Regular expression: Difference between $emptyset$-concate and $lambda$-concate?

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Given the definition below, is that the concatenation $emptyset A$ the same as
$lambda A,$ given $A$ a regular expression? If not, what's the difference? My guess is that if I take concatenation $AB$ as Cartesian product $Atimes B,$ then suppose $A=0,1, emptyset A=emptyset$ but $lambda A=A$. Is this correct?



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automata regular-language regular-expressions




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  • The concatenation of sets is a set of concatenations of elements $AB=ab: ain A wedge bin B$, not pairs. For example for sets $A=a, aa$, $B=b, ab$ we have $AB=ab, aab, aaab$, while $Atimes B = (a,b),(a,ab),(aa,b),(aa,ab)$ (note, that $|AB|=3$, but $|Atimes B|=4$)
    – Jaroslaw Matlak
    Jul 16 at 16:33











  • @JaroslawMatlak: You're right I'm wrong, but is that $emptyset A=emptyset$? Since there are nothing to choose to concatenate?
    – Nong
    Jul 16 at 16:35







  • 1




    According to this definition - yes.
    – Jaroslaw Matlak
    Jul 16 at 16:35














up vote
0
down vote

favorite












Given the definition below, is that the concatenation $emptyset A$ the same as
$lambda A,$ given $A$ a regular expression? If not, what's the difference? My guess is that if I take concatenation $AB$ as Cartesian product $Atimes B,$ then suppose $A=0,1, emptyset A=emptyset$ but $lambda A=A$. Is this correct?



enter image description here




automata regular-language regular-expressions




share|cite|improve this question





















  • The concatenation of sets is a set of concatenations of elements $AB=ab: ain A wedge bin B$, not pairs. For example for sets $A=a, aa$, $B=b, ab$ we have $AB=ab, aab, aaab$, while $Atimes B = (a,b),(a,ab),(aa,b),(aa,ab)$ (note, that $|AB|=3$, but $|Atimes B|=4$)
    – Jaroslaw Matlak
    Jul 16 at 16:33











  • @JaroslawMatlak: You're right I'm wrong, but is that $emptyset A=emptyset$? Since there are nothing to choose to concatenate?
    – Nong
    Jul 16 at 16:35







  • 1




    According to this definition - yes.
    – Jaroslaw Matlak
    Jul 16 at 16:35












up vote
0
down vote

favorite









up vote
0
down vote

favorite











Given the definition below, is that the concatenation $emptyset A$ the same as
$lambda A,$ given $A$ a regular expression? If not, what's the difference? My guess is that if I take concatenation $AB$ as Cartesian product $Atimes B,$ then suppose $A=0,1, emptyset A=emptyset$ but $lambda A=A$. Is this correct?



enter image description here




automata regular-language regular-expressions




share|cite|improve this question













Given the definition below, is that the concatenation $emptyset A$ the same as
$lambda A,$ given $A$ a regular expression? If not, what's the difference? My guess is that if I take concatenation $AB$ as Cartesian product $Atimes B,$ then suppose $A=0,1, emptyset A=emptyset$ but $lambda A=A$. Is this correct?



enter image description here





automata regular-language regular-expressions





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 16 at 16:19
























asked Jul 16 at 16:10









Nong

1,1471520




1,1471520











  • The concatenation of sets is a set of concatenations of elements $AB=ab: ain A wedge bin B$, not pairs. For example for sets $A=a, aa$, $B=b, ab$ we have $AB=ab, aab, aaab$, while $Atimes B = (a,b),(a,ab),(aa,b),(aa,ab)$ (note, that $|AB|=3$, but $|Atimes B|=4$)
    – Jaroslaw Matlak
    Jul 16 at 16:33











  • @JaroslawMatlak: You're right I'm wrong, but is that $emptyset A=emptyset$? Since there are nothing to choose to concatenate?
    – Nong
    Jul 16 at 16:35







  • 1




    According to this definition - yes.
    – Jaroslaw Matlak
    Jul 16 at 16:35
















  • The concatenation of sets is a set of concatenations of elements $AB=ab: ain A wedge bin B$, not pairs. For example for sets $A=a, aa$, $B=b, ab$ we have $AB=ab, aab, aaab$, while $Atimes B = (a,b),(a,ab),(aa,b),(aa,ab)$ (note, that $|AB|=3$, but $|Atimes B|=4$)
    – Jaroslaw Matlak
    Jul 16 at 16:33











  • @JaroslawMatlak: You're right I'm wrong, but is that $emptyset A=emptyset$? Since there are nothing to choose to concatenate?
    – Nong
    Jul 16 at 16:35







  • 1




    According to this definition - yes.
    – Jaroslaw Matlak
    Jul 16 at 16:35















The concatenation of sets is a set of concatenations of elements $AB=ab: ain A wedge bin B$, not pairs. For example for sets $A=a, aa$, $B=b, ab$ we have $AB=ab, aab, aaab$, while $Atimes B = (a,b),(a,ab),(aa,b),(aa,ab)$ (note, that $|AB|=3$, but $|Atimes B|=4$)
– Jaroslaw Matlak
Jul 16 at 16:33





The concatenation of sets is a set of concatenations of elements $AB=ab: ain A wedge bin B$, not pairs. For example for sets $A=a, aa$, $B=b, ab$ we have $AB=ab, aab, aaab$, while $Atimes B = (a,b),(a,ab),(aa,b),(aa,ab)$ (note, that $|AB|=3$, but $|Atimes B|=4$)
– Jaroslaw Matlak
Jul 16 at 16:33













@JaroslawMatlak: You're right I'm wrong, but is that $emptyset A=emptyset$? Since there are nothing to choose to concatenate?
– Nong
Jul 16 at 16:35





@JaroslawMatlak: You're right I'm wrong, but is that $emptyset A=emptyset$? Since there are nothing to choose to concatenate?
– Nong
Jul 16 at 16:35





1




1




According to this definition - yes.
– Jaroslaw Matlak
Jul 16 at 16:35




According to this definition - yes.
– Jaroslaw Matlak
Jul 16 at 16:35















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