# Confusion with the lattice formed by a partition

I was referring to this article here related to the formation of a complete lattice by the partitions of a set. The article has stated that the partitions not only form the lattice for themselves but also for the equivalence relations.

I know that each partition has a corresponding equivalence relation. However, I don't get the derivation here in this article.

Things like the following

$$a \equiv b(modE)$$

I didn't get the things given in the remarks specially after it says

Correspondingly, the partition lattice of S also defines the lattice of equivalence relations $\Delta$ on S

Any guidance pls?

It mentions that

Given a family $\{E_i|i \in I\}$ of equvialence relations on S , we can explicitly describe the join E:=V $E_i$ of $E_i$ , as follows:

$a\equiv b$(modE) iff there is a finite sequence

$a=c_1,c_2,...c_n=b$ such that

$c_k \equiv c_{k+1}(modE_{i(k)})$ for $k=1,..n−1$

I didn't get this one as well. Can anyone please provide some examples so that it is easier for me to visualize.

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$a \equiv b(\text{mod}E)$ means that $a$ and $b$ lie in the same cell of $E$. –  Michael Greinecker Sep 12 '12 at 11:23
@Michael, we have cells only in the partition isn't it? So you mean in the same cell of the partition equivalent to the equivalence relation E isn't it? –  user34790 Sep 12 '12 at 13:27
Yes. Actually, in the present context of equivalence relations it simply means $aEb$. –  Michael Greinecker Sep 12 '12 at 13:49
@Michael, I have also added one more thing that I didn't understand in the original question. Can you give me some examples like with a set A and make it clear to me. I think if I visualize with examples it will be clear to me –  user34790 Sep 12 '12 at 13:53
Please don't change the question. I can explain these things to you on chat: chat.stackexchange.com/rooms/36/mathematics –  Michael Greinecker Sep 12 '12 at 13:54
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Let $\Pi$ be a partition on a set $X$ and $E(\Pi)$ be the corresponding equivalence relation on $X$ given by $x E(\Pi) y$ iff $x$ and $y$ are in the same element of $\Pi$. We say the partition $\Pi_1$ is finer than the partition $\Pi_2$ if every element in $\Pi_2$ is a union of elements in $\Pi_1$. One can then show that $\Pi_1$ is finer than $\Pi_2$ iff $E(\Pi_1)\subseteq E(\Pi_2)$. So the function that maps each partition to the corresponding equivalence relation is an order isomorphism between "finer than" and $\subseteq$.
Since being a complete lattice is preserved under order isomorphisms, it follows from the set of all partitions on $X$ ordered by "finer" being a complete lattice that the set of all equivalence relations on $X$ ordered by $\subseteq$ is a complete lattice too.