Examples of relations that don't satisfy one of the three properties of an equivalence relation while satisfying the other two? Just as a question that I have posed to myself: I want to find three relations $(S, \spadesuit)$, $(R, \clubsuit)$ and $(T, \blacksquare)$ for which $(S, \spadesuit)$ doesn't satisfy the reflexive property but satisfies symmetric and transitive. $(R, \clubsuit)$ doesn't satisfy the symmetric but satisfies the other two. Finally, $(T, \blacksquare)$ doesn't satisfy the transitive but satisfies the other two. Here my notation is that $S$, $R$, and $T$ are sets and $\spadesuit$, $\clubsuit$ and $\blacksquare$ are the relations. So far here is what I have come up with:
$$(S, \spadesuit)=(\mathbf{Z}, \text{x and y are even})$$ $$(R, \clubsuit)=(\mathbf{R},  x \leqslant y)$$ $$(T, \blacksquare )= ([x\, |\, x \text{ is a nonempty set} ] , x\cap y \neq \emptyset )$$
I guess I am asking for people to join me in the creativity here. Are there more interesting relations that satisfy these rules? I predict the question might be deemed inapropriate for the site. So here is a follow up question: I would love to find simple examples of equivalence relations that are provable at a very modest level. For example, I love the relation $x\sim y$ defined to be "$x-y$ is an integer" on the reals or the congruence of line segments. Are there other "easier" examples?

Symmetric, transitive but not reflexive relation
Consider on $\mathbb{R}$ the following relation:
$$x\sim y \iff x\cdot y>0$$
It's symmetric and transitive, but not reflexive because $0 \nsim 0 $.
Reflexive, transitive but not symmetric relation
Let $A$ and $B$ two sets, define the relation as follow:
$$A\sim B \iff A\subseteq B $$
This is clearly not symmetric as $A\subseteq B$ does not imply $B\subseteq A$.
