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Prove: there exists 3 sets: $A, B, C \subseteq \mathbb{N}$ such that: $A\cap B= B \cap C = A \cap C = A\cap B\cap C = \emptyset$ and $|A|=|B|=|C|=\aleph_0$?

also, the sets must exists: $$|\mathbb{N} \setminus {A}| = \aleph_0$$

$$|\mathbb{N} \setminus {B}| = \aleph_0$$

$$|\mathbb{N} \setminus {C}| = \aleph_0$$

in other words, I'm looking for a way to substitute $\mathbb{N}$ into 3 disjoint sets, such that the cardinality of each of them is equal to the cardinality of $\mathbb{N}$

EDIT: the point for this question is to prove a lemma (finding $A, B, C$ as described), as given the set:

$$ M = \{ A \in P(A) \vert \ \ |A| = \aleph_{0} \ \land \ |A^c| = \aleph_{0} \}$$

Prove that: $|M| = \aleph$

by finding 3 sets, such that $|A| = |B|=|C| =|\aleph_0$, then I could determine that: $(B\cup C) \in M$ as $|B \cup C| =\aleph_0$, and: $(B\cup C)^c = \mathbb{N}\setminus(B \cup C) = A$, as $|A| = \aleph_0$.

using the lemma, I'd argue that for every set $\beta \subseteq B: \ (\beta\cup C) \subseteq (B\cup C) \Longrightarrow \ \forall \beta: (\beta \cup C) \subseteq M$.

this is true because $|C| =\aleph_{0}, \ \forall \beta: |\beta \cup C| = \aleph_{0}$ and $ A \subseteq (\beta \cup C)^{c} \Longrightarrow |(\beta \cup C)^{c}| = \aleph_0$.

finally, the collection of all $\beta$ sets is: $\{\beta | \beta \subseteq B \}$, Hence $\{\beta | \ \beta \subseteq B \} = P(B)$.

Notice that $|P(B)| = 2^{\aleph_{0}} = \aleph$

this means that $\left(P(B) \cup C \right) \subseteq M \Longrightarrow \ \aleph =|\left(P(B) \cup C \right)| \leq |M|$.

because $M \subseteq P(\mathbb{N}) \Longrightarrow |M| \leq |P(\mathbb{N})| = \aleph$, hence, by Cantor Bernstein theorem we conclude that $|M| =\aleph$, as wished.

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    $\begingroup$ This is trivial. Let $a\sim b$ for any natural $a, b$, if and only if there exists integer $k$ so that $a-b = 3k$. Then $[1]_\sim$, $[2]_\sim$, $[3]_\sim$ are the sets you search for. $\endgroup$ – Jakobian Aug 13 at 16:28
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    $\begingroup$ The answers below are great, but I encourage you to try to think of an answer on your own that is not the same. A big part of the level of mathematics you are at is learning how to be creative! There are infinitely many different valid examples of a partition that you describe. Try to come up with another. $\endgroup$ – JMoravitz Aug 13 at 16:30
  • $\begingroup$ @JMoravitz thank you. you are very right! I'd give it a try. $\endgroup$ – Jneven Aug 13 at 16:31
  • $\begingroup$ I saw you edit this with more conditions. The additional conditions you just added are unnecessary. Given that $A,B\subseteq \Bbb N$, that $A\cap B=\emptyset$ and that $|A|=|B|=|\Bbb N|$. It follows that $|\Bbb N\setminus A|=|\Bbb N|$ since $(\Bbb N\setminus A)\supseteq B$ $\endgroup$ – JMoravitz Aug 14 at 15:03
  • $\begingroup$ I agree, and yet I rather chosen to go with the clearest explanation as possible $\endgroup$ – Jneven Aug 14 at 15:10
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Since there is no demand on the union of your sets you may consider the following disjoint sets to meet the cardinality conditions

$$A=\{2^K:k=1,2,3,...\}$$

$$B=\{3^k:k=1,2,3,...\}$$

$$C=\{5^k:k=1,2,3,...\}$$

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  • $\begingroup$ I edited it now, but i think that this 3 sets are still existing the needed properties. $\endgroup$ – Jneven Aug 14 at 14:56
  • $\begingroup$ "but i think that this 3 sets are still existing the needed properties". You seem to be under the impression these properties are hard to fulfill. I think that is drawing you back. These properties are easy to fulfill. These three sets have the properties but so does every single other suggestion everyone else has made. I think this is a hard problem to do if you think it is a hard problem. But the only thing actually hard about it is to realize that it is actually very easy. $\endgroup$ – fleablood Aug 14 at 15:32
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Split $\mathbb{N}$ by considering the remainder of the division by $3$ of each integer: $A=\{3k: k\in\mathbb{N}\}$, $B=\{3k+1: k\in\mathbb{N}\}$, and $C=\{3k+2 : k\in\mathbb{N}\}$.

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Pick $A = \{0,3,6,...,3n,...\},\ B = \{1,4,7,...,3n+1,...\}$ and $C = \{2,5,8,...,3n+2,...\}$. Then, we can define a bijective function $f: A \to \mathbb{N}$ with $f(a) = \frac{a}{3}$ (I leave the verification of injectivity and surjectivity to you). Then, you can also define bijective functions $g: B \to \mathbb{N}$ and $h: C \to \mathbb{N}$ in a similar way.

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  • $\begingroup$ Please check the edit I preformed to the question in order the explain the context. $\endgroup$ – Jneven Aug 15 at 11:56
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    $\begingroup$ It is better now, thank you very much. I voted for reopening. $\endgroup$ – ArsenBerk Aug 15 at 12:02
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Perhaps you simply need to be told that this is an easy question and that there isn't actually any trick to it whatsoever?

I suppose the first counter intuitive thing all students of mathematics face is how can it be that a subset can be the same "size" as its superset. Intuition says that if, say, every even number is an whole there must be "fewer" even numbers than whole numbers.

But once we get over that hurdle it becomes very easy to partition an infinite set into two disjoint sets equally infinite. The odd integers and the even integers; ; the primes and the non-primes; numbers that contain the string "$17$" in them and those that don't. etc. etc. etc.

So why would dividing it into three be any harder? After all you could just divide them in two; say, the odd and the even integers, and then divide one of them further it two; say, the odd numbers, the even numbers divisible by $4$ and the even numbers not divisible by $4$.. Or the prime integers, the composite integers that are divisible by $34$ and the composites that are not divisible by $34$ or... pretty much anything you want.

One thing to point out is that there is no requirement that the union of these sets be the entirety of $\mathbb N$. You could do. All whole numbers whose first digit is $2$, all numbers whose first digit is $7$, and all number whose first two digits are $18$, for example.

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