Partial order with an unusual relation.

Question

Determine whether $(V,E)$ is a partial order. $V = \{a, b, c, d\}$ $E = \{(a, b), (b, c), (c, d), (a, c), (a, d), (b, d)\}$

I am confused since the definition of partial order says:

A partial order (PO) $(D,\sqsubseteq)$ consists of a set $D$, called domain, and of a relation $\sqsubseteq \subseteq D \times D$ such that for every $d1, d2, d3 \in D$ we have:

• Reflexivity: $d1 \sqsubseteq d1$
• Transitivity: $d1 \sqsubseteq d2 \land d2 \sqsubseteq d3 \Rightarrow d1 \sqsubseteq d3$
• Antisymmetry: $d1 \sqsubseteq d2 \land d2 \sqsubseteq d1 \Rightarrow d1 = d2$

Now, when I look at the given $E$, I don't understand how is it a relation. How can I use it like if it was let's say relation $<$? Where I can simply see if $d1 < d2$ for example?

Answer 1

A binary relation $E$ on a set $V$ is rigorously defined as a set of pairs of elements of $V$, and the notation $x~E~y$ stands for $(x,y)\in E$.

Said that, now we have $a~E~b~E~c~E~d$, and 3 more: $a~E~c, \ a~E~d, \ b~E~d$, which lead to the transitivity of $E$.
You can also verify antisymmetry of $E$.
However, $E$ is not reflexive, as none of the following holds (though all should): $$a~E~a, \ b~E~b, \ c~E~c, \ d~E~d$$ So, $E$ is not a partial order.

Answer 2

$E$ is the relation $a < b$ if $(a,b) \in E$.

So $a \sqsubseteq b; b\sqsubseteq c; c\sqsubseteq d; a\sqsubseteq c; a\sqsubseteq d; a\sqsubseteq d$.

And that's it. That is a list of every element that is related to any other element.

Is is it reflexive? Does $x \sqsubseteq x$ for all $x\in V$? In other words is $(x,x) \in E$ for all $x\in V$?

Is it reflexive. For an $x\sqsubseteq y$ and $y \sqsubseteq z$ do we always have $x\sqsubseteq z$. That is; if $(x,y)\in E$ and $(y,z) \in E$ do we always have $(x,y) \in Z$?

Is it anti reflexive. If $x \sqsubseteq y$ and $y\sqsubseteq x$ does it always follow that $x =y$. Or if $(x,y),(y,x)\in E$ does that only happen when $x=y$?

Answer:

It's not reflexive. We do not have any of $a\sqsubseteq a$ or $b\sqsubseteq b$ or $c\sqsubseteq c$ or $d\sqsubseteq d$. It does seem to be transitive. Indeed it seems to be $x \sqsubseteq y\iff y$ is higher in the alphabet than $x$. And it is antisymetric as we never have any $x\sqsubseteq y$ and $y\sqsubseteq x$.

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So anyway it isn't a partial as we don't have reflexivity.