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

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.

You need to test if

where are in the set of functions described

There are more than just two elements...

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

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.

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.

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

.

.

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.

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

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.