I'm given $M\subseteq P(\mathbb{N}\times\mathbb{N})$, such that $M$ is countable set.
So: $M=\{A_0 , A_1 ,A_2 , \ldots\}$ .
How can I define $A \in P(\mathbb{N}\times\mathbb{N})$ so that, $\forall i\in\mathbb{N}$, $A\neq A_i$ and $(0,0)\notin A$?
I tried to define $A=\{v_i \mid i\in\mathbb{N}\}$ while:
- if $A_i=\emptyset$ so $v_i=(1,1)$.
- if from all the elements in $A_i$, the minimal coordinate in the $x$ element is equal to $0$, and the minimal coordinate in the $y$ element is bigger than $0$ so: $v_i=(1,\mathrm{minCoordinateY}-1)$.
- if from all the elements in $A_i$, the minimal coordinate in the $x$ element is bigger than $0$, and the minimal coordinate in the $y$ element is equal to $0$ so: $v_i=\mathrm{minCoordinateX}-1,1)$.
- if from all the elements in $A_i$, the minimal coordinate in the $x$ element is bigger than $0$, and the minimal coordinate in the $y$ element is bigger than $0$ so: $v_i=(\mathrm{minCoordinateX}-1,1)$.
But what can I do in the case that from all the elements in $A_i$, the minimal coordinate in the $x$ element is equal to $0$ and the minimal coordinate in the $y$ element is equal to $0$ too?
An alternative approach per Brian M. Scott's answer here would be to instead simply define $A$ by
$$A := \{(1,n) \mid (1,n) \not \in A_n\}$$
This ensures that $A \ne A_i$ for every $i \in \Bbb N$ by a sort of diagonalization argument. If $(1,i) \in A_i$, then $(1,i) \not \in A$, so $A \ne A_i$, and similar occurs the other way around.
Mostly just posting this to get this out of the unanswered queue. Posting as Community Wiki in particular since I have nothing further to add.