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

Answer:

No, it is not.

Step-by-step explanation:

The set $C = \mathbb{N} \times \mathbb{N}$ 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 $[1] \times [2]$ and B doesnt because 1 is not equal to 2.