Ordering with one minimal element, but without the smallest element

Question

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}}.

Answer 1

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.

Answer 2

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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