S = $\{a \in \Bbb{Z} | a < 2+1/2$}
Given that a is in the set of all integers ($\mathbb{Z}$), would the supremum of this set be equal to $(2 + 1/2)$? If so, given an upper bound of $(2 + 1/2)$, how would I prove that for an $\epsilon < (2 + 1/2)$ this couldn't possibly be an upper bound and hence $\sup(S) = (2 + 1/2)$?
On
Just because you can write something doesn't mean that is the must efficient way to write it.
To prove that for any $r < \sup S$ is not an upper bound you prove that for any $r < \sup S$ there must be an element of $s \in S$ so that $r < s \le \sup S$.
If of $\sup S = 2\frac 12$ then that means for any $r < 2\frac 12$ there must be an integer $n$ so that $r < n \le 2\frac 12$.
Is that true. What if $r = 2.49$. Is there an integer $n$ so that $2.49 < n \le 2\frac 12$?
Obviously not. There aren't any integers between $2$ and $3$ so if $n$ is an integer and $n < 2\frac 12 < 3$ then $n \le 2$. So there are no integers and therefore no elements of $S$ that are between $2$ and $2\frac 12$. So we can't have any number between $2$ and $2\frac 12$ actually be the $\sup S$
Can you finish this?
Here's food for thought. Let $S=\{$ all even primes less than $2000\}$. So how does $\sup S$ compare to $2000$. If we notice that $S = \{2\}$ what does that tell us.
Claim 1: $2$ is an upper bound of $S$. Pf: If $s \in S$ then $s\in \mathbb Z$ and $s < 2\frac 12$. So $s < 3$. ANd $s\in \mathbb Z$ so it is not true that $2 < s <3$. And as $s < 3$ and it is not true that $s < $ and $s > 2$ it must not be true that $s > 2$. So $s \le 2$. So $2$ is an upper bound of $S$.
Claim 2: If $r < 2$ then $r$ is not an upper bound. Pf: $2\in \mathbb Z$. And $2< 2\frac 12$. So $2 \in S$. So $r < 2\in \mathbb S$. So $r$ is not an upper bound.....
The set $S$ has $2+\frac12$ as an upper bound, but $2+\frac12$ is not the supremum of $S$. For instance, 2 is an upper bound of $S$ which is strictly smaller than $2+\frac12$.