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Which answers are correct?

  • $| \{1000, 1001 \}| = 2$: True.
  • $| \{1, 2, 2, 3 \}| = 4$: False. There are only $3$ distinct elements.
  • The cardinality of $\mathbb{N} \times \mathbb{Z}$ is $\aleph_0$: I'd say true since each element is -- distinct and countable -- though the counting may go on forever.
David G. Stork
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Travitrinket
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  • Your justification for 5.c is sort of unclear (the other two look fine). It's possible to have uncountable sets where every element is countable. – platty Sep 24 '18 at 18:57

1 Answers1

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As pointed out by platty, apart from 5.c all are fine.

5.c: The anwer is correct, but reasoning is vague.

Let $f: \mathbb{N} \times \mathbb{N} \Rightarrow \mathbb{N}$ be defined by: $f((i,j)) = \frac{(i+j+1)(i+j-2)}{2} + j$. Then you can show that $f$ is a bijection (this map is called Cantor's diagonalization map.) And since $\mathbb{N}$ and $\mathbb{Z}$ have the same cardinality, we are done.

Anon
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