Is it true that $A \in A$? I defined the set $A$ as follow:
\begin{align}
A_0 & =\varnothing \\
A_1 & =\{A_0\}=\{\varnothing\} \\
A_2 & =\{A_1\}=\{\{\varnothing\}\} \\
A_3 & =\{A_2\}=\{\{\{\varnothing\}\}\} \\
& {}\ \ \vdots \\
A & =A_\infty= \{\{\{\cdots\{\varnothing\}\cdots\}\}\}
\end{align}
And now this is clear that $A\in A$.   
Is there any mistake in my conclusion?
 A: You are trying to use intuition to define a set. Which is where your problem is.
It almost seems like you want to write something like this,
$$A_0 = \emptyset$$
$$A_n = \{ A_{n - 1} \}$$
$$A = \lim_{n \to \infty} A_n$$
But what does that limit mean? It needs to be defined itself, and it isn't in your definition.
TLDR: You can't do that! :P
A: The definition you've given of $A$ is doesn't actually say what $A$ is.  $A$ is presumably supposed to be a set.  To say which set it is is to say what things are members of it.  Can you name one thing that is a member of $A$ that you have clearly defined?
There is something somewhat similar to this that is in somewhat standard use: the von Neumann ordinals:
\begin{align}
& \varnothing \\
& \{\varnothing\} \\
& \{\varnothing,\{\varnothing\}\} \\
& \{\varnothing,\{\varnothing\},\{\varnothing,\{\varnothing\}\}\} \\
& {}\ \ \ \vdots
\end{align}
The members of each von Neumann ordinal are the previous von Neumann ordinals.
The set of all finite von Neumann ordinals is the smallest infinite von Neumann ordinals $\omega$.  Then $\omega\cup\{\omega\}$ is the next one, called $\omega+1$ (but not called $1+\omega$ since one defines a non-commutative addition on these.  And so on.
The set of all countably infinite von Neumann ordinals us uncountably infinite, and Cantor defined $\aleph_1$ to be its cardinality.  This is not to be confused with $2^{\aleph_0}$.
A: I think there are two separate questions here:


*

*Why does the sequence of sets $\{\}, \{\{\}\}, \{\{\{\}\}\}$, . . . not "approach a limit"?

*Is it possible to have a set $A$ such that $A\in A$?
The second one is more easily dealt with, since it a precise question. The usual axioms of set theory, ZFC, rule out any such set; specifically, the axiom of Foundation (or Regularity) prevents the existence of such sets. However, if we remove the axiom of Foundation from ZFC, then such sets can exist, and there are even set theories - Quine's NF comes to mind, as well as its better-behaved variant NFU - which require such sets to exist. 
For the former, I think you would find it helpful to try to define in what sense the sequence of sets $\{\}, \{\{\}\}, \{\{\{\}\}\}$, . . . does (in your mind) have a limit, and then try to see what it would take to prove that. Note that there is a clear notion of limit for a monotonic sequence of sets: e.g., if $\alpha>\beta$ implies $A_\alpha\supset A_\beta$ for $\alpha<\kappa$, then it makes sense to define $$\lim_{\alpha\rightarrow\kappa}A_\alpha=\bigcup_{\alpha\in\kappa} A_\alpha,$$ and similarly for descending sequences of sets. But your sequence of sets isn't monotonic, so there's no natural notion of what its "limit" should be. And, even if there were, you would need to argue that properties of terms of the sequence induce properties of the limit, which depends on the properties in question.
A: Set theory forbids the existence of a set $A$ such that $A \in A$, as that kind of construction easily leads to paradoxes, such as Russell's paradox.
In ZFC, the axiom of regularity has this immediate consequence.
