Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
Can you find two incomparable elements in the empty set? – azarel Jul 12 '12 at 5:18
So, as there are no elements in the empty set, there are no incomparable sets. So, it is totally ordered. Is that right? – user33109 Jul 12 '12 at 5:25
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 5:28
up vote 14 down vote accepted

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 $\,\not R\,b$ 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?

share|cite|improve this answer
So, $\varnothing$ is indeed a total order on $\varnothing$.Thanks for the reasoning. – user33109 Jul 12 '12 at 5:34
@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 5:36
I am not convinced that the definition of antisymmetry is quoted correctly. If not mistaken it is : if $aRb$ and $bRa$ then $a=b$. The one quoted is rather the definition of symmetry – Valentin Jul 15 '12 at 22:41
@Valentin: Thanks. You are of course correct; my brain seems to have gone walkabout and come up with an equivalence relation instead of a partial order. I’ve fixed it now. – Brian M. Scott Jul 15 '12 at 22:46

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 $\emptyset$.

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.