1. ## Equivalence relations!

Which of the following relation on the given sets (S) define equivalence relations? Describe the equivalence classes where appropriate:

a) S= R3
x~y if there is a 1 dimensional subspace of R3 which contains x and y

b) S= R3 - {0}
x~y if there is a 1 dimensional subspace of R3 which contains x and y

c) S = R3
x~y if there is a 2D subspace of R3 which contains x and y

I always get confused with equivalence relations so I just wanted to give my thoughts to see if I am on the right lines!

a) I think this is an equivalence relation as it is reflexive, symmetric and transitive.
I think the equivalence classes would be the individual lines that the two points lie on. However this would mean that there would be infinitely many equivalence classes so seems a bit suspect?

b) I dont think this relation is an equivalence relation as it is not transitive. I looked at the case of a triangle with two of the lines going through the origin, but the other therefore wouldnt.

c) I assumed this question would be similar to a. Reflexive, Symmetric and Transitive, with the equivalence classes being the individual planes that the lines lie within.

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2. Be careful with the equivalence classes.

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3. I needed to edit this post, because I was in a hurry last time, and overlooked the zero vector. Here is the corrected solution:

First of all, I will be assuming that your "if"s are "iff"s, and that "subspace" refers to a vector subspace. That said...

(a) Since we can choose $x,y\in\mathbb{R}$ with $x\not\sim y$ (for example $(1,0,0),(0,1,0)$), then $x,y\sim\textbf{0}$, and so transitivity does not hold. So $\sim$ is not an equivalence relation.

(b) Reflexivity and symmetry are obvious here. For transitivity, recall that $(x_1,x_2,x_3)\sim(y_1,y_2,y_3)$ iff $(y_1,y_2,y_3)=a(x_1,x_2,x_3)$ for some $a\in\mathbb{R}$ (the underlying field). Suppose $(x_1,x_2,x_3)\sim(z_1,z_2,z_3)$. Then there are $a,b\in\mathbb{R}$ with $a(y_1,y_2,y_3)=b(z_1,z_2,z_3)$, which means $(z_1,z_2,z_3)\sim(y_1,y_2,y_3)$. So transitivity holds, and $\sim$ is an equivalence relation.

(c) Transitivity is again the issue, here, but this time it might work even with the zero vector. Indeed, for any $x,y\in\mathbb{R}^3$, we have $x,y\in\text{span }\{x,y\}$, where $\text{span }\{x,y\}$ is a two-dimensional vector space. So $x\sim y$, and transitivity holds such that $\sim$ is an equivalence relation (and $\mathbb{R}^3/\sim=\{\mathbb{R}^3\}$).

4. a) S= R3
x~y if there is a 1 dimensional subspace of R3 which contains x and y
First, who would say this instead of something like "the line segment (x, y) is a subset of S"?

Anyway, I agree that (a) and (c) are equivalence relations and (b) is not.

(a) Reflexivity and symmetry are obvious here. For transitivity, recall that iff for some .
I am not sure about this. E.g., (0,0,1) ~ (0,1,0) because they lie on the same line, but there is no such a that (0,0,1) = a * (0,1,0).

(b) Everything is the same from part (a). Please note that wording is important here, since there are plenty of subspaces of , but no subspaces of . In that case, is still an equivalence relation, and indeed .
The relation ~ in (b) is not empty; it is already defined and, for example, (0,0,1) ~ (0,0,2).

Concerning equivalence classes, in (a) it's not lines because any three points are equivalent, and similarly for (c).

5. Originally Posted by emakarov
First, who would say this instead of something like "the line segment (x, y) is a subset of S"?
I think he means vector spaces. So, it is not enough that $x,y$ be in the same line, but they have to be scalar multiples of each other. If I am mistaken in this assumption, then of course my results will be invalid.

6. Originally Posted by hatsoff
I think he means vector spaces. So, it is not enough that $x,y$ be in the same line, but they have to be scalar multiples of each other. If I am mistaken in this assumption, then of course my results will be invalid.