pairwise disjoint subset question

A) Show that $\displaystyle N$ contains infinitely many pairwise disjoint infinite subsets. [Hint: It was shown that $\displaystyle N$~$\displaystyle N$x$\displaystyle N$

what is a pairwise disjoint infinite subset, can someone provide the definition of it?

B) Prove that any infinite set can be written as the countably infinite union of pairwise disjoint infinite subsets.

Re: pairwise disjoint subset question

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**wopashui** A) Show that $\displaystyle N$ contains infinitely many pairwise disjoint infinite subsets. [Hint: It was shown that $\displaystyle N$~$\displaystyle N$x$\displaystyle N$

what is a pairwise disjoint infinite subset, can someone provide the definition of it?

B) Prove that any infinite set can be written as the countably infinite union of pairwise disjoint infinite subsets.

The basic idea is as was stated to you. If $\displaystyle X$ is infinite then $\displaystyle X$ is equipotent to $\displaystyle X\times X$, say that $\displaystyle f:X\times X\to X$ is the bijection. We then have that $\displaystyle \left\{\{x\}\times X\right\}_{x\in X}$ is an infinite collection of pairwise disjoint sets in $\displaystyle X\times X$, thus $\displaystyle \left\{f(X\times\{x\})\right\}_{x\in X}$ is an infinite collection of pairwise disjoint subsets of $\displaystyle X$.

Re: pairwise disjoint subset question

to answer the original poster, a pair-wise disjoint collection of sets is a family $\displaystyle \{A_j\}_{j \in J}$ of sets such that:

$\displaystyle A_i \cap A_j \neq \emptyset \implies A_i = A_j\ \forall i,j \in J$

Re: pairwise disjoint subset question

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**Deveno** to answer the original poster, a pair-wise disjoint collection of sets is a family $\displaystyle \{A_j\}_{j \in J}$ of sets such that:

$\displaystyle A_i \cap A_j \neq \emptyset \implies A_i = A_j\ \forall i,j \in J$

so is this just mean a union of infinite many mutually exclusive sets?

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**Drexel28** The basic idea is as was stated to you. If $\displaystyle X$ is infinite then $\displaystyle X$ is equipotent to $\displaystyle X\times X$, say that $\displaystyle f:X\times X\to X$ is the bijection. We then have that $\displaystyle \left\{\{x\}\times X\right\}_{x\in X}$ is an infinite collection of pairwise disjoint sets in $\displaystyle X\times X$, thus $\displaystyle \left\{f(X\times\{x\})\right\}_{x\in X}$ is an infinite collection of pairwise disjoint subsets of $\displaystyle X$.

this approach seems to work for both question, I do not understand the difference of part A and B? Need some explainations please!

Re: pairwise disjoint subset question

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**wopashui** this approach seems to work for both question, I do not understand the difference of part A and B? Need some explainations please!

In my example there is no need for the union to be countable. Give us an idea how to the second part, a little effort would be nice.

Re: pairwise disjoint subset question

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**Drexel28** In my example there is no need for the union to be countable. Give us an idea how to the second part, a little effort would be nice.

for B) if S is an infinite set, we have 2 cases, S can be uncountable or countably infinite, if S uncountable, this is not possible, so we need to show that S is countable

since S is infinite, then we have$\displaystyle f:N-->S$ , f is injective, so if we show that f is also a surjection, we have f is a bijectiion, i.e. $\displaystyle N~S$ . i.e we need to show that for all$\displaystyle s\in S$ there exisis $\displaystyle n_s\in N$ $\displaystyle s.tf(n_s)=s$

If we have f is a bijection, then by part A, we have part B

so how can I do this?

Re: pairwise disjoint subset question

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**wopashui** for B) if S is an infinite set, we have 2 cases, S can be uncountable or countably infinite, if S uncountable, this is not possible, so we need to show that S is countable?

It is not necessary to consider cases to prove part B.

If $\displaystyle S$ is infinite then there is a countable subset $\displaystyle T\subset S$.

Moreover, there is surjection $\displaystyle f:S\to T$.

Then $\displaystyle \left\{ {f^{ - 1} \left( {\{ x\} } \right):x \in T} \right\}$ is pairwise disjoint countable collection and $\displaystyle S = \bigcup\limits_{x \in T} {f^{ - 1} \left( {\{ x\} } \right)} $.

Re: pairwise disjoint subset question

Hi, Plato, I don't understand what$\displaystyle f^{-1}$ comes from, and can a uncountable set has a countable subset, so if S is uncountable, can T be countable? Do I need part A to show part B?