I would like to ask a question that is a little vague; please bear with me.
Consider the following situation: For each $n\in\mathbb{N}$ we define a countable set $A_n=\{a_{0,n},a_{1,n}\dots\}$. For each $A_n$ and for each element $a_{m,n}$ of $A_n$ we define a countable set $A_{m,n} = \{a_{0,m,n},a_{1,m,n}, a_{2,m,n},\dots\}$
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We continue this process and finally we take the union of all the sets that we have defined. Is this an uncountable set?
EDIT I'll write down some of the steps:
For each $n\in\mathbb{N}$ we define $A_n$ to be a countable set, namely $A_n=\{a_0,a_1,\dots\}$.
For each $n\in\mathbb{N}$ and for each $a_m\in A_n$ we define the set $A_{m,n}=\{a_{0,m,n} , a_{1,m,n}, a_{2,m,n},\dots\}$
For each $(m,n)\in\mathbb{N}^2$ and for each $a_{k,m,n}\in A_{m,n}$ we define the set $A_{k,m,n}=\{a_{0,k,m,n}, a_{1,k,m,n}, a_{2,k,m,n},\dots\}$
...
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At time = $\infty$ we take the Union of all these countable sets. This is the set I'm talking about. I believe that the family of sets can be enumarated in the same way that $\bigcup_{n\in\mathbb{N}}\mathbb{N}^n$ can; so what we actually have is a countable union of countable sets, but it is too counter intuitive for me; anyone can confirm/ disprove?
I have edited my answer based on reinterpretation of the question; I believe the following is correct to the situation now.
At each step $m$, the sets $A_{n_1,n_2,\ldots,n_m}$ are countable, so $$\bigcup_{(n_1,\ldots,n_m)\in\Bbb N^m}A_{n_1,\ldots,n_m}$$is a countable union of countable sets, and is therefore countable. We are interested in the union $$\bigcup_{m=1}^\infty\left(\bigcup_{(n_1,\ldots,n_m)\in\Bbb N^m}A_{n_1,\ldots,n_m}\right)$$which is a countable union of countable sets, and is therefore countable.