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The question :

Let $D$ be a nonempty subset of the reals that is bounded above. Is the supremum of $D$ a limit point of $D$?

My Reasoning: I think this is false for these two cases. Case 1:If I look at $D = \{n \in \mathbb{Z} | n \le 0\}$ the supremum is $0$. And since I need a convergent sequence $\{x_n\} \subset D/\{0\}$ the converges to $0$ for it to be a limit point I can say in this case if I look at $\epsilon = \frac{1}{2}$ for the converges of the sequence it will fail to converge and so $0$ isn't a limit point.

Case 2: Also if I look at $D = {0}$ then the supremum is $0$. And $D$ is a subset of the reals. So if I look for a sequence $\{x_n\} \subset D/\{0\}$ I can't make one because $D/\{0\}$ is the empty set.

My question is this. Since the problem asked about an arbitrary subset of the reals $D$, can I define $D$ to give a counterexample like above or have I misunderstood the question?

--Thanks in advance.

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Your analysis is correct. A number $x$ can be the supremum of a set without having a lot of other points of the set "nearby". – Patrick Mar 30 '12 at 0:37
Your counterexamples are valid, which means that you have a complete answer to the question above. The idea at work here is that in all of your examples, the supremum was an isolated (has a neighborhood containing only itself) element of $D$. Bonus question: Prove that the answer is yes if $D$ does not contain its supremum. – Brett Frankel Mar 30 '12 at 0:37
The real-analysis and general-topology tags I added seem to fit fine. – Brett Frankel Mar 30 '12 at 1:46
up vote 2 down vote accepted

If $D$ is finite of course it does not have any limit points.

If the supremum called $s$ does not belong to $D$, then it will be its limit point, if not one can find a neighborhood of $s$ with no point of $D$ which contradicts the supremum assumption.

Also suppose that $D$ does not have any isolated points. Now if $D$ has a supremum called $s$, pick a neighborhood of $s$. There should be at least a point of $D$, called $d$ which is in this neighborhood otherwise one can find smaller upper bound ($d$ may be equal to $s$). Now because $D$ does not have any isolated points, every neighborhood of $d$ has infinitely many points of $D$. For $d=s$, the supremum has the same property and if not, one can find a neighborhood of $d$ which falls into the neighborhood of $s$ and $s$ is a limit point.

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$\{1\}\cup(3,4)$ has an isolated point, but its supremum is a limit point. – Asaf Karagila Sep 3 '13 at 14:21
Having no isolated point is a sufficient condition and not a necessary one. – Arash Sep 3 '13 at 14:31
Yes. That is true, hence my previous comment. – Asaf Karagila Sep 3 '13 at 14:32

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