Invariant Sets - June 15

[Amit Goyal]

June_15_functions.png

[anon_econ]

I think for the first one if f(x)=x+1 for all x except n, and f(n)=1 will give deg(f)=2..
still thinking about the second one!

[anon_econ]

In the second one do v hv 2 prove it for any arbitrary k or for any one particular k?

[anon_econ]

for k=n f(x)=x is the function. can't figure it out for other values of k.

[Amit Goyal]

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.

[anon_econ]

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.

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NAML

[Amit Goyal]

{2, 3, 4}

[anon_econ]

umm isnt f(4)=5? except when n=4

[Amit Goyal]

{2, 3...., n} that's what i meant.

[anon_econ]

ohh! ;)

Thanks!!

On Fri, Jun 15, 2012 at 1:05 PM, Amit Goyal [via Discussion forum] <[hidden email]> wrote:

{2, 3...., n}

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[tim]

cn u pls explain this questn once again..if u dnt mind..pls :)

[anon_econ]

hey tim. r u hving trouble understanding the solution or the question itself?

[tim]

the solution .... n why did u assume this fn..the one uve taken..cn u assume fn..?

[anon_econ]

we had to prove that out of all one-one and onto functions, there existed some function that satisfied the condition. now u can see that for this function the condition will be satisfied.

[tim]

so u cud choose any fn that satisfied one to one n onto ?

[anon_econ]

in part a, v had 2 find any one-one and onto function that had no invariant subset except for the null set and S itself. so v try 2 define the images of 1,2,..n in such a way that whenever v pick some (not all) elements from them, the image of at least one of them isnt included.

[tim]

ok thnk u :)