why $\prod_{n=1}^{\infty}{A_n}$ with product topology ,where ${A_n=\{0,1\}}$ has discrete topology for n=1,2,3,... is compact set?
I know that $\prod_{n=1}^{\infty}{A_n}$ ,where ${A_n=\{0,1\}}$ is uncountable set and is complete as it do not have cauchy sequence but how to check for totally bounded
I am not getting any direction to solve this problem
This follows from the Tychonoff theorem, which states that the product of compact spaces, endowed with the product topology is compact. Since in this case you have a countable product of metrizable compact spaces, you don't even need the Axiom of choice to prove this.
For each $n\in\mathbb{N},$ $A_n$ is compact (Why?). Now use the Tychonoff theorem to say that the product is compact.
