# Suppose $R$ is a partial order on set $A$, and $B \subseteq A$. Prove theorems about the uniqueness of the smallest element

by Nelver   Last Updated October 18, 2019 09:20 AM

Suppose $$R$$ is a partial order on set $$A$$, and $$B \subseteq A$$. Prove

1. If B has a smallest element, then this smallest element is unique. Thus, we can speak of the smallest element of B rather than a smallest element.

2. Suppose b is the smallest element of B. Then b is also a minimal element of B, and it is the only minimal element.

My attempt:

1.

Suppose $$x \in B$$ and $$x$$ is smallest element. Suppose $$x$$ is not unique. Then there is one more smallest element, call it $$y$$, where $$x≠y$$.

Since $$x$$ is smallest element, $$(x,y) \in R$$

And since $$y$$ is smallest element, $$(y,x) \in R$$

Since $$x ≠ y$$, and $$(x,y), (y,x)$$ both in $$R$$, we conclude that $$R$$ is symmetric. But we know that $$R$$ is partial order, which implies it must antisymmetric. Hence we have a contradiction, which means $$x$$ must be unique.

2.

Suppose $$b$$ is the smallest element.

Suppose $$b$$ is not a minimal element. It means that there is some $$x$$, where $$x ≠ b$$ and $$(x,b) \in R$$. But since $$(b,x) \in R$$, it would imply that $$R$$ is symmetric, but we know it's not. Hence $$b$$ is a minimal element.

Suppose $$b$$ is not the only minimal element. Then there exists one more minimal element, say $$x$$, such that $$x ≠ b$$. It follows that neither $$(x,b) \in R$$ nor $$(b,x) \in R$$. But this is a contradiction, since we know that $$b$$ is a smallest element, and by definition of smallest element we have

$$\forall x \in B (bRx)$$

Which implies that $$(b,x) \in R$$. Hence $$b$$ is the only minimal element. $$\Box$$

Is it correct?

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