I want to prove that $\mathbb{Q}^\omega$is a product of $\sigma$-compacta, but struggling. Is my reasoning correct?
My thoughts: I would just show that $\mathbb{Q}$ (with the standard topology inherited from $\mathbb{R}$ is $\sigma$-compact, which should be enough to see that $\mathbb{Q}^\omega$ is too.
However, I don´t know how to show that for $\mathbb{Q}$. ProofWiki says:
From Rational Numbers are Countably Infinite, $\mathbb{Q}$ is countable. Hence the result from definition of Countable Space is $\sigma$-Compact.
Could anybody clarify this? Thank you.
Definition 1: The space $\mathbb{Q}^\omega$ is defined as a set of all rational sequences endowed with the standard product topology.
Definition 2: A topological space is said to be $\sigma$-compact if it is the union of countably many compact subspaces.
Just use the fact that $\Bbb Q=\bigcup_{q\in\Bbb Q}\{q\}$. And also the fact that each singleton is compact.