Is this ordered set a Lattice?

Question

So I have this problem:

Let $S$ be the set of all reflexive, symmetric relations on $\mathbb N$, $A$ the set of all reflexive antisymmetric relations on $\mathbb N$. Now consider the set $M=S\cup A$. Is $(M,\subseteq )$ a lattice?

Here is what I have come up with:

An ordered set is a lattice if every two elements of the set have an infimum and supremum so i have to show how to calculate them. For the infima:

Let $x,y\in M$: $x\wedge y=x\cap y$, because the result is either a reflexive antisymmetric, reflexive, or empty relation. I am not sure if this is correct.

Now for the suprema, I know it will be some sort of union of the sets depending on their form but I am not sure what properties are kept under the union operation.

The biggest problem that I have with this is that I have a hard time visualizing the set $M$. Any help/tips/whatever is greatly appreciated.

Answer 1

Both $S$ and $A$ are closed under intersection, and $x\cap y\in A$ if $x\in S$ and $y\in A$, so infima aren’t a problem. (Note that reflexivity ensures that the intersection is never empty.) $S$ is also closed under unions, but $A$ is not, and indeed the union of two members of $A$ need not even be in $M$.

Let $X=\{0,1,2\}$, let $D$ be the diagonal in $X\times X$, and let $r=D\cup\{\langle 1,0\rangle,\langle 2,1\rangle\}$ and $s=D\cup\{\langle 0,1\rangle\}$. Then $r,s\in A$, but $r\cup s\notin M$.

In that example the only reasonable candidate for $r\lor s$ is the symmetric closure of $r\cup s$, and it’s not hard to show that this really is $r\lor s$. On the other hand, if we change $s$ to $D\cup\{\langle 0,2\rangle\}$, $r\cup s\in A$, and in that case clearly $r\lor s=r\cup s$. Thus, in general you’ll have to look at the actual form of $r\cup s$ to determine $r\lor s$. I suggest considering the following cases:

• at least one of $r$ and $s$ is equality (i.e., the diagonal)
• $r,s\in S$
• $r\in S\setminus A$ and $s\in A\setminus S$
• $r,s\in A\setminus S$; my example shows why this will have to be split into two subcases.

Answer 2

I assume that the ordering relation on $M$ is that of set inclusion, $\subseteq$.

Hint: If a relation $R$ is an element in $S \cap A$, it is reflexive and symmetric as well as antisymmetric. What does this tell us about $R$?