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.

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

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 $$\mathbb{Z}\cup\{x\}$$.