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Let the Symbol $|\ |$ denote cardinality of an set. Is it possible to construct a family $N_i\subset\mathbb{N}$, $i\in\mathbb{N}$, such that:

1- $|N_i|=\infty$,

2- $\mathbb{N}=\bigcup_{i=1}^\infty N_i$,

3- $|N_1\cap N_i|=i$, $\forall i\geq 2$.

Thanks

Edit: I changed 3.

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Yes it is (replacing for all $i\in\mathbb N$ in 3. by for all $i\geqslant2$). Now, what did you try? –  Did Nov 12 '12 at 16:43
    
You guys are right, its is for $i\geq 2$. I will edit it. –  Tomás Nov 12 '12 at 16:45
    
Hint: split $\mathbb{N}$ into two infinite disjoint sets, one of which will be all of $N_1$, the other which will contain most of the elements of each $N_i$, $i\geq 2$... –  cody Nov 12 '12 at 16:47

2 Answers 2

up vote 5 down vote accepted

You obviously can’t have $N_1$ infinite and $|N_1\cap N_1|=1$, but if you just want $|N_1\cap N_k|=k$ for $k>1$, the answer is yes.

Let $\{M_k:k\ge 2\}$ be any partition of the odd positive integers into infinite sets, and let $N_1$ be the set of even positive integers. Now let

$$\begin{align*} N_2&=M_2\cup\{2,4\}\\ N_3&=M_3\cup\{6,8,10\}\\ N_4&=M_4\cup\{12,14,16,18\}\;, \end{align*}$$

and so on.

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No. Condition 3 implies that $N_1$ has exactly one element ($|N_1\cap N_1|=1$); so $|N_1\cap N_i|\leq1$ for all $i$.

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