Equivalence relation question with functions

Question

We'll define on the set: $A=\Bbb R^{[0,1]}$ the relation $R$ by $fRg$ if $f(0)=g(0)$.

1. Make sure it's an equivilence relation.
2. What is $[\cos x]$ ?
3. Describe all the equivalence classes and the quotient group $A/R$.
4. Describe a subset of $A$, in it only one element exactly from each equivalence class (a system of representatives for the quotient group).

Well, first of all to make sure I understand the notation, $A=\Bbb R^{[0,1]}$ means the set of all the functions from the section $[0,1]$ to the reals right ? And the relation is when two functions are equal when $x=0$.

If it was from the reals to the section then some of these functions would be the trig functions, but the other way around I don't think it would be any elementary function.

1. In order to prove this is an equivalence relation we need to check reflexivity, symmetry and transitivity, so for reflexivity I suppose it's enough to say that $f(0)=f(0)$, symmetry: $f(0)=g(0) \rightarrow g(0)=f(0)$ and transitivity: $f(0)=g(0) \ , \ g(0)=h(0) \Rightarrow f(0)=h(0)$.

2. What is $[\cos x]$ ? I'm not sure I understand the notation here...

3. I suppose the equivalence classes are functions that equal $i\in \Bbb R$ when $X=0$ so there are $i$ equivalence classes. I have no idea how to describe $A/R$.

4. That's basically every function because there can always be $fRf$.

Please share your thoughts on what I did and how to solve this.

Thanks.

Answer 1

Your answer to part 1 is fine.

The notation means "functions from the interval $[0, 1]$ to the reals"; the interval $[0, 1]$ is the set of all real numbers between $0$ and $1$, inclusive.

For part 2, the notation $[\cos x]$ means "the equivalence class of the function $x \mapsto \cos x$. What functions are equivalent (under this relation) to cosine? Ones that have the same value at $x = 0$ as cosine, i.e., all functions $f$ with $f(0) = 1$.

You've observed that for each number $i \in \mathbb R$, there's an equivalence class (consisting of all functions that take the value $i$ at $x = 0$). So the set of equivalence classes is in 1-to-1 correspondence with the real numbers: the functions whose value at $0$ is $i$ correspond to the real number $i$.

For part 4, you need to find one element of each class. So a typical class is the one for $i = 3.5$. Can you think of a function whose value at $0$ is $3.5$? Sure, you can think of millions of them. But a particularly easy one is the constant function $f(x) = 3.5$. And indeed, the set of all constant functions has the property that each function is in a different equivalence class, and each equivalence class contains one of these functions.

You were 90% of the way there with your reasoning. Keep up the good work.

Answer 2

I believe $A$ is equal to the set of all functions that map from $[0, 1] \to \mathbb R $.

$(2)$ The equivalence class of $f(x) = \cos x$ are all those functions $g(x)\in A$ that share the same value at $x = 0$, as does $f(x) = \cos x$. Note that $f(0) = \cos 0 = 1$, and hence, all functions $g(x) \in A$ such that $g(0) = 1$ belong to the equivalence class $[\cos x]$.

$(3)$ You've got the basic idea correct. $R$ partitions $A$ such that for each value $i \in \mathbb R,$ all functions $f$ such that $f(0) = i$, there is associated one and only one equivalence class.

$(4)$ Here, your set can consist of the constant functions $f(x) = i$ for each and every real number $i \in\mathbb R$: We can denote that set as follows $\{f(x) = i, \forall x \in [0, 1]\mid i \in \mathbb R$}

Answer 3

1. Looks good.
2. It is the unique equivalence class containing the function $\cos x \in \mathbb{R}^{[0,1]}$. $[\cos x] = \{f \in \mathbb{R}^{[0,1]} \mid fR\cos x\} = \{f \in \mathbb{R}^{[0,1]} \mid f(0) = \cos 0 = 1\}$.
3. You're close. I think you have the correct equivalence classes, but there are a lot more than you've said.
4. I don't really understand how what you said was relevant. You need to pick a representative of each equivalence class. There is a very simple function in each equivalence class that would be a good representative.