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This is, I believe, a relatively simple set-theoretical question. I am ,however, not sure of the answer. If we take a set, say $A$, and if we call the power set of $A$, $P_{1}(A)$, and we define $$P_{n}(A)=P_{1}(P_{n-1}(A)),$$ and then if we take the limit $$\lim_{n\to\infty} P_{n}(A),$$ is this set actually a set in NBG or does it lead to some sort of contradiction?

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This is a confusing notation, note that you are indexing by ordinals and not by real numbers, and the notation $\infty$ makes so that you think about it as if you are using the real numbers, and the $\infty$ is that formal symbol of "the point beyond the edge of the universe".

In ZF we have the axiom schema of replacement which tells us that definable functions whose domain is a set have a set for an image. In particular the function $n\mapsto\mathcal P^n(A)$. Therefore $\{\mathcal P^n(A)\mid n\in\mathbb N\}$ is a set.

Furthermore the axiom of union tells us that if there is a set, then its union exists, namely if $X$ is a set then $\bigcup X=\{y\mid\exists u\in X: y\in u\}$ is a set.

The limit, if so, can reasonable taken to be $\bigcup\{\mathcal P^n(A)\mid n\in\mathbb N\}$. However we don't necessarily have $X\subseteq\mathcal P^n(X)$, and in particular this is true for power sets. If $A$ is transitive, namely $B\in A\rightarrow B\subseteq A$, then this works out just fine.

You may also want to consider the product, $\prod_{n\in\mathbb N}\mathcal P^n(A)$. This is a non-empty product, in ZF, and again it is a set by similar considerations as with the union. One reason to consider is that if we want to think about this as a directed system with $x\mapsto\{x\}$ as the map from $\mathcal P^n(A)\to\mathcal P^{n+1}(A)$ then the limit must embed $x$ into $\{\ldots x\ldots\}$, that is an infinite number of braces, which is impossible in ZF. But replacing $\mathcal P^n(A)$ by the finite product of $\mathcal P^k(A)$ for $k\leq n$ (cardinality wise this is the same) this is easily corrected.

Where the limit corresponds to the direct limit, this is the inverse limit. And do note that they are very different, for one in their cardinality. The product is larger.

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I'm feeling dirty after talking about direct and inverse limits... :-) But then again, with my recent invasion into the lands of iterated forcing, it seems reasonable to do so! :-) –  Asaf Karagila Dec 30 '12 at 0:35
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A quick summary: "Limit" in the question isn't defined, but any reasonable substitute for it leads to a perfectly good set. –  Andreas Blass Dec 30 '12 at 1:06

Yes, that is a fine set in either NBG or ZF. Though it would be a bit more unambiguous to write $\bigcup_{n\in\omega}P_n(A)$ instead of the limit.

If $A=\varnothing$ then the result is the set of hereditarily finite sets.

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"a bit more unambiguous" ==> A bit less ambiguous? :-) –  amWhy Dec 30 '12 at 0:34
    
@amWhy: Your phrasing is a bit more unambiguous than Henning's! :-P –  Asaf Karagila Dec 30 '12 at 0:35
    
@Asaf: You always see a way to find the humor in things! –  amWhy Dec 30 '12 at 0:36

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