Confusion about whether a supremum of a set can be a limit point if it belongs to the set.

Question

Let $A\subseteq \mathbb{R}$ be a bounded above set of real numbers and $l=SupA$ then which of the following is true?

1. $l$ is a limit point of A.
2. $l\in A$
3. $l\notin A$
4. Either $l\in A$ or $l$ is the limit point of $A$.

Options (1), (2) and (3) can be easily discarded by taking a finite set, an open set and a closed set. For (4) when $l\notin A$, $l\in (l-\epsilon, l+\epsilon)$. Now since $l$ is the least upper bound of $A$ therefore for any $x\in A$, $x>l-\epsilon$ and $x<l+\epsilon$ $\implies$ $x\in(l-\epsilon,l+\epsilon)$ $\implies$ $l$ is a limit point of $A$. Now my doubt starts where $l$ does not belong to $A$, however we can prove it by taking a finite set as a counter example and there are also many infinite sets which can be taken here as a counter example, but is it possible that $l\in A$ and $l$ is also the limit point of $A$. How can we prove it in general? Also we can always find some $x\in A$ in an infinite set (I am looking for an example) which is in the open interval $(l-\epsilon,l)$ and hence in $(l-\epsilon, l+ \epsilon)$. So I am just looking for a general proof here. Please help.

Answer 1

At first we should add an extra condition: $A\neq\varnothing$.

It is immediate that $4$ is true.

For every $\epsilon>0$ point $l-\epsilon$ is not an upperbound of $A$ simply because $l$ is by definition the least upper bound of $A$.

So for every $\epsilon>0$ we can find an element $a\in A$ that satisfies $a>l-\epsilon$.

Does this mean that $l$ is a limit point of $A$?...

Yes if $l\notin A$ (so that $a\neq l$), but not necessarily if $l\in A$.

Examples:

• $A=(0,1)$ so that $l=1\notin A$ and consequently $l$ is a limit point of $A$.
• $A=(0,1]$ where $l=1\in A$ is a limit point of $A$.
• $A=(0,1)\cup\{2\}$ where $l=2\in A$ is not a limit point of $A$.

Answer 2

(1), (2) and (3) are wrong. (4) is obviously true. For all $x\in \mathbb R$, $x\in A$ or $x\in A^c$. Now the fact that for all $\varepsilon >0$ there is $x\in A$ s.t. $l-\varepsilon <x<l+\varepsilon $ doesn't implies that $l$ is a limite point. Just take for example $A=\{2\}$.

Answer 3

$4$ is of the form $P\vee \neg P$, so it is clearly true for any point of $\mathbb R$, not just the supremum of $A$. You have correctly deduced that $1$, $2$, and $3$ are wrong. There is nothing more to say.