Is $\{0,1\}^{\mathbb{N}}$ compact in $\mathbb{N}^{\mathbb{N}}$?

Question

Is $\{0,1\}^{\mathbb{N}}$ compact in $\mathbb{N}^{\mathbb{N}}$ ? How can i proof that?

Answer 1

The topology of $\mathbb N ^{\mathbb N}$ and $\{0,1\}^{\mathbb N}$ are both same as product topologies. Tychonof's Threorem says product of compact spaces is compact. Hence the answer is YES.

Answer 2

$\{0, 1\}^{\mathbb{N}}$ is a product of closed sets of $\mathbb{N}$, and so is closed in $\mathbb{N}^{\mathbb{N}}$. https://proofwiki.org/wiki/Product_of_Closed_Sets_is_Closed

It is compact by Tychnoff's theorem, so clearly the closure is compact. We conclude it is relatively compact in $\mathbb{N}^{\mathbb{N}}$.