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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 :)

CoffeeArabica
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2 Answers2

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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.
Taladris
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  • Ahh I think Ive got it! Correct me if I am wrong: When a relation between a set of elements is symmetric and transitive it has to be reflexive, as long as there is not atleast one element which is not in relation with the other elements. Like in my other example? Because when x,y are even than is y,x even and when when y,z even, we can conclude that x,z has to be even aswell. But as we look at the relation on the set of integers, 1;3;5... wont be elements, which are related in any way with the subset of N, who built a relation. – CoffeeArabica Sep 01 '19 at 08:10
  • @CoffeeArabica: right. This is a good example. The odd numbers are not related to any other integer, so the relation is not reflexive. It is obviously symmetric and transitive. – Taladris Sep 01 '19 at 09:07
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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.

Nitin Uniyal
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