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.

  • $\begingroup$ 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. $\endgroup$ Dec 7, 2012 at 2:18

2 Answers 2


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 $\le \frac{b-s_1}{2^n}$.


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.

  • $\begingroup$ How are you interpreting the word, "distinct" in the OPs question? $\endgroup$ Dec 7, 2012 at 2:21
  • $\begingroup$ how about this, for all $q$ there is strictly increasing $a_{q} \in \mathbb{R} s.t \mid a_{q}-sup(A)\mid < \frac{1}{q}$. Now if there is no sub sequence in $A$ of this sequence the supremum is contradicted. $\endgroup$
    – user123124
    Feb 18, 2020 at 9:23

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .