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Ans should be both false right?


My logic

For the first one say $S=\phi$ then $\phi^*=\epsilon$ which is finite only

and for the second one $S=a$ then $(a)^*$ countably infinite right?

in Theory of Computation by (699 points) | 43 views
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Well, I guess..

In set theory, if asked to pick an element - you could/would definitely pick $\phi$.

I don't know how/if that translates to formal languages also. Especially if you consider $\Sigma$ to be an alphabet - can you really pick $\phi$ ? If a language $L$ over $\Sigma$ was given - you can surely get $\phi$.

I guess the question boils down to - 'If given an alphabet (not language) $\Sigma = \{a, b\}$, is $\phi$ a member of the alphabet?' I don't know the answer to this.

0
ok....
+1

@Debapaul 

As S is singleton set, consider S= {∊} 

S* ={∊} which is a finite set.

Hence S1 is false as it says S* will always be countably infinite set.

However S2 is false as it says S* will always be finite. Consider S= {a} then S* ={ ∊, a, aa, aaa,...}

Dont consider S to be null set as it is mentioned that S is a singleton set

+1
$\phi$ is not a string by the way. It's a symbol used to denote empty set.
String of length $0$ is by definition $\epsilon$
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