Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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


Edit: I changed 3.

share|cite|improve this question
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
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.

share|cite|improve this answer

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 $i$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.