# minimum of this simple set

i need again some help here. i am defining the minimum and max and inf and sup of this set

$A:=(]1,2[ \cup ]2,3]) \cup \{2\}$ which is equal to the interval $(1,3]$

i say, max is 3, and sup is also 3. and 1 is inf but what is minimum? there are many numbers with $1<x$ which can be minimum of $A$, right?

can someone help me to understand the intuitive notion of infimum,supremum, and minimum and maximum. i know that the number which is minimum and maximum must be contained in the set. and every bounded non-empty set has supremum and infimum, but not always minimum and maximum. so in my case, there are maximum, supremum and inf, but no minimum? am i right?

$A:=\left(\; ]1,2[ \;\;\cup\;\; ]2,3]\;\right) \cup \{2\}$ which is equal to the interval $(1,3]$

Conclusions: $\max(A) = \sup(A) = 3$, $\inf(A) = 1$.

You are correct in all your conclusions. The infimum is indeed $1$, but $1 \notin (1, 3]$.

If a minimum exists, it must be IN the set (half-open interval), and it must equal the infimum of the set. As you observed, there is no such minimum value in $A$, so $A$ has no minimum.

(Note, the same is true about the maximum of a set: IF a maximum value of a set exists, it must be IN the set, and it must equal the supremum of the set: in this case, $3$ is the supremum of $A$ and it is also the maximum of $A$ because $3 \in A$.

The infimum and the supremum of a set always exist, but they NEED NOT be IN the set.

• thanks a lot. this is the last time that i tried to understand this logic, because now i undertand :D – doniyor Dec 1 '12 at 15:09
• You're welcome, doniyer! – amWhy Dec 1 '12 at 15:14
• can you please read my assumption under Amr's answer. am i right? – doniyor Dec 1 '12 at 15:18
• Yes, you are right, doniyer. $B:= \bigcup_{z \in \mathbb{Z}} ]z,z+1[$ has a supremum, an infinimum, but no min or max. – amWhy Dec 1 '12 at 15:24
• okay, great, thanks :) – doniyor Dec 1 '12 at 15:28

Not every set has a minimum. Your $A$ is such a set.

• thanks, this sentence gave me the shot! :D great! – doniyor Dec 1 '12 at 15:08

Suppose that a minimum exists.

Let $m$ be a minimum of the set $(1,3]$. let $m'=1+(m-1)/2$.Since $1<m$, It follows that $m'\in(1,3]$ and $m'<m$ (a contradiction)

• great. can you test me with this statement: i say, $B:= \bigcup_{z \in \mathbb{Z}} ]z,z+1[$ has only inf and sup but no max and min, right? – doniyor Dec 1 '12 at 15:15
• $B$ is unbounded – Amr Dec 1 '12 at 15:23