Let $A$ be bounded below, and $B = \{b \in R : b$ is a lower bound for $A\}$. Show that $\sup B = \inf A$. This is   Exercise 1.3.3 (a) from "Understanding Analysis" by Stephen Abbot, , page 17:

Let $A$ be bounded below, and define $B = \{b \in R : b$ is a lower bound for $A\}$. Show that $\sup B = \inf  A$.

Please check my proof:
Suppose $A$ be bound below, it exist number $N\leq n$ for every $n\in \mathbb R$ in the set and $N$ is $\inf{A}$. Then $B=\{b \in \mathbb R:b$ is a lower bound for $A\}$. It exist number $b$ in the set, since it contains only lower bound of $A$ then it has number $b\leq M$ for every $b$ and $M$ is $\sup$ of $B$. But $b$ is lower bound of $A$ and $A$ has $N$ as $\sup$, $N$ is in $B$. $N$ is automatically is sup of $b$ therefore $M=N$ or $\sup {B}=\inf {A}$.
 A: $\inf A$ is defined as the biggest lower bound of $A$. So it is in particular a lower bound of $A$. But that means it is an element of $B$, which is defined as the set of all lower bounds of $A$. Thus $\inf A\in B$. But it is also the biggest lower bound of $A$, so for every $b\in B$ follows $b\leq\inf A$. This gives us $\inf A=\max B$. But if a set has a maximum, this maximum is also its supremum. Thus $\inf A=\max B=\sup B$.
A: Lemma 1
If $Y$ is a subset of partial ordered set $X$ that has maximum/minium then it has supremum/infimum too and it is equal to maximum/infimum.
Lemma 2
If $X$ is a set of real numbers that has an upper/lower bound then it has supremum/infimum too.
Theorem 3
If $X$ is a set of real numbers that has an upper/lower bound then the supremum/infimum of $X$ is the infimum/supremum of the set of the upper/lower bound of $X$.
Proof. If $X$ is a set of real numberst that has an upper/lower bound then by previous lemma it has supremum/infimum too so that we show that the supremum/infimum is the infimum/supremum of the upper/lower bounds. So if $\upsilon:=\sup X$ then it is the minimum of the upper bound of $X$ and if $\lambda:=\inf X$ then it is the maximum of the lower bound so that in any case the theorem follows by first lemma.
