The relation x=1 This one stumped me: Determine if the relations R on $\mathbb{R} $ are reflexive, symmetric, transitive, antisymmetric: $ (x,y) \in R$ iff $x=1$. 
It is reflexive since 1=1. 
It is symmetric since it is the case that for all $x=1, 1=1$
Is it transitive? For it to be transitive, it has to be the case that $\forall x,y,z \in A, (x,y) \in R \wedge (y,z) \in R, \rightarrow (x,z) \in R.$
I want to say yes to that, but I am not certain.
I am sure it is antisymmetric since $(1,1)$ and $(1,1) \in R$ but it is also the case that $1=1$.
How does one handle a case like this?
Thanks!
 A: This is not reflexive, as $(0,0) \not\in R$.
This is not symmetric, as $(1,0) \in R$ but $(0,1) \not\in R$.
This is transitive as if $(x,y) \in R$ and $(y, z) \in R$ then $x=1$ and thus $(x,z)\in R$.
This is antisymmetric because if $(x,y) \in R$ and $(y,x) \in R$ then $x=1$ and $y=1$.
A: It is transitive and antisymmetric, as you claim. However


*

*It is not reflexive, since $(0,0) \not \in R$ for instance. For reflexivity, you need $(x,x) \in R$ for all $x \in \mathbb{R}$.

*It is not symmetric, since $(1,0) \in R$ but $(0,1) \not\in R$.


However, if it was both symmetric and antisymmetric, there would be interesting things to say.
Relations which are both symmetric and antisymmetric are those which are restrictions of the equality relation. Indeed, if $R$ is a relation on a set $X$ that is both symmetric and antisymmetric then, given $x,y \in X$, if $(x,y) \in R$ then $(y,x) \in R$ by symmetry, so $x=y$ by antisymmetry. That is
$$(x,y) \in R \quad \Rightarrow \quad x=y$$
So we must have $R = \{ (x,x) \mid x \in U \}$ for some subset $U \subseteq X$.
For such a relation, $R$ being reflexive is equivalent to $U=X$, in which case $R$ is the equality relation. But $R$ may not be reflexive.
