TITLE: How can I show a supremum of a set is also its limit point?
QUESTION [9 upvotes]: 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 $\epsilon>0$? Or should I use a completely different approach?

REPLY [16 votes]: 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 $\varepsilon > 0$ there exists some $y\ne x$ such that $|y-x| < \varepsilon$ 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 $\sup(A) < \infty$, we can construct a set $A'$ by appending an extra point that is greater than the supremum.  For example, for any $\varepsilon > 0$, 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 $\varepsilon > 0$ there exists some $y$ such that $|y-x| < \varepsilon$ 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 $\varepsilon > 0$ there is a point
$$x_n \in (\sup(A)-\varepsilon, \sup(A)]. $$
In other words, for any $\varepsilon > 0$, we can find a point $x_n \in A$ such that $|x_n - \sup(A)| = \sup(A) - x_n < \varepsilon$.  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.