Prove that if $b\in B$ is the infimum of $B,$ then $b$ is the minimum of $B$.

Question

Let $(A,\le)$ be an ordered set and $B\subset A$. Prove that if $b\in B$ is the infimum of $B$ then $b$ is the minimum of $B$.

Proof

I want to show that $\forall x\in B, b\le x.$

Suppose $b$ is the infimum of $B$.

$\implies$ $b$ is the maximum of all the lower bounds of $B$.

$\implies \forall y\in A: y\le b,$ where each $y$ is a lower bound of $B$.

From here, how do I relate $b\le x$ with $y\le b$?

What can I do now to continue the proof?

Answer 1

I think you're overcomplicating things. Since $b$ is an infimum of $B$, it is a lower bound of $B$. What it means to be a lower bound of $B$ is precisely that $b \le x$ for all $x \in B$. Since $b \in B$, you have your desired result!

Answer 2

Suppose on contrary that $b$ is not minimum. Then, by definition there is an $x\in B$ such that $x<b$. But, for all $y\in B$ we have that $b\leq y$, since $b$ is the infimum of $B$. In particular, $b\leq x$. But this contradicts that $x<b$. Thus, $b$ is minimum.