Every supremum is a limit point. So I found this problem in Abbot's book, introduction to real analysis, it says "show that if A is a bounded set, then supA is a limit point", my argument is as it follows:
This is not true, take for example $A=\left\{\frac1n, n>0\right\}$, then 1 is clearly the supremum of this set, as it's its maximum, but 1 is an isolated point and hence can not be a limit point.
Also take $(0,1)\cup\{4\}$. Clearly bounded and its supremum is also an isolated point, my guess is the book was trying to say "take A as an open and bounded set".
Any help would be appreciated.
Edit: take also {1,2}, clearly bounded by 2, its supremum is 2, but 2 is not a limit point.

Any one-point (or indeed, any non-empty, finite) set provides an instant counterexample to the claim as stated. However, if we add the condition that the set also be open, that fixes it!
More generally, given any bounded set $A$ with non-empty interior, the supremum of the interior of $A$ is a limit point of $A.$
Edit: Even more generally, and possibly what Abbott was going for, if $A$ is a bounded set without a(n isolated) maximum element, then the supremum of $A$ is a limit point of $A.$
