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Let $C$ be a set and let $2^C$ be its power set. Consider the union function $\cup: 2^C\times 2^C\rightarrow 2^C$ such that $\cup (X,Y)=X\cup Y$. For a fixed $Y\in 2^C$, let $\cup_Y$ be the evaluation map $\cup_Y(X)=X\cup Y$.

Are there natural topologies in $2^C$ such that:

  • the union function $\cup$ is continuous;
  • for every singleton ${y_0}$ the evaluation map $\cup_{y_0}$ has dense image?

Any help is welcome.

P.S: I have no concrete problem involving the question above in mind. I'm pursuing a motivation for a nomenclature that I'm using in a not related work.

William Elliot
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Math-Phys-Cat Group
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    The trivial topology (2 open sets) works, but I guess this is not what you want...? – Elle Najt May 19 '20 at 16:53
  • Thank you for the remark. But, yes, I would like to consider a more useful topology. – Math-Phys-Cat Group May 19 '20 at 16:57
  • Do you have some use in mind? Clarifying it would improve the question. – Elle Najt May 19 '20 at 16:58
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    Yes, but is something very particular and not a real problem. I'm working on the classification problem for extensions of Yang-Mills type theories and the best name I find for certain construction is to say that the solution is "dense". I'm trying to find some concrete way to justify the name. Sorry if I was vague. – Math-Phys-Cat Group May 19 '20 at 17:01
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    The most natural topology to put on the power set of $C$ is the usual product topology, viewing $\mathcal{P}(C)$ as $\prod_{i \in C} {0,1}$. In this topology, the union is continuous. – Jonathan Schilhan May 19 '20 at 17:51
  • Thank you for this comment. – Math-Phys-Cat Group May 19 '20 at 17:58
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    Suppose that $\tau$ is a topology on $\wp(C)$ that has the desired properties. Let $y\in C$; $\cup_y$ has a dense image, so each non-empty $\mathscr{U}\in\tau$ must contain a $U_y\subseteq C$ such that $y\in U_y$. In other words, each non-empty $\mathscr{U}\in\tau$ must cover $C$. Closure of $\tau$ under unions will be no problem, since any collection of covers of $C$ is closed under unions, but closure under finite intersections puts real restrictions on $\tau$. It can be done, and $\tau$ can even be Hausdorff, but so far I see no way to make union continuous. – Brian M. Scott May 20 '20 at 02:50
  • Thank you. Thus, are you saying that the second condition is really restrictive and that it is typically not satisfied together with the first condition? – Math-Phys-Cat Group May 20 '20 at 08:13

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