10

I am not sure on how to go about this. Please provide clear explanations.

Martin Sleziak
  • 53,687
  • 2
    Can you find two incomparable elements in the empty set? – azarel Jul 12 '12 at 05:18
  • 1
    So, as there are no elements in the empty set, there are no incomparable sets. So, it is totally ordered. Is that right? –  Jul 12 '12 at 05:25
  • Partially ordered does not mean there exists incomparable elements. Not totally or not linearly ordered is more precise term for this. – William Jul 12 '12 at 05:28

3 Answers3

36

The question as phrased isn’t really meaningful, since you didn’t specify a relation on the empty set. However, there is only one, so I’ll assume that it’s the one that you meant. A relation on a set $A$ is a subset of $A\times A$. Since $\varnothing\times\varnothing=\varnothing$, the only subset of $\varnothing\times\varnothing$ is $\varnothing$. Thus, the question can be interpreted as:

Is $\varnothing$ a partial order on $\varnothing$? Is it a total order?

Let’s recall the definitions:

A relation $R$ on a set $A$ is a partial order if it is reflexive, antisymmetric, and transitive.

  • Reflexive: For each $a\in A$, $a\,R\,a$.
  • Antisymmetric: For all $a,b\in A$, if $a\,R\,b$ and $b\,R\,a$, then $a=b$.
  • Transitive: For all $a,b,c\in A$, if $a\,R\,b$ and $b\,R\,c$, then $a\,R\,c$.

If in addition it’s true that for all $a,b\in A$, $a\,R\,b$ or $b\,R\,a$, then $R$ is a total order on $A$.

The thing to notice here is that each of these is a universally quantified statement: something must be true of every element, pair of elements, or trio of elements of $A$. Thus, in order to show that $R$ is not reflexive, you must find an $a\in A$ such that $a\,\not R\,a$; in order to show that $R$ is not antisymmetic, you must find elements $a,b\in A$ such that $a\,R\,b$ and $b\,R\,a$, but $a\ne b$; in order to show that $R$ is not transitive, you must find elements $a,b,c\in A$ such that $a\,R\,b,b\,R\,c$, and $a\,\not R\,c$; and in order to show that $R$ is not total, you must find elements $a,b\in A$ such that $a\,\not R\,b$ and $b\,\not R\,a$. The crucial point is that in each case you must find elements of $A$ that actually have certain properties. If such elements don’t exist, then $R$ is reflexive (or antisymmetric, transitive, or total).

In your question $A$ is $\varnothing$, the empty set. Is it possible to find elements of $\varnothing$ that have certain properties? Is it possible to find elements of $\varnothing$ at all?

Brian M. Scott
  • 616,228
  • 1
    So, $\varnothing$ is indeed a total order on $\varnothing$.Thanks for the reasoning. –  Jul 12 '12 at 05:34
  • 6
    @AmNatPhil: You’re welcome. This is a good example of what’s meant when people say that something is vacuously true: it’s true simply because there’s nothing there to make it false. – Brian M. Scott Jul 12 '12 at 05:36
  • Does this mean once one wants to enumerate the number of (multi)chain in an interval, one always need "plus 1" considering the empty chain? However, this is contradict to the fact that in incidence algebra the identity element equals zero when the interval is non-trivial i.e. not a single element set. – Peter Whang Sep 02 '23 at 05:02
  • @PeterWhang: In incidence algebra you’re typically looking at the number of (multi)chains from $a$ to $b$ for some $a\le b$, so you’re never going to be counting the empty chain. – Brian M. Scott Sep 04 '23 at 05:53
4

The above answers confirm that $\emptyset$ with the empty ordering is in fact a linear order. I would just like to add that it is in fact also well-ordered under this ordering since it is linearly ordered and every non-empty subset of $\emptyset$ has a least element. This is true vacuously since there are no nonempty subsets at all.

Francis Adams
  • 2,808
2

If a set is totally ordered, then it is already partially ordered.

It is meaningless to say whether any set $A$ is partially or totally ordered. You need to give a set $A$ and a partial ordering $\prec$ on $A$ for this question to make sense.

However, there does exists a total ordering on the $\emptyset$. It is the empty relation. More concretely, you can also think of the $\in$ relation restricted to $\emptyset$.

William
  • 19,935