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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.

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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 $A$.
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  • $\begingroup$ I am totally getting it, but how can we guarantee that if $l\notin A$ then $l$ must be the limit point of $A$. Can we not prove it in general? $\endgroup$ – Huny Apr 4 at 14:38
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    $\begingroup$ Yes. Let it be that $U$ is an arbitrary open set with $l\in U$. Then some $\epsilon>0$ must exists with $(l-\epsilon,l)\subseteq U$. As stated in my answer there will be an element $a\in A$ with $a\in(l-\epsilon,l)$ proving that the set $U\cap(A-\{l\})$ is not empty. This together means exactly that $l$ is a limitpoint of $A$. $\endgroup$ – drhab Apr 4 at 16:23
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(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 $A=\{2\}$.

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$4$ 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.

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  • $\begingroup$ (4) is not of the form $P \land \lnot P$. The statements $l \not\in A$ and $l$ is a limit point of $A$ are not negations of each other. It appears that you may have been responding to an earlier version of the question, but your answer does not make sense given the current state of the question. $\endgroup$ – Xander Henderson Apr 4 at 23:19

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