Re: Invariant Sets - June 15
Posted by
anon_econ on
Jun 15, 2012; 7:30am
URL: http://discussion-forum.276.s1.nabble.com/Invariant-Sets-June-15-tp7577766p7577802.html
oh ok. i had tried this but i cudnt get the 4 invariant sets for k=2. i got the null set, the whole set itself and the set {1}. which is the 4th one?
On Fri, Jun 15, 2012 at 12:38 PM, Amit Goyal [via Discussion forum]
<[hidden email]> wrote:
The idea is exactly the same for any k:
Consider f: {1, 2, .., k, ..., n} -> {1, 2, .., k, ..., n}
f(i) = i for all 1 ≤ i ≤ k-1
= i + 1 for k ≤ i ≤ n-1
= k for i = n
For the above function, there are exactly 2^k invariant sets.