Let $A$ be a set in $\mathbb{R}$, non-empty and bounded above. Prove that $s = \sup A$ if and only if $s$ is an upper bound of $A$ and there exists a sequence $(s_n)$ in $A$ which converges to $s$.

I am a little uncertain, given the fact there is no statement suggesting sequences in $A$ are monotone in anyway, so how can we say they are bounded?

Next, I realised it says "there exists a sequence". So this would mean we need only show that there is at least one sequence in the set $A$ that converges?

I started as:

$\Longrightarrow$ Assume $ s = \sup A$

Then let $\epsilon> 0 $ be arbitrary, meaning $|s - \epsilon| < s$ is no longer an upper-bound for the set $A$ since, there exists some $s_N \in A$ such that $|s - \epsilon| < s_N \leq s$

Now here is where I run into trouble, since I can't make the statement "For some $n \geq N, s_n \geq s_N$ such that $s - \epsilon< s_N < s_n \leq s < s+\epsilon$" I can't make a statement about convergence, since there is no condition that we have an increasing sequence in this set.

But since the question says "there exists a sequence which converges", am I allowed to simply say "take some increasing sequence in $A$ that..."? Or am I missing some sort of idea here, and my method/approach is entirely wrong?

  • 1
    $\begingroup$ You've got the right idea...but need to work it a little. Instead of "let $\epsilon$ be arbitrary", say "for any $\epsilon > 0$", and then say "Now for each $n \in \mathbb N$, consider $\epsilon = \frac{1}{n}$..." You'll get a sequence of points $x_n$ from this. You still have to prove that their limit exists, and that it equals $s$, but those aren't too hard. BTW, the sequence need not be increasing. Think of the case where $A$ is the two element set $\{0, 1\}$, for instance. $\endgroup$ – John Hughes Mar 25 '15 at 10:45
  • $\begingroup$ I'm finding this a little hard to understand. If I only consider $\epsilon = \frac{1}{n}$, then how is this valid, since it doesn't hold for any $\epsilon > 0$? Does this lead back to the fact that we need only find one such sequence? And also, in the example you've provided, how is that not an increasing sequence? since $1>0$ $\endgroup$ – elbarto Mar 25 '15 at 10:53
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    $\begingroup$ When $\epsilon = 1/3$, your "for any $\epsilon$ " claim applies, so there's a number $x_3$, with $|s - x_3| < 1/3$. The same goes for $1/4$, $1/5$, etc. So now you have a sequence of points $x_1, x_2, \ldots$. You want to prove that for these points, we have $\lim x_n = s$. That shouldn't be too hard. As for the not-increasing sequence, if you carry out the argument in the case of that small set, you get $x_2 = x_3 = x_4 = \ldots = 1$, which is not increasing -- it's nondecreasing. And in the case of $[0, 1]$, you could get $x_1 = 1, x_2 = 0.7, x_3 = 1, x_4 = 0.9, ...$; also not increasing! $\endgroup$ – John Hughes Mar 25 '15 at 11:08
  • $\begingroup$ thanks @JohnHughes I finally understand now! $\endgroup$ – elbarto Mar 25 '15 at 13:14

Suppose $s=\sup A$. Then, the supremum, being the least upper bound, is certainly an upper bound. To construct the sequence, note that $s-\frac{1}{n}$ is no longer an upper bound and hence we can select

$$s_n \in (s-\frac{1}{n},s) \cap A .$$

The sequence $(s_n) \subset A$ convergences to $s$ because $|s-s_m| \leq \frac{1}{n}$ for all $m \geq n$.

Conversely, suppose $s$ is an upper bound and that $(s_n) \subset A$ converges to $s$. We just need to prove that no number smaller than $s$ remains an upper bound. So let $\epsilon>0$. We must show that

$$A \cap (s,s-\epsilon) \neq \emptyset .$$

But there exists an $N \in \mathbb{N}$ such that $n \geq N \Rightarrow |s_n - s| < \epsilon.$ In particular

$$s_N \in A \cap (s,s-\epsilon) .$$


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