# very simple statement but how to find inf, sup

i am having trouble with this this. i need to simplify this statement and find, if any, the minimum, maximum, infimum, and supremum of it.

the statement is this:

$A:= (]1,2[ \cup ]2,3]) \cup \{2\}$

i simplified it so much as i could. $A:=\{x \in A | 1<x \leq 3 \}$

now my attempts: minimum is 2 and maximum is 3 and i dont know how to find and define the infimum and supremum.

$Infimum=$ the maximum of all lower bounds
$Supremum=$ the minimum of all upper bounds

thanks for help

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It is clear that $3$ is an upper bound of $A$, for every number in $A$ is $\le 3$. And we cannot get away with a cheaper upper bound than $3$, since $3$ is in our set.

It is also clear that $1$ is a lower bound of $A$. No number $\gt 1$ is a lower bound, for the interval $]1,2[$ contains numbers arbitrarily close to $1$.

(If you want to be very very formal, suppose to the contrary that $1+\epsilon$ is a lower bound of $A$, where $\epsilon$ is positive. We can assume that $\epsilon\lt 1$. The interval $]1,2[$ contains the number $1+\frac{\epsilon}{2}$, which is clearly $\lt 1+\epsilon$, contradicting the assumption that $1+ \epsilon$ is a lower bound.)

Since $1$ is a lower bound of $A$, and no number $\gt 1$ is a lower bound, it follows that $1=\inf A$.

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thanks man, nice! :D –  doniyor Nov 29 '12 at 11:30

You're wrong. A non-empty bounded set of real always has an infimum and a supremum, but not always minimum and maximum (depending on your local customs, you may also accept $-\infty$ as the infimum as a set that is not bounded from below and $+\infty$ as infimum od the empty set, similar for supremum). If it has minimum/maximum, they are the same as infimum/supremum.

Of your set, $2$ is definitely not the minimum as $\frac32\in A$ and $\frac32<2$. The supremum is the least upper bound, the infimum the biggest lower bound. As a warm-up: Can you give a number that is and one that is not an upper bound? Same with lower bound? Can you refine this? Where is the border between being and not being an upper/lower bound? Do the border cases belong to the set $A$ or not?

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thanks Hagen. your warm-up tasks: the number which isnot upper bound is 4 and which is upper bound is 3, right? –  doniyor Nov 30 '12 at 20:16
for lower bound: lower bound is $1<x \leq 2$ am i right? so there is no lower bound? –  doniyor Nov 30 '12 at 20:19