Don't understand how to find the sequence I have this homework problem: Let $A \subseteq \mathbb{R}$ be a set of real numbers which is bounded above, and let $a = \sup A$. Show that there is a sequence $(a_n)$ so that $a_n \in A$ for each $n$ and $a_n \rightarrow a$. [Hint: Use the fact that $a − \frac{1}{n}$ is not an upper bound of $A$ to find $a_n$ , and check that the sequence you get converges to $a$.]
I understand the question what I don't get is how am I to properly use $a-\frac{1}{n}$ to determine $a_n$?  I worked that through to arrive at $\frac{an-1}{n}$ but this seems to produce lunacy.  For example,
$$
\begin{align}
a_1 &= a-1 \\
a_2 &= \frac{a2-1}{2} \\
a_3 &= \frac{a3-1}{3} \\
\end{align}
$$
Which is just gibberish.  I've stared at this long enough; it's time for help.  What simple thing am I not seeing?
 A: Using the hint, as $a$ is $\sup A$, there is at least one element in $A$ that is greater than $a-1$.  If there weren't, $a-1$ would be an upper bound for $A$ that is less than $a$, contradicting the statement that $a=\sup A$.  Pick any element greater than $a-1$ for $a_1$.  Similarly, there is at least one element greater than $a-\frac 12$, so pick one for $a_2$.  Each $a_n$ is chosen to be an element of $A$ that is greater than $a-\frac 1n$.  Now you can do an $\epsilon-N$ proof that the sequence converges to $a$.  If I give you $\epsilon$ you can say that all $a_k$ with $k \gt \frac 1\epsilon$ are within $\epsilon$ of $a$.
A: All you need is the definition: $a = \sup(A)$ is the least upper bound of $A$.
This means that if somebody were to show you an upperbound $b$ of $A$ (so $\forall x \in A: x \le b)$, then you know for sure that $a \le b$.
This implies by taking the contrapositive: any number $b < a$ cannot be an upperbound for $A$! (or it would contradict the above)
Now let $a - \frac{1}{n} <a$ (for any $n$ we substract a strictly positive number). This is not an upperbound for $A$, so there is some $a_n \in A$ such that $a_n > a - \frac{1}{n}$ (or this smaller number would be an upperbound for $A$, which cannot be). You do know that $a_n \le a$, by virtue of $\sup(A)$ being an upperbound for $A$.
So you find a squeezed  sequence $a_n \in (a - \frac{1}{n}, a]$ for every $n$.
Now show $(a_n)_n$ must converge to $a$ (applying the definition of convergence, and using that for every $\varepsilon >0 ,\exists n: \frac{1}{n} < \varepsilon$.
