How can I 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

$a$ 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 $p_{b_n}\in A$.

Answer 2

Hint: let $n>0$. Then there is $x_n\in A$ such as $$ \sup A - \frac 1n < x_n \le \sup A $$

Answer 3

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 $a$.