Then $U$ is a basic open set in the product space $\Sigma_n^ $, $x\in U$, and $U\cap\Sigma_A^ =\varnothing$, so $\Sigma_n^ \setminus\Sigma_A^ $ is open, and $\Sigma_A^ $ is closed. The Tikhonov theorem is the easiest way to prove that $\Sigma_n^ $ is compact, but you could also prove that it’s compact by embedding it in the Cantor set: the middle-thirds Cantor set is compact as a closed, bounded subset of $\Bbb R$, it’s not too hard to show that it’s homeomorphic to $\Sigma_2^ $ and that $\Sigma_2^ $ is homeomorphic to $\Sigma_\, $$ A subshift (Lambda,sigma) is said to be normal if it satisfies a certain synchronizing property called lambda-synchronizing and is infinite as a set.
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