1
$\begingroup$

Having some problems with this question and hoping someone could help.

Let $S$ and $S'$ be the following subsets of the plane: $$S = \{(x,y): y=x+1\text{ and }x\text{ a member of }(0,2)\},$$ $$S'= \{(x,y): y-x\text{ is an integer}\}.$$

A. Show that $S'$ is an equivalence relation on the real line and $S$ is a subset of $S'$. Describe the equivalence classes of $S'$.

B. Show that given any collection of equivalence relations on a set $A$, their intersection is an equivalence relation on $A$.

C. Describe the equivalence relation $T$ on the real line that is the intersection of all equivalence relations on the real line that contain $S$. Describe the equivalence classes of $T$.

C is the problem I am having the most difficulty with.

Answers so far:

a.

  • EDIT: Symmetric: if (x,y) works then, $y-x=n \rightarrow x-y=-n$
  • Reflexive because $x-x = 0$

  • EDIT: Transitive $x-y=n$ and $y-z=k$ $\rightarrow y=k+z \rightarrow x-(k+z)=n \rightarrow x-z = k+n=p$

  • $S: y=x+1 \rightarrow y-x=1$, 1 integer out of the set of all integers in $S'$.

  • Equivalence classes of $S$ would be all diagonal lines with slope 1 through $y=n$ and $x=n$.

b. $E_1$ and $E_2$ equivalence relations. Intersections both contain $(x,y)$ and $(y,x)$ because if they are members of either $E_1$ or $E_2$ they are satisfy equivalence requirements.

c. I am not sure how to answer this. I thought the intersection of all equivalence relations on the real line containing $S$ would be $S$.

Any help would be greatly appreciated.

  • 0
    I'm not sure I understand your proof of symmetry. The point is that if $(x,y) \in S'$, so that $x-y = n$ is an integer, then $y-x=-n$ is an integer, so $(y,x) \in S'$. This shows symmetry. A similar comment can be made for transitivity.2011-07-13
  • 0
    I've added LaTeX formatting to your question. Apologies if I changed any intended meaning; feel free to edit further if you like.2011-07-13
  • 0
    the intersection of all equivalence relations on $\mathbb{R}$ containing $S$ would be $S$ if $S$ were an equivalence relation. but it is not (for example, it's not reflexive since no pair of the form $(x,x)$ is in $S$2011-07-13
  • 0
    I would try to think about the equivalence classes of $S'$ for a little while longer. These are subsets of $\mathbf{R}$, so are probably not diagonal lines!2011-07-13
  • 0
    I think you meant $S$ is a subset of $S'$; the statement "$S'$ is a subset of $S$" is false.2011-07-13
  • 0
    Ah right, that was a mistake.2011-07-13
  • 0
    @Ryan Warnick: The write-up here is far too short. The question is about the intersection of any collection of equivalence relations on $A$. Your write up begins to deal with the intersection of two equivalence relations.2011-07-14
  • 0
    @user6412: Would this be more correct.-- $E_1$ and $E_2$ equivalence relations on A. Intersections both contain $(x,y)$ and $(y,x)$ because if they are members of either $E_1$ or $E_2$ they are satisfy equivalence requirements. Define this intersection to be an equivalence relation V1, then given another equivalence relation E3 on A, the intersection of these two relation is V2 with both containing (x,y) and (y,x) due to both being equivalence relations. Continue until you have the intersection of all E(n) on A, V(n-1).2011-07-14
  • 0
    @Ryan: What you suggest in your response to user6312 won’t work for (B), because the collection of equivalence relations whose intersection you’re taking may be too big to be indexed by positive integers.2011-07-14
  • 0
    @Ryan Warnick: Luckily it is a lot simpler than that. (i) Let $a\in A$. I think you can quickly explain why $(a,a)$ is in every equivalence relation in the collection, and therefore in the intersection of these equivalence relations. (ii) Suppose that $(a,b)$ is in the intersection of the equivalence relations. You can quickly explain why $(b,a)$ also is. (iii) If $(a,b)$ and $(b,c)$ is in the intersection $\dots$. All straightforward, but shows control of the definition.2011-07-14
  • 0
    @Ryan Warnick: I do not understand your identification of equivalence classes. What is the equivalence class of $ln 2$? By definition it is all numbers that differ from $\ln 2$ by an integer. Now can you come up with a "nice" representative for each equivalence class? (We sometimes can't, but here we can.)2011-07-14

1 Answers 1