"For every standard finite set F there is an element g in S such that for all f in F" is false (no such g exists when ), we may use Idealisation to tell us that "There is a G in S such that for all standard f" is also false, i.e. all the elements of S are standard. If S is infinite, then we take for the relation R(g, f): g and f are not equal and g is in S. Since "For every standard finite set F there is an element g in S such that for all f in F" (the infinite set S is not a subset of the finite set F), we may use Idealisation to derive "There is a G in S such that for all standard f." In other words, every infinite set contains a nonstandard element (many, in fact). The power set of a standard finite set is standard (by Transfer) and finite, so all the subsets of a standard finite set are standard. If S is nonstandard, we take for the relation R(g, f): g and f are not equal and g is in S. Since "For every standard finite set F there is an element g in S such that for all f in F" (the nonstandard set S is not a subset of the standard and finite set F), we may use Idealisation to derive "There is a G in S such that for all standard f." In other words, every nonstandard set contains a nonstandard element. As a consequence of all these results, all the elements of a set S are standard if and only if S is standard and finite. Applied to the relation < Since "For every standard, finite set of natural numbers F there is a natural number g such that for all f in F" – say, – we may use Idealisation to derive "There is a natural number G such that for all standard natural numbers f." In other words, there exists a natural number greater than each standard natural number. Applied to the relation ∈ More precisely we take for R(g, f): g is a finite set containing element f. Since "For every standard, finite set F, there is a finite set g such that for all f in F" – say by choosing itself – we may use Idealisation to derive