So 2∈S, then a=2 at this time.
Then1/(1-2) =-1∈ s
Then1/[1-(-1)] =1/2 ∈ s
Then1/(1-1/2) = 2 ∈ s.
So as to enter the cycle.
Because there are no identical elements in the collection.
So there are only three elements in s.
So there must be two other elements.
They are-1 and 1/2 respectively.
If there is only one
Then from a∈S, it must be11-a ∈ s.
There are no identical elements in the collection.
So only a= 1/( 1-a)
Ah-ah? = 1
Answer? -a+ 1=0
Equation has no solution
So there can't be just one.