P is a partition of a set A. Define a relation R on A by declaring $xRy$ iff $x,y\in X$ for some $X\in P$. Prove R is an equivalence relation on A, then prove that P is a set of equivalence classes of R.
I don't know how to prove this
Prove R is an equivalence relation on A. Then Prove a partition is the set of equivalence classes of R
equivalence-relationsset-partition
Related Solutions
The set of partitions is just $V = \{R(x) \mid x \in X\}$.
More formally, let $P_v = v$ for all $v \in V$. Note that $P_v \subseteq X$ for all $v \in V$. I claim that $\{P_v\}_{v \in V}$ is a partition.
There are three conditions that must be checked. First, we must show that $\forall v \in V \exists x \in P_v$. To show this, consider some $v \in V$. Write $v = R(x)$ for some $x$. Then $x \in v = P_v$.
The second condition is that $\bigcup\limits_{v \in V} P_v = X$. To show this, suppose we have some $x \in X$. Then $x \in R(x)$. Let $R(x) = v$. Then $x \in v = P_v$. Therefore, $x \in \bigcup\limits_{v \in V} P_v$. Thus, we see that $\bigcup\limits_{v \in V} P_v = X$.
Finally, the third condition is that for all $v_1, v_2 \in V$, if $\exists x \in P_{v_1} \cap P_{v_2}$, then $v_1 = v_2$. Suppose there is some $x \in P_{v_1} \cap P_{v_2}$. Then we see that $R(x) = v_1$. And we see that $R(x) = v_2$. Therefore, $v_1 = v_2$.
You have the right idea, but I feel some explanation is lacking.
I would rewrite the definition of $R$ a bit more formally, for the sake of making it perfectly clear. I would say $$(x,a) \in R \iff (\exists !\,m \in I)(x \in A_m \text{ and } a \in A_m)$$ This makes it clearer to me that $x,a$ must lie in the same partite set. While clarity is certainly a subjective issue, I think it would also make proof-writing for this easier.
You claim the result for reflexivity without proving or explaining anything. Let's focus in a little. Reflexivity would mean that, $\forall x$, we have $(x,x) \in R$. We have an underlying partition, so given $x \in A$, for a unique $i \in I$, $x \in A_i$. Well, $x \in A_i$ and $x \in A_i$ (for that $i$), however redundant-sounding on the tongue, is a true statement, so $(x,x) \in R$.
You claim the result of symmetry likewise: you state it, but argue nothing for it. So, symmetry requires that $$ (x,a) \in R \implies (a,x) \in R $$ We begin by assuming, then $(x,a) \in R$ as a given. Well, $x \in A_m$ and $a \in A_m$ (for the relevant $m$) is given by definition, and logically equivalent to $a \in A_m$ and $x \in A_m$, which means $(a,x) \in R$, giving the symmetry.
You do at least explain the argument a bit with the transitive case, and have the right idea. $(x,a),(a,y) \in R$ means that, for some $m,n$, $x \in A_m, a \in A_m, a \in A_n, y \in A_n$. But since $\{A_i\}_{i \in I}$ partitions $A$, then $A_m = A_n$, i.e. $m=n$. Thus, $x \in A_m, a \in A_m, y \in A_m$, and by the first and last, $(x,y) \in A_m$, as desired.
Best Answer
The definitions of partition and equivalence classes are very closely related.
The definition of a partition is a collection of disjoint subsets $X_\alpha$ of $A$ so that each and every $a\in A$ is in precisely one $X_\alpha$.
So why does that imply an equivalence relation and why are the equivalence classes precisely the sets $X_\alpha$?
Why Reflexivity: $a R a$ for all $a\in A$. That is precisely the condition that every $a\in A$ is in some subset $X_a$. There is an $X_a$ so that $a \in X_a$. So $a \in X_a$ and $a \in X_a$ so $a R a$.
Why Symmetry: If $a R b$ then $b R a$. Well, that's impossible not to be true! If $a,b$ are both in the same subset $X_i$ then... they are both in the same subset $X_i$. So $a R b$ means there is an $X_i$ so that $a,b \in X_i$ and so $b,a\in X_i$ and $b R a$.
Why trasitivity: If $a R b$ and $b R c$ then $a R c$. Why? Well, this is where the $X_{\alpha}$ are distinct comes into play. $a$ must by is some $X_a$. But $X_a$ is the only $X_a$ so that $a \in X_a$ because that is what a partition is; each element of $A$ is in precisely one set. So if $a R b$ then they are in the same set which means $b \in X_a$. And that's the only set $b$ is in. And if $b R c$ then $b$ and $c$ are in the same set. And that set is $X_a$ because that's the only set $b$ is in. So $a,b$ and $c$ are all in $X_a$. So $a,c \in X_a$. Which meas $a R c$.
No an equivalence class is collections of subsets $E_a$ of $A$ so that for each $a\in A$ then $E_a = \{b \in A|a R b\}$. In other words $E_a = \{b\in A| a,b \in $ the same subset partion$\}$. But there is only one partion $X_a$ that contains $a$ and $E_a$ is the set that has precisely all the elements that also in the same partition and nothing else; the $E_a$ is the set that contains all the same elements of $X_a$. Which means $E_a$ is equal to $X_a$. And that is true for every element and for every subset of the partition. So $\{E_a\}$ is the exact same subset as the partition.