I thought that ($\mathcal{P}$ is the powerset) $$\mathcal{P}(\mathbb{N}) = \lim_{n \to \infty} \mathcal{P}(\{0, 1, 2, ..., n\}) = \bigcup_{n = 1}^{\infty} \mathcal{P}(\{0, 1, 2, ..., n\})$$
But recently found out this is wrong.
So I wonder, does $$\lim_{n \to \infty} \{0, 1, 2, ..., n\} =^{?} \bigcup_{n = 1}^{\infty} \{0, 1, 2, ..., n\} =^{?} \mathbb{N}$$
And if yes, how come the first equation does not hold?
Yes it is.
If $A_n:=\{0,1,\dots,n\}$ then evidently: $$\limsup A_n=\bigcap_{n=1}^{\infty}\bigcup_{k=n}^{\infty}A_k=\{0,1,2,\dots\}=\bigcup_{n=1}^{\infty}\bigcap_{k=n}^{\infty}A_k=\liminf A_n$$ so that $\lim A_n$ is defined and takes value $\{0,1,2,\dots\}$.
edit concerning comparing with $\wp(A_n)$
If $A_1\subseteq A_2\subseteq\cdots$ and $f$ operates on sets on a monotonic way (i.e. $A\subseteq B\implies f(A)\subseteq f(B)$) then we have $\lim_{n\to\infty} A_n=\bigcup_{n=1}^{\infty} A_n$ and of course also $\lim_{n\to\infty}f(A_n)=\bigcup_{n=1}^{\infty}f(A_n)$.
But we cannot just state that $\bigcup_{n=1}^{\infty}f(A_n)=f(\bigcup_{n=1}^{\infty} A_n)$.
So it does not have to be true that $\lim_{n\to\infty}f(A_n)=f(\bigcup_{n=1}^{\infty} A_n)$