pairwise disjoint subset question

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October 10th 2011, 06:11 PM
wopashui

pairwise disjoint subset question

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

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.
October 10th 2011, 10:23 PM
Drexel28

Re: pairwise disjoint subset question

Quote:

Originally Posted by wopashui

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

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 $X$ is infinite then $X$ is equipotent to $X\times X$ , say that $f:X\times X\to X$ is the bijection. We then have that $\left\{\{x\}\times X\right\}_{x\in X}$ is an infinite collection of pairwise disjoint sets in $X\times X$ , thus $\left\{f(X\times\{x\})\right\}_{x\in X}$ is an infinite collection of pairwise disjoint subsets of $X$ .
October 10th 2011, 10:48 PM
Deveno

Re: pairwise disjoint subset question

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

$A_i \cap A_j \neq \emptyset \implies A_i = A_j\ \forall i,j \in J$
October 11th 2011, 05:28 PM
wopashui

Re: pairwise disjoint subset question

Quote:

Originally Posted by Deveno

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

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

so is this just mean a union of infinite many mutually exclusive sets?
October 11th 2011, 05:36 PM
wopashui

Re: pairwise disjoint subset question

Quote:

Originally Posted by Drexel28

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

this approach seems to work for both question, I do not understand the difference of part A and B? Need some explainations please!
October 11th 2011, 07:35 PM
Drexel28

Re: pairwise disjoint subset question

Quote:

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!

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.
October 12th 2011, 01:01 PM
wopashui

Re: pairwise disjoint subset question

Quote:

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.

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

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

so how can I do this?
October 12th 2011, 01:32 PM
Plato

Re: pairwise disjoint subset question

Quote:

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?

It is not necessary to consider cases to prove part B.
If $S$ is infinite then there is a countable subset $T\subset S$ .
Moreover, there is surjection $f:S\to T$ .
Then $\left\{ {f^{ - 1} \left( {\{ x\} } \right):x \in T} \right\}$ is pairwise disjoint countable collection and $S = \bigcup\limits_{x \in T} {f^{ - 1} \left( {\{ x\} } \right)}$ .
October 12th 2011, 06:34 PM
wopashui

Re: pairwise disjoint subset question

Hi, Plato, I don't understand what $f^{-1}$ 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?