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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.

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  • $\begingroup$ The statement moves from set-theoretic to group-theoretic structure without saying anything about the group being considered. This context should be added to the question (it can be inferred, but it should be stated explicitly). $\endgroup$
    – Bill Dubuque
    Dec 19, 2013 at 14:18
  • $\begingroup$ @BillDubuque this question is from intro to set theory course, I'm not sure I have the knowledge to what you mean (group theory). $\endgroup$
    – GinKin
    Dec 19, 2013 at 14:22
  • $\begingroup$ The exercise explicitly mentions the quotient group in a couple places. Are you sure that quotient groups have not already been introduced at this point? $\endgroup$
    – Bill Dubuque
    Dec 19, 2013 at 14:46
  • $\begingroup$ @BillDubuque Yeah it was introduced? (sort of), I just didn't know that there exists group theory. $\endgroup$
    – GinKin
    Dec 19, 2013 at 14:55
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    $\begingroup$ Often set theory is take after (or in parallel) with other courses such as abstract algebra, so then it makes sense to mention examples like the above. Did your teachers present quotient sets, and their equivalent view as partitions, etc. $\endgroup$
    – Bill Dubuque
    Dec 19, 2013 at 22:58

3 Answers 3

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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.

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  • $\begingroup$ Thanks a lot. But how do you describe $A/R$ ? $\endgroup$
    – GinKin
    Dec 19, 2013 at 13:55
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    $\begingroup$ My fourth paragraph above describes $A/R$ -- the set of equivalence classes is in 1-1 correspondence with the reals. To the degree that you meant "describe the quotient group" rather that "describe the quotient", I assume that the group structure on $[0, 1]^{\mathbb R}$ is "addition of functions"; in that case, the group structure on the quotient (once you identify it with the real numbers) is "addition of numbers", because $(f+g)(0) = f(0) + g(0)$, by definition of the sum of functions. $\endgroup$
    – John Hughes
    Dec 19, 2013 at 16:15
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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 $\{f(x) = i, \forall x \in [0, 1]\mid i \in \mathbb R$}

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  • $\begingroup$ I have only ever seen $X^Y$ to mean the set of all functions $Y \to X$. The benefit of this slightly counterintuitive notation is that $|X^Y| = |X|^{|Y|}$. Is there a reason why you think $A$ does not follow this convention? $\endgroup$
    – Michael Albanese
    Dec 19, 2013 at 13:51
  • $\begingroup$ Thanks for the answer amWhy. But how do you describe $A/R$ ? $\endgroup$
    – GinKin
    Dec 19, 2013 at 13:59
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    $\begingroup$ You've pretty much described $A/R=\{[f(x)]\mid f(0) = i,\; i\in \mathbb R\}$ $\endgroup$
    – amWhy
    Dec 19, 2013 at 14:05
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  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.
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