- Show that if \(n > 1\) there is bijective correspondence of
\[ A _ {1} \times \dots \times A _ {n} \quad \text { with } \quad (A _ {1} \times \dots \times A _ {n - 1}) \times A _ {n}. \]
- Given the indexed family \(\{A_{1}, A_{2}, \ldots\}\), let \(B_{i} = A_{2i-1} \times A_{2i}\) for each positive integer \(i\) . Show there is bijective correspondence of \(A_{1} \times A_{2} \times \cdots\) with \(B_{1} \times B_{2} \times \cdots\) .
Solution
Define \[ \Phi:A_1\times\cdots\times A_n \longrightarrow (A_1\times\cdots\times A_{n-1})\times A_n \] by \[ \Phi(a_1,\dots,a_n)=((a_1,\dots,a_{n-1}),a_n). \] Its inverse is \[ ((a_1,\dots,a_{n-1}),a_n)\longmapsto(a_1,\dots,a_n), \] so \(\Phi\) is bijective.
For \(B_i=A_{2i-1}\times A_{2i}\), define \[ \Psi(a_1,a_2,a_3,a_4,\dots) =((a_1,a_2),(a_3,a_4),\dots). \] This maps \(\prod_{j\ge1}A_j\) into \(\prod_{i\ge1}B_i\). Its inverse simply flattens the sequence of ordered pairs: \[ ((b_1,b_2),(b_3,b_4),\dots) \longmapsto (b_1,b_2,b_3,b_4,\dots). \] Thus the two infinite products are in bijective correspondence.