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a) $[0,1] \cap \mathbb Q$

b) $P(\mathbb Q)$

c) $\mathbb R \setminus \mathbb Q $

d) $\{(a, b) ∈ \mathbb R\times\mathbb R | a, b \in\mathbb N\}$

I answered a) and d)

a) any intersection between two sets where one if finite must be countable

b) by definition, any power set of $\mathbb Z, \mathbb Q, \mathbb N$ is not countable

c) $\mathbb R\setminus \mathbb Q$ does not remove the irrational numbers from $\mathbb R$ hence it remains uncountable.

d) Wasn't too sure about this one to be honest, the $a,b\in\mathbb N$ implies that $a$ and $b$ must be aligned with the cardinality of N, which is a countable set? But the domain is the plane of real numbers..

Am I wrong?

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Eddard
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  • note that D is really just the set $\Bbb N \times \Bbb N$ which has the cardinality of $\Bbb N$ – pancini Aug 14 '15 at 07:09
  • In a) none of the two sets are finite. – Seven Aug 14 '15 at 07:09
  • @Seven woops, but if there exists an intersection between the two then must there not be a finite realm? – Eddard Aug 14 '15 at 07:13
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    No, the intersection of two infinite sets can be infinite. It is still countable, but because of other reasons. And any power set of $\mathbb N,\mathbb Z,\mathbb Q$ is not countable, but that is not by definition. – Hirshy Aug 14 '15 at 07:18

1 Answers1

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a) any intersection between two sets where one if finite must be countable

Irrelevant. While the statement is correct, it does not apply to the set $[0,1]\cap\mathbb Q$ since none of the two sets is finite.

b) by definition, any power set of Z,Q,N is not countable

Wrong. That is not true "by definition". There is a theorem that shows that.

c) $\mathbb R\setminus \mathbb Q$ does not remove the irrational numbers from $\mathbb R$ hence it remains uncountable.

Circular. Your statement "proves" that the set of irrational numbers ($\mathbb R\setminus\mathbb Q$) is irrational because the set of irrational numbers is irrational. Did you prove that irrational numbers are uncountable in your class before?

d) Wasn't too sure about this one to be honest, the a,bEN implies that a and b must be aligned with the cardinality of N, which is a countable set? But the domain is the plane of real numbers..

The fact that the domain is the real numbers means nothing. For example, $\{a\in\mathbb R: a\in\mathbb N\}$ is a countable set.

5xum
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  • Alright then, so I'm assuming that my answers (a & d) are wrong? It appears that only d) is a countable set? – Eddard Aug 14 '15 at 07:51
  • @Rozan I never said any of your claims are wrong. I only said they are not well justified. For example, $\mathbb N\cap \mathbb N$ is a countable set. However, it is not countable because the intersection of a finite set with any set is countable. That is not a correct justification, even though the statement I am trying to justify is correct. – 5xum Aug 14 '15 at 08:01
  • I see, my justifications stem from a hazy amalgamation of ideas taken from various youtube videos, thank you for correcting me! – Eddard Aug 14 '15 at 21:14