1. ## Converging Subsequence

Let x_n be a bounded sequence and let x=sup {x_n: n is a natural number} and let x_n < x for all n. Prove there is a subsequence convergent to x.

2. ## Re: Converging Subsequence

Originally Posted by veronicak5678
Let x_n be a bounded sequence and let x=sup {x_n: n is a natural number} and let x_n < x for all n. Prove there is a subsequence convergent to x.

First shoe that $\displaystyle \forall M \in \mathbb{Z}^+ \, \exists k \in (x_n)$ such that $\displaystyle k \in (x-\frac{1}{N},x)$. Denote this $\displaystyle k$ as $\displaystyle x_{n_M}$.

now show that the subsequence $\displaystyle (x_{n_M})$ converges to $\displaystyle x.$

3. ## Re: Converging Subsequence

Thanks for the help. That makes sense now, but how do I come up with things like that? I never know where to start these proofs.

4. ## Re: Converging Subsequence

You should have noticed that the key to the condition "x_n< x for all n" is that no member of the sequence is equal to x. For example, if {x_n}= {1/n} then x= sup {1/n}= 1 but {1/n} is a sequence converging to 0 so has no subsequence converging to 1. Now, the fact, that x is a sup but NOT a maximum should lead you to what abhishekkgp did.

5. ## Re: Converging Subsequence

I see. I'm trying to think more like that. Thanks for the answers.