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.

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Originally Posted by 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.

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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 $\displaystyle X$.

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:

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

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Originally Posted by 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:

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

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

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Originally Posted by 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 $\displaystyle X$.

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

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Originally Posted by wopashui

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

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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.

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Originally Posted by 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.

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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?

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Originally Posted by 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?

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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)} $.

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?