2026-07-072026-07-07http://salesiana.dossiersoluciones.com/handle/123456789/212274We work in set-theory without choice $\ZF$. Given a closed subset $F$ of $[0,1]^I$ which is a bounded subset of $\ell^1(I)$ ({\em resp.} such that $F \subseteq \ell^0(I)$), we show that the countable axiom of choice for finite subsets of $I$, ({\em resp.} the countable axiom of choice $\ACD$) implies that $F$ is compact. This enhances previous results where $\ACD$ ({\em resp.} the axiom of Dependent Choices $\DC$) was required. Moreover, if $I$ is linearly orderable (for example $I=\IR$), the closed unit ball of $\ell^2(I)$ is weakly compact (in $\ZF$).Functional AnalysisGeneral TopologyLogic03E25, 54B10, 54D30, 46B26text