Can a relation be transitive when it is symmetric but not reflexive? [duplicate]

Solution 1:

Symmetric means we can represent the relation as an undirected graph. Transitive means this graph is composed of connected components which either look like a point or $K_n$ with each point connected to itself. Reflexivity means each point is connected to itself.

Thus, a necessary and sufficient condition is that one component is a point (i.e.- one element is not related to any other). In exactly these cases your proof fails because it requires each $x$ to be related to another $y$.

Solution 2:

The answer is yes; the empty relation (on any nonempty set $X$) is transitive, symmetric, and not reflexive.

Solution 3:

If there exists $(x,y)$ in a symmetric relation, there also exists $(y,x)$. If this relation is transitive, then $(x,x)$ and $(y,y)$ must exist in the relation. However, this is not sufficient to make the relation reflexive.   That would further require that every $x$ in the underlying set has some $y$ where $(x,y)$ is in the symmetric-and-transitive relation.

A symmetric and transitve relation, $R$, over set $A$, is also reflexive exactly when $\forall x{\in}A~\exists y{\in} A :(x,y)\in R$.

A symmetric and non-reflexive relation, $R$, over set $A$, may be transitive only if there is something (in $A$) not related to anything.   A symmetric $R$ where something is not related to anything is not neccessarily transitive, but it may be.  

$${\begin{Bmatrix}(0,0),&(0,1),&&(0,3)\\(1,0),&(1,1),&&(1,3)\\~\\(3,0),&(3,1),&&(3,3)\\&&&&(4,4)\end{Bmatrix}{\text{ is a symmetric, non-reflexive, transitive relation}\\\text{over the set }\{0,1,2,3,4\}}\\\begin{Bmatrix}&(0,1)\\(1,0),&(1,1),&&(1,3)\\~\\&(3,1)\\&&&&(4,4)\end{Bmatrix}{\text{ is a symmetric, non-reflexive, non-transitive relation}\\\text{where something (2) is not related to anything}}}$$