Real analysis proof 
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*(a) Let $X$ be a non-empty subset of $\mathbb R$.


(i) What does it mean to say that $b\in\mathbb R$ is an upperbound of $X$?
(ii) What does it mean to say that $b \in \mathbb R$ is the least upper bound of $X$?
(iii) Suppose $X$ contains a largest element,$x \in X$.
Show that $x$ is the least upper bound of $X$
I am confused as to how to define the upperbound! any ideas anyone!
 A: (i) An upper bound for $X\subseteq\mathbb{R}$ is a number $b\in\mathbb{R}$ so that $b\geq x$ for all $x\in X$.
Intuitively a number is an upper bound for $X$ if that number is larger than or equal to any of the numbers in $X$.
(ii) The least upper bound of $b^*$ of a set $X\subseteq\mathbb{R}$ is an upper bound such that $b^*\leq b$ for any upper bound $b$ of $X$.
The intuition here is that the least upper bound is the smallest possible upper bound. You can think of the least upper bound as an upper bound for $X$ that is also a lower bound for the set of upper bounds of $X$.
(iii) If $x^*=\max X$, then clearly $x^*\geq x$ for all $x$ in $X$. Therefore $x^*$ is an upper bound. Clearly, any upper bound $b$ should satisfy  $b\geq x^*$ (since $x^*\in X$), and therefore $x^*$ is the least upper bound.
In this case, if it turns out that $X$ contains a largest element then this largest element is also an upper bound. And not only that, it is also the smallest possible one.
A: Suppose $S$ is a nonempty subset of the real
numbers. An element $a \in \mathbb{ R}$ is termed an upper bound for $S$ if $x \leq  a$ for all $x  \in S$. 
An element $b \in \mathbb{ R}$ is a lower bound for S if $b \leq x$ for all $x \in S$.
. An element $a \in \mathbb{ R}$ is termed a least upper bound for $S$ if $a$ is an upper bound for $S$ and, for any upper bound say $a'$ of S,
$a \leq a'$. In other words, $a$ is the least possible element we can choose as an upper bound. The least upper bound of $S$ is an upper bound that is less than or equal to any other upper bound for $S$.
For a least upper bound to exist, the set must be bounded from above.
The following are equivalent for a nonempty subset $S$ of $\mathbb{R}$:
(a) $S$ has a maximum element – an element that is larger than every other element of S.
(b) The least upper bound of $S$ is in S.
Moreover, if these equivalent conditions hold, the maximum element equals the least upper bound.
Similarly, The following are equivalent
(a) $S$ has a minimum element – an element that is smaller than every other element of $S$.
(b) The greatest lower bound of $S$ is in $S$.
Moreover, if these equivalent conditions hold, the minimum element equals the greatest lower bound.
A finite set has a maximum element and a minimum element. Thus, any finite set contains its least
upper bound and greatest lower bound.
Following simple example may clear your doubt.
Consider a set $S = \{1, 2, 3 ,4\}$, we may take its upper bound any number greater than or equal to 4 say, $4.5, 10, 1000$, all these numbers are the upper bound of the set $S$ but least upper bound is only 4.
A: (i) $b$ is an upper bound of $X$ if for every $a\in X$, $a\le b$ holds.
(ii) Let $U$ denote the set of all upper bounds for $X$.  $b$ is a least upper bound of $X$ if $b$ is minimal in $U$.  Minimal means it is in $U$ and for all $c\in U$, $b\le c$.
(iii) Suppose $x\in X$ is largest.  Then for every $a\in X$, $a\le x$.  Hence $x\in U$, the set of upper bounds.  I leave for you the proof that $x$ is minimal in $U$.
A: About the upper bound, we can define the upper bound to be $ UpperBound(X)=[y\in \mathbb{R}: \forall x \in X, y\geq x] $
Therefore, we can define the least upper bound of X to be the minimum of the set UpperBound(X), like so:
$ sup(X)=min(UpperBound(X))=min(y\in \mathbb{R}: \forall x\in X, y\geq x)$
About the third question, let's assume that x is the largest element in X. This means that $ \forall a \in X, a \leq x$. Therefore, according to the definition of the Upper Bound, x is an upper bound for X. Let's show that x is the least upper bound.
Let's take $ y\in UpperBound(X)$. Let's show that $y\geq x $. Let's falsely assume that $y <x $. Let's note that $ y<x \in X$, and therefore, we found an element in X (x), which is greater than y, and therefore, y isn't an upper bound.
Therefore, x must be the least upper bound.
