# Genrate sequence thats its limit is the supremum of a set

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?

• My guess is that you meant “sequence”, not “series”. Nov 16 '19 at 18:31

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 $$\lim a_n = \sup A$$.

.

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 $$a_n$$.)

• 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 $\in$ Nov 16 '19 at 19:03
• 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. Nov 16 '19 at 20:52

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

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