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In a topological group, for any neighborhood $U$ of the origin, there is another such neighborhood with the property that $V.V\subseteq U.$ I conjecture a similar property for topological inverse semigroups.

Conjecture: Suppose for each idempotent $e$ we have an open set $U_e$ containing it. Then there is an open set $V_e$ such that $e\in V_e\subseteq U_e$ and $V_e. V_f\subseteq V_{ef}$ for all idempotents $e, f$.

Is this seem like a reasonable conjecture? I need this property in a very special case of a result that I am working on. I am looking for proof or related references.

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  • $\begingroup$ If you have finitely many idempotents this seems like it should be ok but if you have infinitely many it would seem you would need to take an infinite intersection of open sets and so it seems less hopeful $\endgroup$
    – Benjamin Steinberg
    Commented Mar 14, 2023 at 10:06
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    $\begingroup$ @BenjaminSteinberg: That is exactly the concern. Can you see any conditions, other than finiteness, that would force this condition? $\endgroup$
    – Bumblebee
    Commented Mar 14, 2023 at 10:20
  • $\begingroup$ Can you prove it for an infinite topological semilattice? $\endgroup$
    – Benjamin Steinberg
    Commented Mar 14, 2023 at 10:48
  • $\begingroup$ Maybe you might try compactness where uniform continuity could help $\endgroup$
    – Benjamin Steinberg
    Commented Mar 14, 2023 at 10:50
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    $\begingroup$ Can you do this for the power set 2^X under intersection for X infinite with the product topology? $\endgroup$
    – Benjamin Steinberg
    Commented Mar 14, 2023 at 12:38

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