Real Analysis – How to Show a Supremum of a Set is Also Its Limit Point

Question

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?

Answer 1

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

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

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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.