Let { }, , be a bounded sequence of real numbers and let x := sup{ | }.

If for all n in N , < x, prove that x is an accumulation point of { }

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

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- Oct 9th 2008, 08:10 AMCatoShow that x is an accumulation point
Let { }, , be a bounded sequence of real numbers and let x := sup{ | }.

If for all n in N , < x, prove that x is an accumulation point of { }

This question is given me alot of trouble, anyone can help solve this plz? - Oct 9th 2008, 08:28 AMwatchmath
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, contains for some and .

Hint: you need to use the fact that supremum is the least upperbound - Oct 9th 2008, 08:39 AMCato
can it be proven using sub-sequences?

- Oct 9th 2008, 09:05 AMwatchmath
- Oct 9th 2008, 10:02 AMCato
In the book, it says,

A point s, is called an accumulation point of a sequence { } if there exists a sub-sequence of { } that converges to s. - Oct 9th 2008, 10:39 AMbkarpuz
In the case where is true for all , I dont believe that defined as in your definition is not always an accumulation point for all bounded sequences.

. Let for .

Counter Example

It is obvious that is bounded.

It is easy to see that (this value holds with ) and for all .

But tends to as tends to .

I feel that the claim is true when holds for all , I will be thinking on it. - Oct 9th 2008, 10:46 AMwatchmath
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 is not a lower bound, i.e, there are terms in the sequence which are in . Choose such element with the least index, say

Choose 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 AMbkarpuz
Here is the proof you want.

**Proof**.

Let be a bounded sequence and hold for all , where .

On the contrary, suppose that is not an accumulation point for .

In this case, we may find a sufficiently small such that for all or equivalently for all .

In this present case, we arrive at the following contradiction:

The proof is hence completed. - Oct 9th 2008, 11:09 AMCato
thanks alot guys, you are both very helpful. Much appreciated.