I have a specific example below and i think my proof is wrong because it seems too simple, it would work for the general case which I doubt is true. I would also be interested in the general answer as well; when is the product of closed sets closed in the product topology?

Show that the set $ \prod_{\alpha\in\mathbb{R}} \mathbb{N} $ is a closed subset of the topological space $ \prod_{\alpha\in\mathbb{R}} \mathbb{R} $ (endowed with the product topology)

My attempt: $\mathbb{N}$ is a closed subspace of $\mathbb{R} $ with the canonical topology. In particular, $\overline{\mathbb{N}}=\mathbb{N}$. Also since we are in the product topology, it a known fact that closure of a product is the product of the closures. These 2 facts together mean $$ \overline{ \prod_{\alpha\in\mathbb{R}} \mathbb{N} }= \prod_{\alpha\in\mathbb{R}} \overline{\mathbb{N}} = \prod_{\alpha\in\mathbb{R}} \mathbb{N} $$ Since $ \prod_{\alpha\in\mathbb{R}} \mathbb{N} $ contains its closure, it is closed


1 Answer 1


If that fact about closures is known to you (and it is indeed true), then this is a valid argument to show $\prod_{\alpha \in \mathbb{R}} \mathbb{N}$ is closed in $\prod_{\alpha \in \mathbb{R}} \mathbb{R}$.

  • 1
    $\begingroup$ Thank you! In my experience, I am wrong anytime I think something has a 2 line proof.. By missing some important detail or assumption $\endgroup$
    – NazimJ
    Mar 19, 2019 at 22:02

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