Prove that the sequence does not have a convergent subsequence

[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

 

[PeterDonis]

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

 

[martinbn]

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.

 

[pasmith]

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?

 

[Mark44]

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.