Let \(X\) be the two-element set \(\{0,1\}\) . Find a bijective correspondence between \(X^{\omega}\) and a proper subset of itself.
Solution
Let \[ P=\{(y_1,y_2,\dots)\in X^\omega:y_1=0\}. \] This is a proper subset of \(X^\omega\), since for example \((1,0,0,\dots)\notin P\).
Define \[ F:X^\omega\longrightarrow P, \qquad F(x_1,x_2,x_3,\dots)=(0,x_1,x_2,x_3,\dots). \] The inverse map is the left shift \[ G(0,y_2,y_3,\dots)=(y_2,y_3,\dots). \] Then \(GF=\mathrm{id}_{X^\omega}\) and \(FG=\mathrm{id}_P\). Thus \(F\) is a bijection of \(X^\omega\) with the proper subset \(P\).