This question comes from Serge Lang's Undergraduate Analysis section II-1 problem 11.
Let $S$ be a bounded set of real numbers. Let $A$ be the set of its points of accumulation. That is, $A$ consists of all numbers $a \in \mathbb{R}$ such that $a$ is a point of accumulation of an infinite subset of $S$. Then $A$ is bounded. Assume that $A$ is not empty. Let $b$ be its least upper bound.
In part (a) of this question, I proved that $b \in A$ and thus $b$ is a point of accumulation denoted $limsup(S)$.
Part (b) states:Let $c$ be a real number. Prove that $c$ is the limit superior of $S$ if and only if $c$ satisfies the following property. For every $\epsilon$ there exists only a finite number of elements $x \in S$ such that $x > c + \epsilon$, and there exits infinitely many elements $x$ of $S$ such that $x > c - \epsilon$.
I am trying to prove the forward direction of this claim, specifically, I want to prove that that assuming $limsup(S) = c$ that there exist only finitely many $x \in S$ such that $x > c + \epsilon$.
My initial thought was that if we suppose the contrary - that there are in fact infinitely many points greater than $c + \epsilon$, then that would make $c + \epsilon$ a point of accumulation, contradicting that $limsup(S) = c$. However, I don't think there's any way to guarantee that those points will be clustered in a way that we can assert that $c + \epsilon$must be a point of accumulation.
So instead I tried this approach:Since $S$ is bounded, there must be some bound $M$ such that $M \geq x$$\forall x \in S$. First, if $M = c$ this trivially proves there must be finitely many elements greater than $c + \epsilon$ (because there are zero). If $M < c$ this leads to a contradiction, because since $c$ was shown to be an accumulation point, there must be infinitely many points of $S$ greater than $M$ for $c$ to remain an accumulation point.
Thus assume $M > c$. Let us assume for purpose of contradiction that there are in fact infinitely many points $x \in S$ such that $x > c + \epsilon$. Then the interval $(c + \epsilon, M)$ contains infinitely many points, where we can select $\epsilon$ such that $c + \epsilon$ can get arbitrarily close to $M$. Thus, any open neighborhood around $M$ with $c + \epsilon$ as its left endpoint will have infinitely many points of $S$ enclosed.
So, $M$ must be a point of accumulation, contradicting that $limsup(S) = c$, thus there may only be finitely many points $x > c + \epsilon$.
I have one concern about this, though. For $M$ to be a point of accumulation, every open neighborhood around it must contain infinitely many points of $S$. If we chose the left endpoint of the open neighborhood to be less than or equal to $c$ itself, could we still assume there exist infinitely many points of $S$ in such an interval? In every previous case, we assumed there was a buffer of size $\epsilon$ between $c$ and $M$, but I think it would still work, because for some arbitrary $\epsilon$ in this new interval, we would still have the requisite infinitely many points. I'm not sure if this is the case though, and it feels shaky. Can anyone confirm if this is a valid assumption to make based on the preceding work?