Talk:Sequence on Product Space Converges to Point iff Projections Converge to Projections of Point

One wonders: Why the restriction to finite products? The power of the Tychonoff topology is precisely that it makes this theorem true for arbitrary products. — Lord_Farin (talk) 20:34, 20 July 2015 (UTC)

At the very end of the current version of the sufficiency proof, we need the maximum $M$ to be finite, which is only guaranteed if there is a *finite* number of spaces. However, I confess that I cooked up this proof myself and did not yet consult the literature. There may well be a better argument, which does not have this bottleneck. --S.anzengruber (talk) 08:26, 21 July 2015 (UTC)

The Tychonoff topology has all projections equal to the whole space except for finitely many factors, meaning the proof carries over without difficulty. — Lord_Farin (talk) 19:16, 21 July 2015 (UTC)

Rename suggestion

No brainer. Good job. We might want to review the links to make sure we have the correct (non-finite) version of the results and definitions linked.

Might be worth making sure there is genuinely no reason to keep the finite version up (e.g. it rests on fewer assumptions and axioms, for example some version of AoC), but I haven't taken time to check this. Someone who knows their way around better than self-taught me might be better placed. --prime mover (talk) 11:48, 8 April 2022 (UTC)

Can confirm that the general version is a theorem of $\sf {ZF}$ alone, even if one considers the more general notion of nets rather than just sequences. (Of course if Choice fails the product space would be empty, but this simply means the theorem is vacuously true). Jumpythehat (talk) 12:13, 8 April 2022 (UTC)

Solid. No dependencies on any other pages, no issues with source reviews, job done. --prime mover (talk) 12:35, 8 April 2022 (UTC)