For a set M with x ∈ M being the only minimal element on that set, is x the least element as well? Show your answer for finitie and infinite set M.

Question

I came across this question in a refrence book for discrete mathematics:

For a partially ordered set M with x ∈ M being the only minimal element on that set, is x the least element as well? Show your answer for finitie and infinite set M.

After long thought about this question, I believe x is indeed the least element in case of finite set M. In case of infinite set M, x isn't the least element. I am unable to find the appropriate approach to prove this though.

Any help would be greatly appreciated!

We have a partial order on the set, I guess. Do we assume anything more about this ordering? — Berci, Nov 22 '18 at 23:08
Yes, there is a partial order on the set. This is the only available piece of information. Forgot to mention, sorry! — Harold J. Fike, Nov 22 '18 at 23:16

score 0 · Accepted Answer · answered Nov 22 '18 at 23:51

In a finite set, every element is greater than or equal to some minimal element. Applying this to an arbitrary element $y$, shows that there is some minimal element below $y$. Since there is but one minimal element $x$, this proves that $x\le y$. Hence $x$ is least.

In an infinite set, this property need not hold. Consider the set of integers $\mathbb{Z}$, ordered in the usual way. It has neither minimal nor least elements. Let's also include one more element, $x$, that has the property that it incomparable to every integer. One can check that $x$ is minimal, but not least, in this partially ordered set $\mathbb{Z}\cup\{x\}$.

For a set M with x ∈ M being the only minimal element on that set, is x the least element as well? Show your answer for finitie and infinite set M.

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