prove that if the poset L has a least element, then that element is unique.

Question

I am trying to work through this question to study for Discrete Structures 2 but I am struggling as I am new to the concept of least upper bound and greatest lower bound if someone could please prove this and elaborate by explaining the proof.

Answer 1

The least element (if it exists) is precisely the supremum of $\emptyset$: indeed, the supremum of $\emptyset$ (if it exists) is an element $x$ of the poset such that:

• $x$ is greater than or equal to every member of $\emptyset$ - this is true for all $x$!
• whenever $y$ is greater than or equal to every member of $\emptyset$ (that is, whenever $y$ is in the poset), we have $x \leq y$.

More simply put, it is an element $x$ of the poset such that for every element $y$, we have $x \leq y$. But that's exactly what it means for $x$ to be the least element of the poset.

So it's enough to prove the more general (and more useful) fact that the supremum of a subset $A$ of the poset is unique, if it exists.

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To do this, let $x$ and $y$ both be suprema of $A$; we aim to show $x=y$. What do we know?

• $x$ and $y$ are both upper bounds for $A$;
• whenever $z$ is an upper bound for $A$, we have $x \leq z$ (since $x$ is a least upper bound for $A$)
• whenever $z$ is an upper bound for $A$, we have $y \leq z$ (since $y$ is a least upper bound for $A$)

Given these facts, there's basically only two deductions we can make:

• we have $x \leq y$ (deduced from the second bullet point, letting $z=y$)
• we have $y \leq x$ (deduced from the third bullet point, letting $z=x$)

By antisymmetry (since we're in a poset), then $x=y$.