I want to prove that if $S$ is a nonempty set of $\mathbb{R}$ that is bounded from above but has no greatest element, then l.u.b. $S$ is a cluster point of S.
I know a direct proof might be simpler, but I want to try a proof by contradiction where you initially assume that l.u.b. $S$ is not a cluster point of $S$. Here's what I have so far:
Let $x =$ l.u.b.$S$. Assume that x is not a cluster point of S. By definition, for every $\epsilon > 0$, $B(x, \epsilon)$ contains only finitely many points. Since $x =$ l.u.b.$S$, there are no points of $S$ greater than $x$, and $x \notin S$ (trivial because no greatest element in $S$). Therefore, every point $p_i\in B(x, \epsilon)$ satisfies $p_i < x$. Notice each point $p_i$ is a distance $\epsilon_i$ away from $x$, where $\epsilon_i = d(x, p_i)> 0$:$${x-\epsilon_1, x-\epsilon_2, \ldots, x-\epsilon_n}$$Let $\alpha = \min(\epsilon_1, \ldots, \epsilon_n)$. $x - \alpha$ is strictly less than x. Moreover, each $p_i = x - \epsilon_i\le x-\alpha$ meaning $x - \alpha$ is an upper bound. This contradicts the definition of $x$ as a least upper bound of $S$. Thus, $x =$ l.u.b.$S$ is a cluster point of S.
The problem I have with this proof is that once you get to the later steps and narrow it down to a finite number of points $p_i$, it makes no sense for there NOT to be a greatest element. If anyone could help clarify this point I would greatly appreciate it. Thank you so much!