# Show that x is an accumulation point

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

This question is given me alot of trouble, anyone can help solve this plz?
• Oct 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, $\displaystyle [x-\epsilon,x+\epsilon]$ contains $\displaystyle s_n$ for some $\displaystyle n$ and $\displaystyle s_n\neq x$.

Hint: you need to use the fact that supremum is the least upperbound
• Oct 9th 2008, 08:39 AM
Cato
can it be proven using sub-sequences?
• Oct 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?
• Oct 9th 2008, 10:02 AM
Cato
In the book, it says,

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

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

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

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

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

I feel that the claim is true when $\displaystyle x_{n}{\color{red}{<}}x$ holds for all $\displaystyle n\in\mathbb{N}$, I will be thinking on it.
• Oct 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 {$\displaystyle s_n$} if there exists a sub-sequence of {$\displaystyle s_n$} 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 $\displaystyle x-\frac{1}{n}$ is not a lower bound, i.e, there are terms in the sequence which are in $\displaystyle (x-1/1, x)$. Choose such element with the least index, say $\displaystyle x_{n_1}$
Choose $\displaystyle x_{n_2}$ 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.
• Oct 9th 2008, 10:58 AM
bkarpuz
Here is the proof you want.

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