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?
 A: $a$ 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 $p_{b_n}\in A$.
A: 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.
A: Hint: let $n>0$. Then there is
$x_n\in A$ such as
$$
\sup A -  \frac 1n < x_n \le \sup A
$$
A: 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.
A: You can construct a senquence which converges to $a$.
For $a-1$, then there exists $a_1 \in A$ such that $a-1<a_1<a$.
Then for $a_1<a$, there exists $a_2 \in A$ such that $\max\{a_1, a-\frac12\}<a_2<a$
Then for $a_2<a$, there exists $a_3 \in A$ such that $\max\{a_2, a-\frac13\}<a_2<a$
$\cdots$
Then there is a sequence $\{a_n\}\in A$ which converges $a$.
