If R is a total order and b is a minimal element of B, then b is the smallest element of B.

Question

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

Prove that, if $R$ is a total order and $b$ is a minimal element of $B$, then $b$ is the smallest element of $B$.

I attempted an indirect proof of the above but I am not sure if it's correct, as the provided solution is A LOT longer and more complicated...

My attempt: The goal is $\forall x\in B(bRx)$, so let's do an indirect proof and assume $\sim bRx$.

Since this is a total order, for all $x$ and $y$ in $A$, either $xRy$ or $yRx$. But we have $\sim bRx$, so we get $xRb$.

And since $b$ is a minimal element, that means that nothing in $B$ is such that it is both $xRb$ and is not identical to $b$. But this just means $\forall x\in B(xRb \to x=b)$. So $x$ is identical to $b$.

But if that's the case, the assumption $\sim bRx$ is the same as $\sim xRb$. This contradicts with two things:

1) All relations in a total order are reflexive, but if $x=b$, surely this also means $\sim bRb$.

2) We derived $xRb$ and this contradicts with $\sim xRb$.

Hence $\sim bRx$ cannot be true and $bRx$ must be true.

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So this is my proof; but the provided solution seems a lot more complicated and it proceeded by proving case by case (namely assuming $x=b$ and $\sim x=b$). So I am not sure if my proof is right. Could anyone please check my proof and see if it's ok please?

Thank you!

Answer 1

It's easy. Let $\leq$ denote the total order. Let $b$ be any minimal element of $B$, and $u$ another element of $B$ different from $b$. Then $u\leq b$ or $b\leq u$. If $u\leq b$ then you get $u=b$, a contradiction. Thus, $b\leq u$, so $b$ is minimum.

Answer 2

Let $x$ be a arbitrary element of $B$ and let $b$ be a minimal element of $B$. The goal then is to show that $(b,x)\in R$. By the definition of total order, we must have either $(b,x)\in R$ or $(x,b)\in R$. If $(b,x)\in R$ we are done, so let's assume that $(x,b)\in R$. Because $b$ is a minimal element of $B$ and $(x,b)\in R$ we can conclude that $x=b$. But we know that $(x,x)\in R$ since $R$ is reflexive, so substituting $b$ for $x$ it follows that $(b,x)\in R$. Since these cases are exhaustive, and because $x$ was arbitrary, it follows that $\forall x\in B(b,x)\in R$. Thus, $b$ is the $R$-smallest element of $B$.