# Thread: Determine if this set of continuous functions is a group

1. ## Determine if this set of continuous functions is a group

Problem: The set of continuous functions $\displaystyle f:[0,1] \rightarrow [0,1]$ 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 $\displaystyle f:[0,1] \rightarrow [0,1]$ 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.

$\displaystyle a*(b*c)=(a*b)*c$

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.

$\displaystyle a*b = a(b(x))$ ??

Any help is appreciated.

2. ## Re: Determine if this set of continuous functions is a group

You need to test if $\displaystyle (f \circ (g\circ h))(x) = ((f\circ g) \circ h )(x)$

where $\displaystyle f,g,h$ are in the set of functions described

3. ## Re: Determine if this set of continuous functions is a group

There are more than just two elements...

For instance, the functions
x
x^2
x^3
etc
Are all in your supposed group.

4. ## Re: Determine if this set of continuous functions is a group

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?

5. ## Re: Determine if this set of continuous functions is a group

Originally Posted by tangibleLime
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?
The group is a SET (see the first line of your first post). Each element in the set is a function from [0, 1] to [0, 1]

The first is the domain, the second is the range.

6. ## Re: Determine if this set of continuous functions is a group

Ah okay. So I'm testing for associativity like this...

$\displaystyle f(g(h(x))) = h(f(g(x)))$

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,
$\displaystyle f(g(h(0))) = f(g(0)) = f(0) = 0$
$\displaystyle h(f(g(0))) = h(f(0)) = h(0) = 0$

Is that incorrect?

Assuming it isn't, I'm now trying to come up with the unique identity element...

$\displaystyle a*e=e*a=a$

In other words,
$\displaystyle f(g(x)) = g(f(x)) = f(x)$

Which is true if g(x)=x... or f(x)=x?? So the identity is... what? I think I'm on the wrong track.

7. ## Re: Determine if this set of continuous functions is a group

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.

8. ## Re: Determine if this set of continuous functions is a group

Originally Posted by TheChaz
I'm pretty sure this set is not closed under composition.
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

$\displaystyle (f \circ g) \circ h = f \circ (g \circ h)$.

$\displaystyle ((f \circ g) \circ h)(x) = (f \circ g)(h(x)) = f(g(h(x)) = f((g \circ h)(x)) = f \circ (g \circ h)(x)$.

9. ## Re: Determine if this set of continuous functions is a group

Originally Posted by Deveno
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)....
I didn't see the f(1) = 1 condition until later.

10. ## Re: Determine if this set of continuous functions is a group

and apparently, you edited your post, as i was writing mine. it's all good ^^.

11. ## Re: Determine if this set of continuous functions is a group

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 $\displaystyle 0\le x\le 1$ to a number in that interval.