Remark
62
[27F]In the exercise [029] the elements are identified using variables \(a_ 1,\ldots a_ k\) that we may have denoted using other letters such as \(a,b,c,d,\ldots \). If we instead think of \(a_ 1,\ldots a_ k\ldots \) as values of a function \(a_ i=a(i)\), \(a:I\to X\) then the set \(\{ a_ 1,\ldots a_ k\ldots \} \) always exists (for any choice of \(I\)) since it is the image of the function \(\{ a_ 1,\ldots a_ k\ldots \} =\{ x\in X:\exists i\in I, x=a_ i\} \).