Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

Give an example of a set with partial order, that has a single minimal element but no first element.

The example that was given is $\mathbb Z \cup \{2.5\}$, with relation $<$. I can't see why the totality property doesn't apply here, and why is this a valid example...

Thanks for any assistance!

share|cite|improve this question
I think they might mean for $2.5$ not to be related to any of the integers (since then it becomes a valid example). – Tobias Kildetoft Jun 20 '14 at 20:46
up vote 1 down vote accepted

This is the canonical example, but it is a poor choice of additional element.

Since $\Bbb Z\cup\{2.5\}$ is a subset of $\Bbb Q$ which has a natural $<$ defined, it means that we can re-interpret the usual $<$ as the one induced from $\Bbb Q$. In that case, yes, this is a totally ordered set.

So let us take, instead, something which is not part of the real, or rational numbers. Let's take $i$ to be the imaginary unit of $\Bbb C$. Now there's no natural way to interpret $0<i$ or $i<3$, so it is less dangerous to confuse these two.

Consider now $<$ to be the usual order on $\Bbb Z$. It is irreflexive and transitive. I claim that the very same order, without additional pairs, is also irreflexive and transitive on $\Bbb Z\cup\{i\}$.

  1. It is irreflexive, because certainly it is true that for every $k\in\Bbb Z$, the pair $(k,k)\notin <$; but since $<\subseteq\Bbb{Z\times Z}$, it also follows that $(i,i)\notin <$.

  2. It is transitive because if $x<y$ and $y<z$, then certainly neither $x,y$ or $z$ can be $i$, since it doesn't appear in any of the ordered pairs. So transitive follows since we know that $<$ is transitive on the integers.

Therefore $(\Bbb Z\cup\{i\},<)$ is a partial order without a minimum or maximum, that has a single minimal element and a single maximal element. What is that minimal/maximal element? $i$, of course. There is no ordered pair of the form $(i,x)$ nor an ordered pair of the form $(x,i)$. So it has to be minimal and maximal.

My suspicion is that the original example was intended to be like this. If $<$ was explicitly said to be the same $<$ on the integers, then it works just fine, but without pointing that out, this becomes confusing and ambiguous.

share|cite|improve this answer

Consider the following diagram, oriented left to right

  ...--o----o----o----o---- ...

The $0$ is minimal, but there is no least element. This is in the spirit of the answer in your book, but maybe the book is unclear about where they are putting $2.5$. It is also unclear what they mean by "first" element.

share|cite|improve this answer

Your Answer


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

Not the answer you're looking for? Browse other questions tagged or ask your own question.