Cardinality of sets ($\aleph_0$, distinct elements)

Question

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.

Answer 1

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.