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.
