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What can be the example of an ordering with one minimal element, but without the smallest element? Can this be an example: Proper subset relation defined on the set {{1}, {1,2}}.

Kambar
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2 Answers2

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If $P$ is a finite partial order, and $p\in P$, there is always a minimal element $q\le p$. Thus, if a finite partial order has exactly one minimal element, that element is automatically the smallest element of the partial order. Thus, if you want a partial order with exactly one minimal element and no smallest element, you’ll need an infinite partial order. Here’s the Hasse diagram of one that works:

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The element $\circ$ in the upper left is the unique minimal element, and there is clearly no smallest element. A slightly more interesting example:

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Brian M. Scott
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  • I don't understand your diagram. For example. in the first diagram, isn't the upper most start a minimal element? – Sunghee Yun Apr 12 '20 at 19:00
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    @SungheeYun: None of the stars in either diagram is a minimal element: each has other elements strictly below it. The only element in either diagram that has no other elements strictly below it is the circle. – Brian M. Scott Apr 12 '20 at 19:08
  • Got you! This is indeed great illustration of the examples. Thanks, Brian! – Sunghee Yun Apr 12 '20 at 19:19
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    @SungheeYun: You’re welcome! – Brian M. Scott Apr 12 '20 at 19:32
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    just upvoted your answer :) – Sunghee Yun Apr 12 '20 at 19:33
  • this answer is super useful, but a warning to beginners like myself: this discussion assumes that a p.o. is by definition reflexive. I am reading a book now in which a p.o. is merely transitive and anti-symmetric. With this book's definition, I think the first claim would have to read "If $P$ is a finite p.o. and $p \in P$, there is always a minimal element $q$". Aha, but despite this, your second claim holds even under my definition. Ok i feel better now. Even with my def., there can be no finite-sized example. – 311411 Jul 05 '20 at 17:53
  • @311411: What your book calls a partial order is in my experience more often called a strict partial order; the term partial order most often refers to my kind. – Brian M. Scott Jul 05 '20 at 17:55
  • @BrianM.Scott As an example, given $\mathbb{N} = {0,1,2,...}$ and $Y=\mathcal{P}(\mathbb{N})$ with $\subset$, does $E = {\mathbb{N}, \mathbb{N}\setminus {0}, \mathbb{N}\setminus {1}, \mathbb{N}\setminus {1,2}, \mathbb{N}\setminus {1,2,3}...}$ work? :) – no lemon no melon Sep 21 '21 at 20:55
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    @nolemonnomelon: Yes: its Hasse diagram looks like my second eample. – Brian M. Scott Sep 22 '21 at 21:32
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You will need a partially, not totally ordered, set to get an example. You may take a (silly) example like: $$\{(0,y)\in {\Bbb R}^2: 0<y<1\} \cup \{(1,y)\in {\Bbb R}^2: 0\leq y<1\}$$ with the partial ordering $(x,y) \prec (x',y')$ iff $x=x'$ and $y\leq y'$. Then $(1,0)$ is a minimal element but not a smallest element (because it is not related to the elements with $x=0$) and there are no other minimal elements.

H. H. Rugh
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