Relations with Combinations of Reflexivity, Symmetry and Transitivity Properties

Theorem
Let $S$ be a set which has at least $3$ elements.

Then it is possible to set up a relation $\circledcirc$ on $S$ which has any combination of the $3$ properties:
 * Reflexivity
 * Symmetry
 * Transitivity

but this is not possible for a set which has fewer than $3$ elements.

Proof
In the following:
 * $(1): \quad S_n$ the set $S_n = \set {s_1, s_2, \ldots, s_n}$ of cardinality $n$, where $n \in \Z_{\ge 0}$ is a non-negative integer.


 * $(2): \quad \circledcirc$ denotes an arbitrary relation on $S_n$.

Let:
 * $\map R \circledcirc$ denote that $\circledcirc$ is reflexive
 * $\map S \circledcirc$ denote that $\circledcirc$ is symmetric
 * $\map T \circledcirc$ denote that $\circledcirc$ is transitive

Set of Cardinality $0$
When $S = S_0 = \O$, the result Relation on Empty Set is Equivalence applies.

Hence $\circledcirc$ is always reflexive, symmetric and transitive:


 * $\map R \circledcirc$
 * $\map S \circledcirc$
 * $\map T \circledcirc$

Set of Cardinality $1$
When $S = S_1 = \set {s_1}$, there are $2$ relations on $S_1$:

$(1): \quad \circledcirc := \O$, which is the null relation on $S_1$.

From Null Relation is Antireflexive, Symmetric and Transitive, $\circledcirc$ is always antireflexive, symmetric and transitive, and so:


 * $\neg \map R \circledcirc$
 * $\map S \circledcirc$
 * $\map T \circledcirc$

$(2): \quad \circledcirc := \set {\tuple {s_1, s_1} }$, which is the trivial relation on $S_1$.

From Trivial Relation is Equivalence, $\circledcirc$ is always reflexive, symmetric and transitive.


 * $\map R \circledcirc$
 * $\map S \circledcirc$
 * $\map T \circledcirc$

Thus when $n = 1$, it is not possible for a relation on $S_n$ to be either non-symmetric or non-transitive.

Set of Cardinality $2$
Let $S = S_2 = \set {s_1, s_2}$.

From Reflexive Relation on Set of Cardinality 2 is Transitive, it is not possible for $\circledcirc$ to be both reflexive and non-transitive.

From Relation on Set of Cardinality 2 cannot be Non-Symmetric and Non-Transitive, it is not possible for $\circledcirc$ to be neither symmetric nor transitive.

Thus the following combinations of $R$, $S$ and $T$ are not possible with $S_2$:


 * $\map R \circledcirc, \neg \map S \circledcirc, \neg \map T \circledcirc$
 * $\map R \circledcirc, \map S \circledcirc, \neg \map T \circledcirc$
 * $\neg \map R \circledcirc, \neg \map S \circledcirc, \neg \map T \circledcirc$

However, we note the following examples of the possible combinations of properties for relations on $S_2$:

$(1): \quad \circledcirc := \O$, which is the null relation on $S_2$.

From Null Relation is Antireflexive, Symmetric and Transitive, $\circledcirc$ is always antireflexive, symmetric and transitive, and so:


 * $\neg \map R \circledcirc$
 * $\map S \circledcirc$
 * $\map T \circledcirc$

$(2):$
 * $\circledcirc := \set {\tuple {s_1, s_1}, \tuple {s_1, s_2}, \tuple {s_2, s_1}, \tuple {s_2, s_2} } = S \times S$

that is, the trivial relation.

From Trivial Relation is Equivalence:.


 * $\map R \circledcirc$
 * $\map S \circledcirc$
 * $\map T \circledcirc$

$(3): \quad \circledcirc := \set {\tuple {s_1, s_2} }$:

By inspection:


 * $\neg \map R \circledcirc$
 * $\map S \circledcirc$
 * $\neg \map T \circledcirc$

$(4):$
 * $\circledcirc := \set {\tuple {s_1, s_1}, \tuple {s_1, s_2} }$

By inspection:


 * $\neg \map R \circledcirc$
 * $\neg \map S \circledcirc$
 * $\map T \circledcirc$

$(5):$
 * $\circledcirc := \set {\tuple {s_1, s_1}, \tuple {s_2, s_2}, \tuple {s_1, s_2} }$

By inspection:


 * $\map R \circledcirc$
 * $\neg \map S \circledcirc$
 * $\map T \circledcirc$