Bounded Sequence having a convergent subsequence

**kathrynmath** · November 12th 2008, 08:32 PM

Ok, so i did this proof and I'm unsure about it?

I wnat to prove that ${a_n}$ has a convergent subsequence.

Proof:
Since ${a_n}$ is a bounded sequence of real numbers, it must have at least one accumulation point. Let A be that accum pt. By definition, A is an accum pt of ${a_n}$ iff each neighborhood contains points of ${a_n}$ which are not A. Let $\varepsilon>0$ be arbitrary. So, we can let $a_n$ $_k$ elemnet of (A- $\varepsilon$ ,A+ $\varepsilon$ ). As $\varepsilon--->0$ ,(A- $\varepsilon$ ,A+ $\varepsilon$ ) accumulates at A. Therefore, A is an accum pt of $a_n$ $_k$ .

I'm afraid I wasn't specific enough and should have included the case for finitely many distinct terms.

**ThePerfectHacker** · November 13th 2008, 08:45 AM

Originally Posted by kathrynmath

Ok, so i did this proof and I'm unsure about it?

I wnat to prove that ${a_n}$ has a convergent subsequence.

Proof:
Since ${a_n}$ is a bounded sequence of real numbers, it must have at least one accumulation point. Let A be that accum pt. By definition, A is an accum pt of ${a_n}$ iff each neighborhood contains points of ${a_n}$ which are not A. Let $\varepsilon>0$ be arbitrary. So, we can let $a_n$ $_k$ elemnet of (A- $\varepsilon$ ,A+ $\varepsilon$ ). As $\varepsilon--->0$ ,(A- $\varepsilon$ ,A+ $\varepsilon$ ) accumulates at A. Therefore, A is an accum pt of $a_n$ $_k$ .

I'm afraid I wasn't specific enough and should have included the case for finitely many distinct terms.

If you can find a monotomic subsequence then that would complete the proof. See this and post #9.

**kathrynmath** · November 13th 2008, 09:54 AM

Originally Posted by ThePerfectHacker

If you can find a monotomic subsequence then that would complete the proof. See this and post #9.

.
So this proves that the sequences accumulates at A? therefore, it converges?

**Plato** · November 13th 2008, 10:14 AM

Originally Posted by kathrynmath

By definition, A is an accum pt of ${a_n}$ iff each neighborhood contains points of ${a_n}$ which are not A.

The above is the key to this proof.
There is a point of the sequence $a_{n_1 } \in \left( {A - 1,A + 1} \right)$
Then $\lambda _2 = \min \left\{ {\frac{1}{2},\left| {A - a_{n_1 } } \right|} \right\}\; \Rightarrow \;\left( {\exists a_{n_2 } \in \left( {A - \lambda _2 ,A + \lambda _2 } \right)} \right)$
Now if $k \geqslant 3\;\;\& \;\lambda _k = \min \left\{ {\frac{1}{k},\left| {A - a_{n_{k - 1} } } \right|} \right\}\; \Rightarrow \;\left( {\exists a_{n_k } \in \left( {A - \lambda _k ,A + \lambda _k } \right)} \right)$

Now clearly you have a sequence of distinct terms and $\left( {a_{n_k } } \right) \to A<br />$ .

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