Prove that R is an equivalence relation on S.

Question

let $S$ be the union of disjoint sets $A_1, \dots , A_k$. Let $R$ be the relation consisting of pairs $(x,y) \in S \times S$ such that $x,y$ belong to the same member of $\{A_1, \dots, A_k\}$. Prove that $R$ is an equivalence relation on $S$.

To answer the question I did the following, I am not sure if it is correct.

Reflexive

For any $x \in A_i$, $(x,x) \in R$ and this also hold for any $y \in A_i$, so therefore reflexivity holds

symmetry

For any $x,y \in A_i$ where $x = y$, then $(x,y)=(y,x) \ in R$ therefore symmetry holds

Transitivity

given $(x,y),(y,z) \in R$ then $x=y$ and $y=z$ so therefore $x=z$ and Transitivity holds

I don't know if the above counts as a valid proof but that is what i was able to come up with.

Answer 1

I think you're confusing some things. The relation is "belonging to the same $A_i$", not "being equal".

Reflexive You have to show that $(x,x)\in R$ for all $x\in \bigcup A_i$. This is obvious since $x$ belongs to the same $A_i$ as $x$.

Symmetric You have to show that, if $(x,y)\in R$, then $(y,x)\in R$. This is also obvious, because if $x$ belongs to the same $A_i$ as $y$, then $y$ belongs to the same $A_i$ as $x$.

Transitive You have to show that, *if $(x,y)\in R$ and $(y,z)\in R$, then $(x,z)\in R$. Again, this is very simple, since $x$ belongs to the same $A_i$ as $y$ and $y$ belong to the same $A_j$ as $z$, which is $A_i$, $x$ belongs to the same $A_i$ as $z$.

I am implicitly using that the union is disjoint several times, and in particular, transitivity absolutely requires it in my sketch above.

Answer 2

No, that's not a good proof.

For symmetry and transitivity you should not assume equality of the objects involved. For example, when you assume that $xRy$ ... you know that $x$ and $y$ are from the same set $A_i$ (for some $i$), but that does not mean that $x=y$.

OK, I'll do a proper proof of symmetry, and then you try transivitity:

So, for symmetry we need to show that for any $x,y$: if $xRy$ then $yRx$. OK, so assume $xRy$. that means that $x$ and $y$ are both elements from the same set $A_i$ for some $i$. But that means that $y$ and $x$ are both elements from the same set $A_i$ for some $i$. And therefore $yRx$.

Your justification for reflexivity is very poor. You pretty much just immediately say that $xRx$, but you need to actually show this. So: start out by saying that $x$ is some arbitrary element from $S$, and then show that therefore $xRx$. One step you need to do is: given that $x \in S$, it must be true that $x \in A_i$ for some $i$ ...