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Given a set $\Sigma$ of letters, apply the Kleene star operation to it, and we get $\Sigma^*$, the set of all finite-length sequences from $\Sigma$, called strings (allowing a letter appearing more than once in a string).

If $\Sigma$ is empty, then $\Sigma^*$ consists only one string, the empty string.

If $\Sigma$ is not empty, then $\Sigma^*$ has an infinite cardinality.

I wonder if $\Sigma^*$ can be countably infinite? When will that be true? When $\Sigma$ is finite?

When will its cardinality is uncountably infinite?

Thanks!

Tim
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  • $\Sigma^*$ is countably infinite iff $\Sigma$ is countable and nonempty (we need some form of axiom of choice to show that countable union of countable sets is countable). – Marcin Łoś May 27 '14 at 05:54
  • For finite alphabet $\Sigma$, list the empty word, then the words of length $1$, alphabetically, then the words of length $2$, alphabetically, and so on. That gives you an enumeration of all words, as a sequence, so if $\Sigma$ is finite non-empty, the set of words is countably infinite. – André Nicolas May 27 '14 at 06:11
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    @MarcinŁoś You need the axiom of choice to show that a countable union of countable sets is countable, but you don't need any choice to prove that the set of all finite sequences from the set $\mathbb N$ (or any other countable set $\Sigma$) is countable. – bof May 27 '14 at 06:17
  • This answer: http://math.stackexchange.com/a/107428/138178 – askyle May 27 '14 at 07:13
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    @Marcin: You don't need the axiom of choice to prove that the set of finite sequences of a countable set is countable. Moreover, you don't need the axiom of choice to prove that the set of finite sequences from a well-orderable set are well-orderable, and if the alphabet set is infinite, then the two have the same cardinality. – Asaf Karagila May 27 '14 at 10:58
  • Oh, ok, so we don't need it here - admittedly, I haven't really thought about this special cases when I wrote it. Luckily, there are people here who know better than that ;) – Marcin Łoś May 27 '14 at 12:40

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$\sum^{\star} = \bigcup_{k=0}^{\infty} \sum^{k}$, where $\sum^{k}$ is a set of strings of letters from $\sum$ which have a length is equals to k.

So if alphabet $\sum$ is finite or countable then $\sum^{\star}$ is countable.

On the other hand if a cardinality of $\sum^{\star} = \bigcup_{k=0}^{\infty}(\sum^{k}\setminus\sum^{k-1})$, let $\sum^{-1}=\emptyset$, is uncountable then some of $\sum^{k}\setminus\sum^{k-1}\sim \sum\times...\times\sum$ (k times) must be uncountable, hence $\sum$ must be uncountable.

DenisMath
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