# Prove that the sequence does not have a convergent subsequence

• fabiancillo
In summary: Some might say that it is bad practice, and I don't know if you can get banned for it, but I find it easier to type and easier to read.
fabiancillo
Hello i have problems with this exersice

Let $$\{X_{\alpha}\}_{\alpha \in I}$$ a collection of topological spaces and $$X=\prod_{\alpha \in I}X_{\alpha}$$ the product space. Let $$p_{\alpha}:X\rightarrow X_{\alpha}$$, $$\alpha\in I$$, be the canonical projections

a)Prove that a sequence $$\{a_n\}$$ converges on $$X$$ if and only if the sequence $$\{p_{\alpha}(a_n)\}$$ converges on $$X_{\alpha}$$ for all $$\alpha \in I$$.

b) Let $$I$$ the set of all sequences $$\alpha:\mathbb{Z}_{\geq 1}\rightarrow \{-1,1\}$$. Let the sequense $$a_n=\prod_{\alpha \in I}\alpha(n)\in [-1,1]^{I}$$. Prove that $$\{a_n\}$$ does not have a convergent subsequence. Is $$[-1,1]^{I}$$ sequentially compact? Is $$[-1,1]^{I}$$ firts contable?

My attempt:
a) Take $$I = \mathbb{N}$$ and $$X_i = \mathbb{R}$$ for all $$i \in \mathbb{N}$$. Now the elements in $$X$$ are real sequences and the goal is to prove that if $$a_n$$ is a sequence of these sequences (i.e. a bi-infinite real sequence), then it converges if and only if the real sequence $$a_n^{(k)}$$ converges for all $$k \in \mathbb{N}$$.

How would the demonstration of the general case be?

b) I don't know

Delta2
Moderator's note: Thread moved to Calculus & Beyond homework forum.

Delta2
May be you can start with reminding yourself what it means for a sequence in a topological space to converge, and what the product topology is.

If you put your LaTeX code in $ tags, it will be set inline. This takes up less space and makes your post easier to read: Let [itex]\{X_{\alpha}\}_{\alpha \in I}$ a collection of topological spaces and $X=\prod_{\alpha \in I}X_{\alpha}$ the product space. Let $p_{\alpha}:X\rightarrow X_{\alpha}$, $\alpha\in I$, be the canonical projections

a)Prove that a sequence $\{a_n\}$ converges on $X$ if and only if the sequence $\{p_{\alpha}(a_n)\}$ converges on $X_{\alpha}$ for all $\alpha \in I$.

b) Let $I$ the set of all sequences $\alpha:\mathbb{Z}_{\geq 1}\rightarrow \{-1,1\}$. Let the sequense $a_n=\prod_{\alpha \in I}\alpha(n)\in [-1,1]^{I}$. Prove that $\{a_n\}$ does not have a convergent subsequence. Is $[-1,1]^{I}$ sequentially compact? Is $[-1,1]^{I}$ firts contable?

Keith_McClary
pasmith said:
If you put your LaTeX code in [itex] tags, it will be set inline.
... or between pairs of ## tags, which I find easier to type. I don't use the itex ... /itex or tex ... /tex tags at all.

## 1. What is a convergent subsequence?

A convergent subsequence is a sequence of numbers within a larger sequence that approaches a specific limit as the sequence progresses. In other words, the numbers in the subsequence get closer and closer to a particular value as more terms are added to the sequence.

## 2. How do you prove that a sequence does not have a convergent subsequence?

To prove that a sequence does not have a convergent subsequence, you must show that no matter how you choose a subsequence from the original sequence, it does not approach a specific limit. This can be done by finding a specific value or range of values that the subsequence does not approach, or by showing that the subsequence becomes increasingly erratic or divergent as more terms are added.

## 3. What are some common methods for proving the non-existence of a convergent subsequence?

Some common methods for proving that a sequence does not have a convergent subsequence include the use of the Monotone Convergence Theorem, the Bolzano-Weierstrass Theorem, and the Cauchy Criterion. These theorems provide conditions under which a sequence must have a convergent subsequence, so if these conditions are not met, it can be concluded that the sequence does not have a convergent subsequence.

## 4. Can a sequence have more than one convergent subsequence?

Yes, a sequence can have multiple convergent subsequences. This can occur when there are multiple values or limits that the subsequence approaches as more terms are added to the sequence. However, it is also possible for a sequence to have no convergent subsequences at all.

## 5. How does the existence of a convergent subsequence affect the overall convergence of a sequence?

If a sequence has a convergent subsequence, it does not necessarily mean that the entire sequence is convergent. However, if a sequence does not have a convergent subsequence, it is impossible for the sequence to be convergent. In other words, the existence of a convergent subsequence is a necessary but not sufficient condition for the convergence of a sequence.

• Calculus and Beyond Homework Help
Replies
15
Views
537
• Calculus and Beyond Homework Help
Replies
1
Views
561
• Calculus and Beyond Homework Help
Replies
6
Views
3K
• Calculus and Beyond Homework Help
Replies
34
Views
2K
• Calculus and Beyond Homework Help
Replies
3
Views
1K
• Calculus and Beyond Homework Help
Replies
3
Views
930
• Calculus and Beyond Homework Help
Replies
7
Views
1K
• Calculus and Beyond Homework Help
Replies
4
Views
631
• Calculus and Beyond Homework Help
Replies
3
Views
1K
• Calculus and Beyond Homework Help
Replies
2
Views
470