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Let $a_1, a_2, a_3$,.... be a sequence so that $a_1<a_2<a_3<...$ and so that the sequence is bounded. (The set of points is bounded.) Let $u$ denote the least upper bound of ${[a_n]}$ from n=1 to infinity. Show that $u$ is a limit point of the set of points ${[a_n]}$ from n=1 to infinity.

Since $u$ is the least upper bound and I want to prove that u is the limit point, I know need to prove that $u$ is not in the set. I just am unsure of how to begin.

Let M= ${[a_n]}$ from n=1 to infinity.

First, I will assume that $u\in M$. (Proving by contradiction.)

I also know that I want to show that the points in the set "limit" to $u$, I'm just struggling.

user102839
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  • You know that the sequence is bounded. So $a_n \leq M$ fore some $M \in \mathbb{R}$. Also we know that $u = \sup a_n$. So what doe s that mean? – user112167 Dec 05 '13 at 02:16
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    Hint: What's the definition of limit point? – Carl Love Dec 05 '13 at 02:19
  • @CarlLove The definition of a limit point: u is a limit point of M if (a,b) is a segment containing u, then (a,b) contains a point of M distinct from P. – user102839 Dec 05 '13 at 02:22

2 Answers2

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To prove that $u$ is not in the set, assume the opposite. Then $u = a_n$ for some $n$. Then $u < a_{n+1}$. So $u$ is not an upper bound. Contradiction.

Carl Love
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Use the definition of limit point and supremum!

For $u=\sup_n a_n$, it means that this is the smallest upper bound of the sequence.

Let $\varepsilon>0$ be arbitrary, then the interval $(u-\varepsilon,u+\varepsilon)$ must contain elements of the sequence $a_n$, because otherwise $u-\varepsilon$ would be a smaller upper bound.

Berci
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