# Show that x is an accumulation point

• October 9th 2008, 08:10 AM
Cato
Show that x is an accumulation point
Let { $x_n$}, $\ n \geq 1$, be a bounded sequence of real numbers and let x := sup{ $x_n$ | $\ n \geq 1$}.
If for all n in N , $x_n$ < x, prove that x is an accumulation point of { $x_n$}

This question is given me alot of trouble, anyone can help solve this plz?
• October 9th 2008, 08:28 AM
watchmath
To show that it is an accumulation point you need to show that for every interval containing x, the interval also contains some terms from the sequence different from x.

Mathematically you need to show for every epsilon, $[x-\epsilon,x+\epsilon]$ contains $s_n$ for some $n$ and $s_n\neq x$.

Hint: you need to use the fact that supremum is the least upperbound
• October 9th 2008, 08:39 AM
Cato
can it be proven using sub-sequences?
• October 9th 2008, 09:05 AM
watchmath
Quote:

Originally Posted by Cato
can it be proven using sub-sequences?

I am not sure what you mean by that?
So what is the accumulation point definition given in your book?
• October 9th 2008, 10:02 AM
Cato
In the book, it says,

A point s, is called an accumulation point of a sequence { $s_n$} if there exists a sub-sequence of { $s_n$} that converges to s.
• October 9th 2008, 10:39 AM
bkarpuz
Quote:

Originally Posted by Cato
Let { $x_n$}, $\ n \geq 1$, be a bounded sequence of real numbers and let x := sup{ $x_n$ | $\ n \geq 1$}.
If for all n in N , $x_n$ < x, prove that x is an accumulation point of { $x_n$}

This question is given me alot of trouble, anyone can help solve this plz?

In the case where $x_{n}{\color{red}{\leq}}x$ is true for all $n\in\mathbb{N}$, I dont believe that $x$ defined as in your definition is not always an accumulation point for all bounded sequences.

Counter Example
. Let $x_{n}:=\frac{2n}{n^{2}+1}$ for $n\in\mathbb{N}$.
It is obvious that $x_{n}$ is bounded.
It is easy to see that $x=1$ (this value holds with $n=1$) and $x_{n}<1$ for all $n\in\mathbb{N}\backslash \{1\}$.
But $x_{n}$ tends to $0$ as $n$ tends to $\infty$. $\rule{0.3cm}{0.3cm}$

I feel that the claim is true when $x_{n}{\color{red}{<}}x$ holds for all $n\in\mathbb{N}$, I will be thinking on it.
• October 9th 2008, 10:46 AM
watchmath
Quote:

Originally Posted by Cato
In the book, it says,

A point s, is called an accumulation point of a sequence { $s_n$} if there exists a sub-sequence of { $s_n$} that converges to s.

Ok

Since x is the supremum, by definition if we substract a little bit from x, then x is no longer a lower bound.

So for every n we have that $x-\frac{1}{n}$ is not a lower bound, i.e, there are terms in the sequence which are in $(x-1/1, x)$. Choose such element with the least index, say $x_{n_1}$
Choose $x_{n_2}$ in (x-1/2, x) with the same properties.
If you continue, then you have a subsequence. Then you need to prove that this subsequence converges to x too.
• October 9th 2008, 10:58 AM
bkarpuz
Here is the proof you want.

Proof.
Let $\{x_{n}\}$ be a bounded sequence and $x_{n} hold for all $n\in\mathbb{N}$, where $x:=\sup\{x_{n}:n\in\mathbb{N}\}$.
On the contrary, suppose that $x$ is not an accumulation point for $\{x_{n}\}$.
In this case, we may find a sufficiently small $\varepsilon>0$ such that $x_{n}\not\in(x-\varepsilon,x+\varepsilon)$ for all $n\in\mathbb{N}$ or equivalently $x_{n}\leq x-\varepsilon$ for all $n\in\mathbb{N}$.
In this present case, we arrive at the following contradiction:
$x=\sup\{x_{n}:n\in\mathbb{N}\}\leq\sup\{x-\varepsilon:n\in\mathbb{N}\}=x-\varepsilon.$
The proof is hence completed. $\rule{0.3cm}{0.3cm}$
• October 9th 2008, 11:09 AM
Cato
thanks alot guys, you are both very helpful. Much appreciated.