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 $\alpha \in I$. 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 dont know
 A: I think I can sketch a proof for part (a):
($\implies$) Let $a := \lim_{n\to\infty}a_n \in X$. Fix some $\alpha \in I$. Let $U$ be a neighborhood of $p_\alpha(a)$ in $X_\alpha$. Then, $p_\alpha^{-1}(U) \cong U \times \prod_{\alpha' \in I\backslash\{\alpha\}}X_\alpha$ is a neighborhood of $a \in X$, so all but finitely many $a_n$ lie in $p_\alpha^{-1}(U)$. Therefore, all but finitely many $p_\alpha(a_n)$ lie in $U$, so $\{p_\alpha(a_n)\}_{n \in \mathbb{N}}$ converges to $p_\alpha(a) \in X_\alpha$. Since $\alpha \in I$ was arbitrary, the claim follows.
($\impliedby$) Let $y_\alpha := \lim_{n\to\infty}p_\alpha(a_n) \in X_\alpha$. We know that if $U_\alpha$ is a neighborhood of $y_\alpha$, that all but finitely many $p_\alpha(a_n)$ lie in $U_\alpha$. Pick some neighborhood $U$ of $a:=(y_\alpha)_{\alpha \in I}$ in $X$. For all $\alpha \in I$, we have that $p_\alpha(U)$ is a neighborhood of $y_\alpha$, since projections are open maps. So all but finitely many $p_\alpha(a_n)$ are contained in $p_\alpha(U)$. Now, I think it suffices to consider a neighborhood system (similar to bases of topological spaces), i.e. we can consider $U$ to take the form
$$
U = \left(\prod_{\alpha \in J}U_\alpha \right)\times \left( \prod_{\alpha \in I \backslash J} X_\alpha \right),
$$
where $\#J<\infty$ and $U_\alpha$ are neighborhoods of $y_\alpha$ respectively. So we see $p_\alpha(U) = U_\alpha$ if $\alpha \in J$. If $\alpha \in I \backslash J$, then we get $X_\alpha$, so the sequence is entirely contained in there anyway. Then, because $J$ is finite, we get that all but finitely many $a_n$ are contained in $U$.
I hope this was helpful!
