Is 0* decidable?

Problem Detail: 

I found a statement (without explanation) that a language $A = 0^*$ is decidable. How is that possible? I mean, how would we build a Turing machine that would accept (or reject) a possibly infinite string of 0's? I also thought that maybe we could create an enumerator that would create all words from $0^*$ with increasing length, but I am not sure if we can.

So is $0^*$ a decidable language? And if so, why?

Asked By : 3yakuya

Answered By : Jake

$0^*$ is the set of finite strings consisting only of $0$. There are no possibly infinite strings in $0^*$. It is trivially regular because the regex $0^*$ accepts exactly $A$ by definition. All regular problems are computable so we can definitely create a Turing machine for it (look up NFA's and DFA's for more info on the connection of Turing machines to regular languages).

This is just a confusion in what is meant by Kleene closure. If you look here you can see that it is the union of all strings of length 1, 2, 3, ... and so on for all natural numbers. Infinity is not a natural number so there are no strings of infinite length in $A$.

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Question Source : http://cs.stackexchange.com/questions/35155