You need to test if
where are in the set of functions described
Problem: The set of continuous functions such that f(0)=0, f(1)=1, and f is increasing. Binary operation is composition of functions.
First off, I'm a little confused about the part. Is that just saying the domain and range of the function is from 0 to 1?
Now, I trying to test this for associativity.
But I started to run into problems. There are only two elements in this set, but I need three to test for associativity, don't I? Or do I just use one twice? But what if one works twice, but the other doesn't?
Regardless, I tried using composition to test for associativity, and got stuck.
??
Any help is appreciated.
I still don't understand. Where are x, x^2 etc coming from?
Is the first [0,1] in the problem specifying the domain of the functions and the second is specifying the range? And therefore, is x, x^2 etc coming from the fact that putting 0 or 1 into any of those will result in 0 or 1?
Ah okay. So I'm testing for associativity like this...
Which is true since if x=0, they all return 0 and if x=1 they all return one, so it ends up as 0=0 or 1=1, right?
If x=0,
Is that incorrect?
Assuming it isn't, I'm now trying to come up with the unique identity element...
In other words,
Which is true if g(x)=x... or f(x)=x?? So the identity is... what? I think I'm on the wrong track.
f(x) = x is the identity, since when you compose this with any function g, gf = g.
Your "associativity" is wrong. It should be
f(gh) = (fg)h. You don't want to move them around, just change the way they are grouped.
suppose f,g are in the set under consideration. then fg(0) = f(g(0)) = f(0) = 0, while fg(1) = f(g(1)) = f(1) = 1.
clearly, fg is continuous if f and g are (standard proof from first-year calculus).
so the main thing is: is fg increasing if f and g are?
suppose 0 ≤ x < y ≤ 1.
then 0 = g(0) ≤ g(x) < g(y) ≤ g(1) = 1, because g is increasing.
so 0 = f(g(0)) ≤ f(g(x)) < f(g(y)) ≤ f(g(1)) = 1, because f is increasing.
@tangibleLime: for any functions f,g,h: we can show that if the compositions are defined, then
.
.
You may be confusing "[0, 1]" with "{0, 1}". The first is an interval containing an infinite collection of numbers, the second is a set containing only two numbers. Saying that f:[0, 1]->[0, 1] means that f maps all numbers in the closed interval to a number in that interval.