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Prove that the sequence does not have a convergent subsequence

cristianoceli

New member
Jun 16, 2016
23
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 $k \in \mathbb{N}$.



b) I don't know