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Let $A \subseteq \mathbb{R}$ be a bounded above set.

Prove that exists a sequence $\{a_n\}_{n=1}^{\infty}$ such that $\forall n \in \mathbb{N}: a_n \in A$ and $$\lim_{n \to \infty}a_n = sup(A)$$.

I thought about: for every $n \in \mathbb{N}$ taking $a_n \in A - (- \infty, a_{n-1}]$ Is that a good idea?

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    $\begingroup$ My guess is that you meant “sequence”, not “series”. $\endgroup$ Nov 16 '19 at 18:31
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Close but not quite.

There are a few things you overlooked.

1) Its possible that $(a_n, \sup A)$ need not have any values at all. This can only happen if $\sup A \in A$. For example take a finite set $A$ or the set $(0,1) \cup \{2\}$.

In the case that $\sup A \in A$ then $A$ has a max element. There's nothing in the definition of sequence that the terms be unique. In this case just take $a_k = \sup A$ for all $k$. That'll do. This is not an interesting case.

MUCH more interesting is when $\sup A \not\in A$. In this case $\max A$ does not exist and $A$ is infinite.

In this case $(a_n, \sup A)$ will never be non empty and you can inductively find $a_n < a_{n+1} < a_{n+2} < .......... < \sup A$.

But

2) there's no reason to assume $\lim a_n = \sup A$. Instead you might have $\lim a_n = m$ for some $m < \sup A$.

For example. Suppose $A = (0, 2)$ and you chose $a_k = 1 - \frac 1k$. Then $\lim a_n = 1 < \sup A = 2$.

Can you work around this?

Let $a_k \in (\sup A - \frac 1k)$. Because $\sup A$ is least upper bound $\sup A - \frac 1k$ is not an upper bound and so an $a_k\in (\sup A- \frac 1k, \sup A); a_k\in A$ will always exist. And for any $\epsilon > 0$ if $n > \frac 1\epsilon$ then $\sup A - \epsilon < \sup A - \frac 1n < a_n <\sup A$. so $\lim a_n = \sup A$.

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Note: your idea that $a_{n} < a_{n+1} < \sup A$, was an EXCELLENT idea and very good first step to solving. But ultimately wasn't necessary. But could still be used. Maybe, let $a_{n+1} \in (\frac {\sup A + a_n}2, \sup A)$ (that is, $a_{n+1}$ is between $\sup A$ and the midpoint between $\sup A$ and $a_n$.)

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  • $\begingroup$ I dont understand what do you mean by$a_k \in (\sup A - \frac{1}{k})$ the right hand side is a number, not a group so why you using in $\in$ $\endgroup$ Nov 16 '19 at 19:03
  • $\begingroup$ Oops, a typo. I meant $a_k \in (\sup A -\frac 1k, \sup A)$. I was trying to use the OP's language and notation. $\endgroup$
    – fleablood
    Nov 16 '19 at 20:52
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Put $\;M:=\sup A\;$ , then by the very definition of supremum of a real set, we have that for any natural number $\;n\in\Bbb N\;$ there exists $\;a_n\in A\;$ s.t.

$$M-\frac1n<a_n\le M$$

Now make your choice and use the squeeze theorem...

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