# Uniqueness of minimum and maximum of set

i am stuck in this simple but foggy problem. i need to prove or show that the min and sup are unique if they exist.

$A$ is a nonempty set and $B$ is nonempty subset of $A$. i am trying to show that the $minB$ and $supB$ are unique once they exist.

my start is this:
$minB \Longleftrightarrow \forall_{x \in B} \exists_{m \in B} m \leq x$
if there is $m' \in B$ then $m \leq m' \leq x$

but i have bad feeling doing this, it seems to be not enough proved.

Can someone help me please with this? Thanks alot

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Your definition of minimum doesn't look quite right. First $\min B$ is not a sentence, so you can't write $\min B \iff \text{something}$. What you mean (I suppose) is the following:
$m$ is a minimum of $B$ iff (1) $m \in B$ and (2) $m \le x$ for all $x \in B$.
Now let's prove uniqueness: Suppose $m$ and $m'$ are minima of $B$. Then we have $m, m' \in B$ by (1) applied to $m$ and $m'$. Now by (2) for $m$ (with $x = m' \in B$) we have $m \le m'$, by (2) for $m'$ we have analogously $m' \le m$. As $\le$ is antisymmetric, it follows $m = m'$. That is: A mininum of $B$ is unique, if it exists.
Now try to do the same for a maximum of $B$.
m is maximum of B iff (1) $m \in B$ and (2) $x \leq m$ for all $x \in B$. assume, $m$ and $m'$ are maxima of B. then by (1) $m,m' \in B$, okay. by (2) for $m$ with $x=m' \in B$ we have $m' \leq m$ or for $m'$ with $x=m \in B$ we have $m \leq m'$. antisymmetry gives us that $m=m'$. – doniyor Nov 17 '12 at 11:54
Exactly ${}{}{}{}$ – martini Nov 17 '12 at 12:12