Let $A$ be a set and $s$ its supremum. Given an $\epsilon>0$, can I assure that I can find and $a\in A$ that meets: $s-\epsilon<a<s$?

Question

Let $A$ be a set and $s$ it's supreme. Given an $\epsilon>0$, can I assure that I can find and $a\in A$ that meets: $s-\epsilon<a<s$ ?

I tried saying that $s>a\,\, \forall a\in A$, and also, seeing that we don't know if the set A is lower bounded if I fix and $a$ I'll always be able to pick and $\epsilon$ big enough to mate it true, but... that's enough?

I believe I need something more to demonstrate (or negate) it, any help?

Answer 1

No, actually. Let for example $A=[0,1]\cup\{2\}$. Then for $\epsilon=\tfrac12$, such element $a$ does not exist. If there is a sequence of elements in $A$ converging to $s$, then you could prove the statement you conjectured.

Answer 2

No.

Let $A=\{42\}$. Then $s=\sup A = 42$ and for $\epsilon:=1$, there is no $a\in A$ with $s-\epsilon<a<s$.