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 $\epsilon>0$? Or should I use a completely different approach?

Solution 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 $A'$.

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.

Solution 2:

$a$ is the least upper bound for $A$. What does that mean?

First, it's an upper bound. That means each point $b\in A$ satisfies $b\leq a$.

Second, it's the least number that is an upper bound for $A$. This means that any other number $b$ with $b<a$ is not an upper bound for $A$, so there is at least one point $p_b$ of $A$ which exceeds this $b$--that is, $b<p_b$. This means that for each $b<a$, we can demonstrate a $p_b\in A$ with $b<p_b<a$ (the last inequality is due to the fact that $a$ is an upper bound for $A$).

Now define the sequence $b_n=a-\frac1n$. Since $b_n\rightarrow a$, and $b_n<p_{b_n}<a$, we must have that $p_{b_n}\rightarrow a$.

Thus $a$ is a limit point of $A$ since each $p_{b_n}\in A$.