To prove $\sup B = \inf A$ Let $A$ be bounded below and $B = \{b \in \mathbb{R} : b$ is a lower bound for $A\}$. 
My Work 
Now to understand I assumed some values, like set $A = (1,4)$ and $\inf A=1$ and so set $B = (-\infty , 1]$.
To prove $ \sup B=1$
I assume that $\sup B$ is bigger than $1$, say $\sup B= 1.5$ so we can find some $b \in (1, 1.5)$. But then set $B$ is set of lower bound for set $A$ which is $(1,4)$. So there is contradiction as the lower bound is bigger than element of set
Now let us say that $\sup B < 1$, say $-0.5$. So we will not have any $b$ between $(-0.5 , 1)$. But then $ \inf A = 1$ is contradicted because lower bound of $A$ is in set $B$.
My Doubt : Is this the correct way to do this ? Of course we require formal proofs, but will formal proof go along the same lines ?
Thanks
 A: You have exactly the right idea, but need to be a little more flexible because your sets might not be intervals. A good way to prove an equality like this is to show it by combining two inequalities; i.e. by proving separately that $\sup B \leq \inf A$ and $\sup B \geq \inf A$.
I'll leave the first of these inequalities to you, because both parts are very similar. For the second, note that since $\inf A$ is the greatest lower bound of $A$ we just need to show that $\sup B$ is at least as great as any lower bound $c$ of $A$.  But now by the definition of $B$ we have that $c \in B$, and because $\sup B$ is at least as great as any element of $B$ we have $\sup B \geq c$, as desired.
A: Notice that $B$ is bounded from above by any $a \in A$ so there exists $s = \sup B$. This means that:


*

*$s \ge b, \forall b \in B$

*$\forall \varepsilon > 0\, \exists b \in A$ such that $s-\varepsilon < b$
We claim that $s = \inf A$. To show this, we need:


*$s \le a, \forall a \in A$

*$\forall \varepsilon > 0\, \exists a \in A$ such that $s+\varepsilon > a$
For $3.$, assume that $s > a$ for some $a \in A$. Then for $\varepsilon = s- a> 0$ we get from $2.$ in the definition of supremum that there exists $b \in B$ such that $$a = s-\varepsilon < b$$
which is a contradiction because $b$ is supposed to be a lower bound for $A$.
For $4.$, take $\varepsilon > 0$ and notice that $s + \varepsilon > s$ so $s+\varepsilon \notin B$ (since $s+\varepsilon \in B$ would contradict $1.$ from the definition of supremum). Hence $s+\varepsilon$ is not a lower bound for $A$ so there exists $a \in A$ such that $s+\varepsilon > a$.
We conclude $s = \inf A$.
