Question to an exercise about relations which should be not reflexive but symmetric and transitive

Question

First I thought about a certain relation, but wasnt sure about the transitivity in the end, so I went a save route with:

For $x,y\in$ N :

$x\sim y \Longleftrightarrow x,y $: even

But the question was about my first thought! At first I thought I could get a not-reflexive, symmetrical, transitiv relation with the following:

For $ x \in M$:

$x\sim y \Longleftrightarrow | x-y|> 0$

But in this case I wasnt sure if it is transitive, since $x\sim y, y \sim z \Rightarrow x\sim z$. Can there be the case that $x = z$? So that we get the case of $x\sim x$, which isnt possible since the relation is not reflexive?

This is my first question, and as you can see I am just getting started with my math education.. I hope such questions dont bother the more experienced to much :)

Answer 1

Assume that $\sim$ is a relation that is symmetric and transitive.

Then, if $x\sim y$, we have $y\sim x$ by symmetry, so $x\sim x$ by transitivity. So, if $x$ is related to any element of the set, it has to be related to itself.

So, for the relation to not be reflexive, you need at least one element that is not related to any other element.

For example:

1. The empty relation on a nonempty set $X$.
2. Take any set $B$ with a symmetric, transitive (reflexive or not) relation and arbitrarily add one or more point with the property that the new points are not related to any point of the set.
3. A concrete example: the set of vertices $V$ of a nonoriented graph $G$ with the relation $\sim$ defined by $v\sim w$ if and only if there is a path between $v$ and $w$ in $G$. The relation $\sim$ is clearly symmetric and transitive, since the graph is nonoriented. If there are isolated vertices, then it is not reflexive.

Answer 2

Answering the title of ur problem ,"A relation which is symmetric and transitive but not reflrexive" is like this one.

On set $A=${$1,2,3$} consider the relation $R=${$(1,2),(2,1),(1,1),(2,2)$}. It is not reflexive as $(3,3)$ is missing.