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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 $\sup {B}=\inf {A}$.

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    wait ,why edited my question ? – 2017-01-16
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    Is this an exercise in a book or you just come up with it? – 2017-01-16
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    @juniven exercise from Understanding Analysis by Stephen Abbot – 2017-01-16
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    What page please? – 2017-01-16
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    it is in page 17 – 2017-01-16
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    There I see Exercise 1.3.3 (a) Thank you. – 2017-01-16
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    I think it doesn't happen $N$ is in $B$ necessarily. – 2017-01-16
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    I see fine this answer: http://math.stackexchange.com/questions/1461732/proof-that-inf-a-sup-b – 2017-01-16
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    @joseabp91 Thank ^ ^ – 2017-01-16
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    @Fourier I was about to answer your question, but as joseabp91 pointed out, I think you need to look it out. In case you can't follow the proof there, I am willing to post my answer. Just chat with me. – 2017-01-16
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    You question was edited as your typesetting was a bit unusual. Moreover, the body of the post should be self-contained and not depend on the content of the title (which in turn was/is a tad on the too complicated side for a title). – 2017-01-16

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$\inf A$ 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 $\inf A=\max B=\sup B$.