How can I show a supremum of a set is also its limit point? It just seems like I can't wrap my head around this. For a set $A$ how can I show that the supremum of that set $\sup(A)$ is also a limit point of said set?
Can I just set
$$a=\sup(A)$$
$$b\le a \: \:  \forall \: b \in A$$
and then show that there's a point different from $a$ in any $\epsilon$-neighbourhood for any $\epsilon&gt;0$? Or should I use a completely different approach?
The other answers to this question are incorrect.  It is not generally true that the supremum of a set $A$ in $\mathbb{R}$ is a limit point of that set.  For example, as pointed out in the comments by Daniel Fischer, for any $a \in \mathbb{R}$, we have $\sup\{a\} = a$, but $a$ is not a limit point of $\{a\}$.  To explain this, first recall the definition:


  Definition: Let $A \subseteq \mathbb{R}$.  A point $x\in\mathbb{R}$ is called a limit point of $A$ if for any $\varepsilon &gt; 0$ there exists some $y\ne x$ such that $|y-x| &lt; \varepsilon$ and $y\in A$.  In other words, every $\varepsilon$-neighborhood of $x$ contains a point of $A$ which is different from $x$.


If we take $A = \{a\}$ then, as noted above $\sup(A) = a$.  But no neighborhood of $a$ contains any point of $A$ other than $a$.  Hence, by the definition above, $a = \sup A$ is not a limit point of $A$.  Indeed, given any set $A$ such that $\sup(A) &lt; \infty$, we can construct a set $A'$ by appending an extra point that is greater than the supremum.  For example, for any $\varepsilon &gt; 0$, the set
$$A' = A \cup \{ \sup(A) + \varepsilon \} $$
satisfies the properties


$\sup(A') = \sup(A) + \varepsilon$, and
$\sup(A')$ is not a limit point of $A'$.




If, instead we ask "Is the supremum of a subset of $\mathbb{R}$ an adherent point of $A$?", then the answer is "Yes", subject to appropriate caveats.


  Definition:  Let $A \subseteq \mathbb{R}$.  A point $x\in\mathbb{R}$ is called an adherent point of $A$ if for any $\varepsilon &gt; 0$ there exists some $y$ such that $|y-x| &lt; \varepsilon$ and $y\in A$.  In other words, every $\varepsilon$-neighborhood of $x$ contains a point of $A$.


Note here that we do not require $y$ to differ from $x$.  Hence $a$ is an adherent point of $\{a\}$.  Note that every limit point of a set is also an adherent point of that set, and that all the points of $A$ are also adherent points.  To show that the supremum of a bounded-from-above set is an adherent point, we can argue as follows:


If $\sup(A) \in A$, then $\sup(A)$ is an adherent point of $A$, and we are done.
If $\sup(A) \not\in A$, then it follows from the definition of the supremum that for any $\varepsilon &gt; 0$ there is a point
$$x_n \in (\sup(A)-\varepsilon, \sup(A)]. $$
In other words, for any $\varepsilon &gt; 0$, we can find a point $x_n \in A$ such that $|x_n - \sup(A)| = \sup(A) - x_n &lt; \varepsilon$.  This implies that $\sup(A)$ is an adherent point of $A$.


Note that the second case is (in every way that matters) the argument given by MPW, mookid, and Paul in their answers.
The assertion is only true for open sets in R. Any neighborhood of supA except itself intersects A must be nonempty otherwise there is a smaller upper bound in that region. Also supA is not in A.