Question

Let V be an alphabet, and let A and B be subsets of \({\bf{V*}}\) Show that \({\bf{|AB}}\left| {{\rm{ }} \le {\rm{ }}} \right|{\bf{A||B|}}\).

Short answer

V be an alphabet, and A and B be subsets of \({\bf{V*}}\) this Show that \({\bf{|AB}}\left| {{\rm{ }} \le {\rm{ }}} \right|{\bf{A||B|}}\).

Step-by-step solution

Step 1:Information given in the question

Here V be an alphabet, and let A and B be subsets of \({\bf{V*}}\) and AB represents the concatenation of A and B.

\({\bf{AB = \{ xy|}}\,{\bf{x}} \in {\bf{Aandy}} \in {\bf{B\} }}\)

Show the result.

Since\({\bf{AB = \{ (xy)|x}} \in {\bf{A}}{\rm{ and }}{\bf{y}} \in {\bf{B\} }}\). AB contains at least as many elements as \({\bf{A \times B = \{ (a,b)|x}} \in {\bf{Aandy}} \in {\bf{B\} }}\).

\(\left| {{\bf{AB}}} \right| \le \left| {{\bf{A \times B}}} \right|\)

Since A contains \(\left| {\bf{A}} \right|\) elements and B contains \(\left| {\bf{B}} \right|\)elements,\({\bf{A \times B}}\) contains \(\left| {\bf{A}} \right|\left| {\bf{B}} \right|\) elements. Then \(\left| {{\bf{AB}}} \right| \le \left| {{\bf{A \times B}}} \right|{\bf{ = }}\left| {\bf{A}} \right|\left| {\bf{B}} \right|\).

Therefore, therequired result is|AB| ≤ |A||B|.