Show what order relations apply:

Set $X = \mathbb{C}$.

$(z_1,z_2) \Leftrightarrow Re(z_1) \leq Re(z_2)$

"($z_1$ in relation to $z_2$) is equivalent to ((the real part of $z_1$) $\leq$ (the real part of $z_2$))."

I know most of the order relation rules, like reflexive, symmetric, antisymmetric and transitive.

I was thinking:

$z_1 = a+bi$ and $Re(z_1) = a$
$z_2 = c+di$ and $Re(z_2) = c$


$(a+bi,c+di) \Leftrightarrow a \leq c$

But I do not know if this is useful or how to go from there. I can't see any of the relation rules applying here, since most of our other examples have been along the lines of
$X = \left \{a,b,c\right \},$ $R=\left \{(a,a),(a,b),(b,c)\right \}$


The usual method with these questions is, for a given property (reflexivity, etc.), either prove it false by a specific counter-example, or prove it true by working on non-specific members of the set ($x,y$, etc.) that can take any value in that set.

Call the relation $R$.

$R$ is reflexive: Let $x\in\mathbb{C}$ with $\;x=a+bi,$ where $a,b\in\mathbb{R}$. We have $\;a\leq a$ so $(x,x)\in R$.


$R$ is transitive: Also let $y,z\in\mathbb{C}$ with $\;y=c+di,\;z=e+fi,$ where $c,d,e,f\in\mathbb{R}$ and suppose $(x,y)\in R$ and $(y,z)\in R$. Then $a\leq c\leq e$ so $(x,z)\in R$.


$R$ is not symmetric: $0,1\in\mathbb{C}$ and $(0,1)\in R$ since $0\leq 1$ but $(1,0)\notin R$.


$R$ is not antisymmetric: $i,2i\in\mathbb{C}$ and we have both $(i,2i)\in R$ and $(2i,i)\in R$ (because the real part of both $i$ and $2i$ is $0$) but $i\neq 2i$.


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