- #1

Eclair_de_XII

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- 91

## Homework Statement

"Let ##S## be a non-empty set of real numbers bounded from above, and let ##x=sup(S)##. Prove that ##x \in S## or ##x## is an accumulation point of S."

## Homework Equations

__Neighborhood:__"A subset ##O## of ##ℝ## is a neighborhood of ##x\in ℝ## iff there exists an ##\epsilon>0## such that ##(x-\epsilon,x+\epsilon) \subseteq O##."

__Accumulation point:__"A number ##x \in ℝ## is an accumulation of a set ##S## if any neighborhood of ##x## contains an infinite number of elements in ##S##."

## The Attempt at a Solution

"Since ##S## is a non-empty set of real numbers bounded from above, there exists a least-upper bound of ##S##, by the least-upper-bound property of ##ℝ##. Let ##x=sup(S)##. Then we wish to prove that ##x\in S## or ##|(x-\epsilon,x+\epsilon) \cap S|=\infty##.

First, we note that the interval, ##(x,x+\epsilon)\cap S## is empty, because there exist no ##a\in S## such that ##a > x=sup(S)##. Thus, we need to consider the cardinality of the set ##(x-\epsilon,x]\cap S##.

We start by observing that ##x-\epsilon## is not an upper bound for ##S##, since ##x-\epsilon<sup(S)##. Hence, there exists an ##a \in S## such that ##a>x-\epsilon##. Consequently, since ##x## is an upper bound of ##S##, ##x \geq a > x-\epsilon##. Thus, the interval ##(x-\epsilon,x]\cap S## is non-empty.

If ##S## was a finite set of real numbers, then it has a most element. And that most element was ##x=sup(S)=max(S)\in S##."

Then I wanted to claim that if ##S## was infinite, then ##|(x-\epsilon,x] \cap S|=\infty##. But then I thought of a set where ##S## could be infinite, and yet the latter statement about its intersection with ##(x-\epsilon,x]## was untrue.

Suppose ##S=\{x-\frac{\epsilon}{2}\}\cup (-\infty,x-\epsilon]##. Then ##|(x-\epsilon,x] \cap S|=1##, but ##S## would still be an infinite set, and ##x## still couldn't be an accumulation point of ##S##. Additionally, I'm wondering why ##x## can't be in ##S## and be an accumulation point for ##S## at the same time. For example, the interval ##[0,1]## contains the element ##1##, which is also its supremum, and an accumulation point for the interval.

Can anyone tell me any flaws or leaps in logic I may have made? Thank you...