$Sup(E) – \epsilon \lt a \leq Sup(E)$ where $\epsilon$ is any number greater than zero and $a \in E$
If I understand this correctly it is saying that there is always an element in the set that falls between the supremum and the supremum minus some positive number no matter how big or small that number is. If it is very small it equals the suprema itself. If it is sufficiently large then it could be another element a of the set. It is proved by contra-positive if I am understanding this correctly an $\epsilon$ is chosen so that no element in $A$ falls between the two. This suggests then that the element $a$ must be less than or equal to $Sup(E) – \epsilon_0$ then we use this fact to show $sup(E)$ is actually not the supremum and with some algebra arrive to a contradiction with $\epsilon$
I tried to do this with an exercise in the book. Show that if $E$ is a non-empty bounded subset of $\mathbb{Z}$, then $inf(E)$ exists and it belongs to $E$ So i tried to use approximation property to show that inf(E) belongs to $E$ i.e $\mathbb{Z}$
Choose an $x_0$ such that
let $i=infimum(E)$ then we have:
Since it is an infimum it should look like:
$i \leq x_0 \lt i + 1$ Well if $x_0 = i$ then we are done
$i \lt x_0 \lt i + 1$
$i \lt x_1 \lt x_0$ Through some manipulation we get:
$i – x_0 \lt x_1 – x_0 \lt 0$ and $-x_0 \gt -i-1$
but something is off here
Best Answer
Let $E$ be a non-empty subset of $\Bbb Z$ with infimum $i$. Then by the infimum approximation property with $\varepsilon=1$, there exists $a \in E$ such that
$$ i \leq a < i + 1 $$
Note that $a$ is unique, since the interval $[i, i+1)$ has length $1$ (with only one endpoint) so it must contain exactly one integer. We want to show $i=a$. Suppose for a contradiction $i<a$. Then, we can choose $\varepsilon < a-i$ so that
$$ i < i + \varepsilon < a < i + 1 $$
which implies $a \notin [i, i+\varepsilon)$. But since $a$ is the only integer in $[i,i+1)$, this gives
$$ [i, i+\epsilon) \cap \Bbb Z = \emptyset $$
which violates the approximation property.