# Question about limit points of a Subset of $\mathbb{R}$

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?

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