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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!

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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 at 20:46

2 Answers 2

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.

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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.

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