Prove that the sequence does not have a convergent subsequence

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Homework Help Overview

The discussion revolves around a problem in topology, specifically concerning the convergence of sequences in product spaces. The original poster is tasked with proving that a particular sequence does not have a convergent subsequence and is exploring the properties of the space involved.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to relate the convergence of sequences in product spaces to their projections in individual topological spaces. They express uncertainty about how to demonstrate the general case for part (a) and indicate a lack of understanding for part (b).

Discussion Status

Some participants suggest that the original poster should clarify their understanding of convergence in topological spaces and the product topology. There is an acknowledgment of the need to explore foundational concepts before proceeding with the proof.

Contextual Notes

The problem involves sequences defined in a specific topological context, and there are indications of formatting issues with LaTeX in the original posts. The discussion may be constrained by the original poster's current understanding of the relevant topological properties.

fabiancillo
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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
 
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Moderator's note: Thread moved to Calculus & Beyond homework forum.
 
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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 [itex] tags, it will be set inline. This takes up less space and makes your post easier to read:

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?
 
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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.
 

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