TITLE: Is it true that $A \in A$?
QUESTION [5 upvotes]: 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?

REPLY [2 votes]: 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 $2^{\aleph_0}$.