Q8. Let . Let be functions from S to S (one-one and onto). For any function f, call D, subset of S, to be invariant if for all x in D, f(x) is also in D. Note that for any function the null set and the entire set are ‘invariant’ sets. Let be the number of invariant subsets for a function.
a) Prove that there exists a function with .
b) For a particular value of k prove that there exist a function with =
Consider the function defined piecewise as f(x) = x – 1 is and f(x) = n if x = 1
Of course null set and the entire sets are invariant subsets. We prove that there are no other invariant subsets.
Suppose be an invariant subset with at least one element.
Since we are working with natural numbers only, it is possible to arrange the elements in ascending order (there is a least element by well ordering principle).
Suppose after rearrangement where is the least element of the set
If then is not inside D as is the smallest element in D. Hence D is no more an invariant subset which is contrary to our initial assumption.
This must equal to 1.
As D is invariant subset must belong to D. Again f(n) = n-1 is also in D and so on. Thus all the elements from 1 to n are in D making D=S.
Hence we have proved that degree of this function is 2.
For a natural number ‘k’ to find a function with = define the function piecewise as
f(x) = x for
= n for x=k
= x-1 for the rest of elements in ‘n’
To construct an invariant subset the ‘k-1’ elements which are identically mapped, and the entirety of the ‘k to n’ elements considered as a unit must be considered. Thus there are total k-1 + 1 elements with which subsets are to be constructed. There are subsets possible.