R \subseteq E[/math], either [math]R \in R[/math], or [math]R \in E \setminus R[/math].
That is, either [math]R \in R[/math], or [math]R \notin R[/math].

Case 1: [math]R \in R[/math]: This means, [math]R[/math] satisfies the property: "[math]x \notin x[/math]", i.e., [math]R \notin R[/math].

Case 2: [math]R \notin R[/math]: This means, [math]R[/math] doesn't satisfy the property: "[math]x \notin x[/math]", i.e., [math]R \in R[/math].

It's clear from cases 1 and 2 that [math]R \in R \implies R \notin R[/math] and [math]R \notin R \implies R \in R[/math].
That is, we have a contradiction here.
So, our assumption that [math]R \in E[/math] is wrong!

Therefore, [math]R \notin E[/math].

It's clear from claim 2, that we have constructed a set ([math]R[/math]) that doesn't belong to [math]E[/math]. This means [math]E[/math] is not the set of everything. So, finally we conclude that there is no set that contains everything.

Hence, your set [math]F := \{x:S(x)\}[/math] is not a set.