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I'm trying to prove the following,

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

Prove that if $b$ is the smallest element of $B$, then it is also the greatest lower bound of $B$.

My givens and goals so far are:

Givens

$b$ is the smallest element of $B$

Goals $$ (x\in A) \land (\forall y \in B (xRy) \land (\forall z \in A (\forall w \in B (zRw))\rightarrow xRz) $$ I don't know how to proceed from here. Could someone help me with this?

Thanks.

Siminore
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ChemDude
  • 421

2 Answers2

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Take $L(B)$ as the set of the Lower bounds of $B$, defined by: $$L(B)=\left\{x\in A\Big/\forall y\in B \ \ \ xRy\right\}$$

  • Givens $b$ is the smallest element of $b$ is equivalent to : $$\begin{align}b&\in B&\text{ and } &&\forall x\in B\ \ \ bRx \end{align}$$
  • Goal $b$ is the greatest element of lower bounds of $B$: $$\begin{align}b&\in L(B) &\text{ and }&& \forall t\in L(B)\ \ \ tRb \end{align}$$

Because $b$ is the smallest element of $B$ then $\forall x\in B\ \ \ bRx$ then $$b\in L(B)\tag1$$

Now given an element of $t\in L(B)$ we have $tRx$ for all $x\in B$ in particular $tRb$ because $b\in B$, so : $$\forall t\in L(B)\, \ \ tRb \tag2$$

$(1)$ and $(2)$ means exactly that $b$ is the greatest element of $L(B)$

Elaqqad
  • 13,725
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Since $b$ is the smallest element of $B$, we have that

  1. $b \in B$ and
  2. for every $x \in B$, $b \leq x$.

Hence $b$ is a lower bound for $B$. If $b'$ is another lower bound, then $b' \leq b$ (since $b \in B$). Hence $b = \sup B$.

Siminore
  • 35,136