Given that s=lub(A) and s is not in A, then s-1 is not an upper bound for A

Then there is an element a_1 is A and s-1<a_1<s.

Let e_2=max{s-(1/2),a_1}; so e_2<s.

Then e_2 is not an upper bound of A, so there is an a_2 in A and e_2<a_2<s.

If n>2, let e_n=max{s-(1/n),a_n-1}; so e_n<s.

Then e_n is not an upper bound of A, so there is an a_n in A and e_n<a_n<s.

Note that a_1<a_2<…<a_n and s-(1/n)<a_n<s.