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If $L_1 = \emptyset$ , $L_2= \{a\}$ then what is $$L_1\cdot L_2^* \cup L_1^*$$

The answer given is $\{\epsilon\}$ but I think it should be $\{\epsilon,a\}$. My Approach :
$L_1^* = \{\epsilon\}$
$L_2^* = \{\epsilon,a\}$
$L_1\cdot L_2^* = \{\epsilon,a\}$
$L_1\cdot L_2^* \cup L_1^* = \{\epsilon,a\}$.

Where's my mistake?

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    $\begingroup$ The mistake is in $L_2^*$, which is infite: $L_2^* =\{a^n\mid n\in \mathbb N\}$. Also, another mistake is in the final part. For any language $L$, $\emptyset\cdot L=\emptyset$. $\endgroup$
    – R B
    Dec 21, 2014 at 15:01
  • $\begingroup$ Got my mistake ... Please post the same comment as Answer.. would like to mark it as answer $\endgroup$
    – Jay Teli
    Dec 21, 2014 at 15:08
  • $\begingroup$ $L_1\cdot L_2^* = \emptyset \cdot L_2^* = \emptyset$ And thus $L_1\cdot L_2^*\cdot L_1^* = \emptyset \cdot L_1^*= \emptyset$. $\endgroup$
    – R B
    Dec 21, 2014 at 15:08

2 Answers 2

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For any language $L$, $$\emptyset⋅L=\emptyset$$.

Therefore, $$L_1\cdot L_2^* = \emptyset \cdot L_2^* = \emptyset$$

And thus $$L_1\cdot L_2^*\cup L_1^* = \emptyset \cup L_1^*= L_1^* = \{\epsilon\}$$.

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  • $\begingroup$ I meant question is L1⋅L∗2 U L∗1 and not L1⋅L∗2⋅L∗1.. ans will be set of empty set.. and not empty set... $\endgroup$
    – Jay Teli
    Dec 21, 2014 at 15:14
  • $\begingroup$ kindly edit ur solution accodingly.. so that i can mark it as solution. $\endgroup$
    – Jay Teli
    Dec 21, 2014 at 15:14
  • $\begingroup$ Could you please write it in Latex? I'm having hard times reading your comments (not sure what you mean). $\endgroup$
    – R B
    Dec 21, 2014 at 15:15
  • $\begingroup$ something else.. $\endgroup$
    – Jay Teli
    Dec 21, 2014 at 15:16
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Your problem is in applying the definitions of the concatenation and Kleene-star operators. The exercise was probably deliberately chosen to test your understanding of the difference between $\emptyset$ (the language containing no strings at all) and $\{\epsilon\}$ (the langauge containing only the empty string), and how the operators apply to them, but you also tripped up on $\{a\}^*\!$.

  • For any language $L$, $L^*$ is the language formed by taking zero or more strings from $L$ and concatenating them. In particular, if $L=\emptyset$, the only way that you can choose zero or more strings from $L$ is to choose none. If you start with no characters and concatenate nothing, you still have no characters: the empty string. So $L_1^* = \emptyset^* = \{\epsilon\}$. On the other hand, if you start with $L_2=\{a\}$ and choose zero or more strings from that language, you have chosen zero or more $a$s. Therefore, $L_2^* = \{a^n\mid n\geq 0\} = \{\epsilon, a, aa, aaa, \dots\}$.

  • For any languages $L$ and $L'\!$, $L\cdot L'$ is the set of all strings that can be made by choosing one string from $L$ and one from $L'$ and concatenating them. You can never choose a string from the empty language, so you can never make any strings by concatenating something in $\emptyset$ and something in $L2^*$, so $L_1\cdot L_2^* = \emptyset\cdot L_2^* = \emptyset$.

So the answer is $\emptyset \cup \{\epsilon\} = \epsilon$.

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  • $\begingroup$ I am tempted to mark ur ans as the best solution but since the previous ans was also precise and quick response to my query i am keeping it as final ans.. $\endgroup$
    – Jay Teli
    Dec 21, 2014 at 17:03
  • $\begingroup$ Thanks for such a beautiful explanation and soorry that i am not marking it as final ans .. thanks a lot.. :) $\endgroup$
    – Jay Teli
    Dec 21, 2014 at 17:05

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