Question

For each $n \in \mathbb{N}$, let $A_n = [n] \times [n]$. Define $B = \bigcup_{n \in \mathbb{N}} A_n$. Does $B = \mathbb{N} \times \mathbb{N}$? Either prove that it does, or show why it does not.

Answer 1

No, it is not.

Explanation:

The set
`C = \mathbb{N} * \mathbb{N}` contains every ordered pair of Natural numbers, while B only contains those pairs in which both values in each entry are the same. Therefore, C is a bigger set than B, but B is not equal to C because for example C contains
`[1] * [2]` and B doesnt because 1 is not equal to 2.