Question

#3 R Show that a continuous increasing function f: R Cannot have a 3-cycle. First show that it cannot have a 2-egple.

Answer 1

first start with that a continuous increasing function
f:R →R
cannot have a two cycle.

let assume that it have a two cycle

(i, j) that means f(i) = j and f()= i

without loss of generality assume i < j .

but as the function is increasing hence contradiction.

i<j= f(i) <f() i.e. j<i that means i<j i <i (which is very absurd)

Note:we do not need continuity on the above arguement.

next let's move to the next part,a continuous increasing function
f:R →R
cannot have a three cycle.

let assume it have a three cycle with 3 points a,b,c where a<b<c .(without loss of generality )

then there can be only 2 possible configuration of the cycle.

1)

(a, b,c) which means f(a) = b; f(b) = c; f(c) = a.

2)

(a, c,b) which means f(a) = c; f(0) = a; f(c) = b.

in the case 1) as the function is increasing,

a<b<c= f(a) f(b)<f(c), that is b<c<a so, a <e c <a (contradiction)

in the 2) case,as the function is increasing,

a<b<c= f(a) f(b)<f(c), that is c<a<b so, a <c>c<a b< c>c<b(contradiction)

Note: here the contradiction is to the law of trichotomy of real numbers which says given two real nubers a,b only one and exactly one happen out of these 3 relations

1) a<b 2)a=b 3)a>b

a <bce (a, b, c) (a, e, b) t. +3 b c (a, b, e) (a, e, b)