Prove $a \leq b$, $\mbox{GLB}(\{a,b\}) = a$ , $\mbox{LUB}(\{a, b\}) = b$ are equivalent

Question

It seems really obvious to be as a set containing $a$ and $b$ have GLB as $a$ and LUB as $b$. But how can it be proved mathematically?

Answer 1

Let $S=\{a,b\}$. Let $a\ne b$, otherwise, it is easy to verify that they are same.

Claim 1: $a< b$ is equivalent to $GLB(S)=a$

Proof: Let $a< b$ but $GLB(S)=b$. Then, $b\le a$ since $a\in S$, which is a contradiction.

Claim 2: $GLB(S)=a$ is equivalent to $LUB(S)=b$

Proof: Assuming $a,b\in \mathbb{R}$, by completeness of $\mathbb{R}$, $S$ has LUB and GLB. Since $a\ne b$, $a=LUB(S)\implies b=GLB(S)$.

Answer 2

If $a\le b,$ then clearly $a$ is a lower bound of $\{a,b\},$ and if $c$ is a lower bound of $\{a,b\},$ then in particular $c\le a,$ so what can we conclude?

If $a$ is the greatest lower bound of $\{a,b\},$ then $a\le b,$ and clearly $b\le b,$ so $b$ is an upper bound of $\{a,b\}.$ Furthermore, if $c$ is an upper bound of $\{a,b\},$ then in particular, $b\le c,$ so what can we conclude?

Finally, if $b$ is the least upper bound of $\{a,b\},$ then what can we conclude?