Here is the question from Master entry exam of computer science called KONKUR in IRAN:
What is cardinality of this:
$$\bigcup\limits_{i=1}^{\infty} N^i$$
As far as I concerned the right answer is aleph zero . But we are talking about union of countably infinite sets, infinite time and not finite. Please someone explain it to me the answer and the reason behind that.
Thank you.
Say WLOG $0\not\in\mathbb N$ (I know, I'm sorry).
Notice that $\mathcal F:=\displaystyle\bigcup_{i=1}^\infty\mathbb N^i$ is the set of the finite sequences of natural numbers.
This set has the same cardinality as the set $\mathcal G$ of the strictly increasing sequences of natural numbers. One can see this through the bijection $\mathcal F\to\mathcal G$ given by $$(n_1, n_2, \dots, n_k)\mapsto (n_1, n_1+n_2, \dots, n_1+n_2+\dots+n_k)$$
Now the map $\mathcal G\to\mathbb N$ given by $$(n_1, n_2, \dots, n_k)\mapsto 2^{n_1}+2^{n_2}+\dots+2^{n_k}$$ is, of course, an injection, since binary representation is unique. Therefore, $\mathcal G$ and $\mathcal F$ are countable.