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If $S$ is a bounded set of real numbers, how can we prove that there are distinct sequences in $S$ that converge to $\sup S$ and $\inf S$?

I'm not even sure how to begin on this problem. I know the set $S$ is bounded so it has a supremum and an infimum.

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It's not clear what you mean by "distinct sequences." But if $S=\{0\}$, it's not true under the meanings I might guess you intended. –  Thomas Andrews Dec 7 '12 at 2:18

3 Answers 3

Let $a=\sup S$. Let $n$ be any positive integer. Then $a-\frac{1}{n}$ is not an upper bound of $S$, so there is some $a_n\in S$ with $a\geq a_n>a-\frac{1}{n}$. Now the sequence $(a_n)$ converges to $a$. The case for the infimum is similar.

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How are you interpreting the word, "distinct" in the OPs question? –  Thomas Andrews Dec 7 '12 at 2:21

Let's start the sup. First recall the definition. Let $S$ be a non-empty set. We have $b=\sup S$ if $s\le b$ for every $s\in S$, and there is no $b'\lt b$ with this property.

Let $s_1$ be any element of $s$. Perhaps $s_1=b$, in which case we can take $s_2=s_3=s_4=\cdots =b$, and we have found a suitable sequence.

If $s_1\ne b$, let $m_1=\dfrac{b+s_1}{2}$. There is an element $s_2\in S$ such that $s_2\ge m_1$, else $m_1\lt b$ would be an upper bound for $S$.

If $s_2=b$, let $s_3=s_4=\cdots =b$. Otherwise, let $m_2=\dfrac{b+s_2}{2}$, and let $s_3$ be any element of $S$ that is $\ge m_2$.

Continue forever. To show that the sequence we obtain has limit $b$, draw a picture, and observe that the distance from $s_n$ to $b$ is $\le \frac{b-s_1}{2^n}$.

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Here is an example to understand what's going on, and then try to prove the assertion. Let $S$ be the open interval $(0,1)$ with $\inf{S}=0,\, \sup{S}=1$, and consider the sequences

$$ \left\{x_n\right\}_{n=1}^{\infty}= \left\{\frac{1}{n+2}\right\}_{n=1}^{\infty},\quad \left\{y_n\right\}_{n=1}^{\infty} = \left\{1-\frac{1}{n+2}\right\}_{n=1}^{\infty}.$$

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But we were not told that $S$ was so nice. It might be that your numbers are in in $S$. For example let $S=\mathbb R \setminus \mathbb Q \cap (0,1)$ –  Ross Millikan Dec 8 '12 at 2:42

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