Timeline for Existence of a *really* nice topology on the powerset of a topological space
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| Mar 3, 2023 at 20:16 | comment | added | Emily | Thank you so much, I'm finally able to understand this point as well! | |
| Mar 3, 2023 at 17:12 | comment | added | James E Hanson | @Emily If $\tau$ is non-trivial, then there is a non-empty closed proper subset $G \subset \mathcal{P}(X)$. Since $\mathcal{P}(X)$ has a transitive homeomorphism group, this implies that there is a non-empty closed proper subset $G'$ containing $\varnothing$. Therefore, since $F \subseteq G'$, $F \neq \mathcal{P}(X)$. | |
| Mar 3, 2023 at 15:36 | comment | added | Emily | Thank you very much, I see it now! Could I ask one last question? In the end of the proof of the lemma, I understand the implication ($\tau$ trivial $\Rightarrow$ $F=\mathcal{P}(X)$), but I haven't yet been able to understand why we also have $\tau$ non-trivial $\Rightarrow$ $F\neq\mathcal{P}(X)$. Could you please explain why this hols? (And sorry if this is a silly question!) | |
| Mar 3, 2023 at 6:36 | comment | added | James E Hanson | Regarding 1), the inverse of a homeomorphism that fixes $\varnothing_2$ is also a homemomorphism that fixes $\varnothing_2$, so you also have that $F_2 \subseteq f^{-1}(F_2)$, which implies $f(F_2) \subseteq F_2$. I would need to think about 2). | |
| Mar 1, 2023 at 20:33 | history | edited | Emily | CC BY-SA 4.0 |
typos
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| Mar 1, 2023 at 20:29 | comment | added | Emily | In this direction, do you know how restrictive just 1–4 + ($f$ continuous $\Rightarrow$ $f_*$ continuous) would be? (where $f_*$ continuous is either as a function or as relation, I'd love to know either answer) | |
| Mar 1, 2023 at 20:29 | vote | accept | Emily | ||
| Mar 1, 2023 at 20:29 | comment | added | Emily | (although the notion of continuity for $f^{-1}$ here is the relational one, i.e. that whenever $V$ is open in $\mathcal{P}(Y)$, the sets \begin{gather*}\{U\in\mathcal{P}(X)\ |\ U\cap f^{-1}(V)\neq\emptyset\},\\\{U\in\mathcal{P}(X)\ |\ U\subset f^{-1}(V)\}\end{gather*} are also open in $\mathcal{P}(X)$). | |
| Mar 1, 2023 at 20:29 | comment | added | Emily | 2) I was looking further at the case of the Vietoris topology these days and realised that my assumption for $f^{-1}$ to be continuous when $f$ is so might not be so natural, as in that context a closed and open (but not necessarily continuous) map $f$ already induces a continuous $f^{-1}$ | |
| Mar 1, 2023 at 20:28 | comment | added | Emily | 1) I see why a homeomorphism $f$ of $\mathcal{P}(\mathcal{P}(X))$ fixing $\emptyset_2$ should satisfy $F_2\subset f(F_2)$, but why should we also have $F_2=f(F_2)$? | |
| Mar 1, 2023 at 20:27 | comment | added | Emily | Wow, this is amazing, thank you so much! (Sorry also for taking a little while to finally reply, I was carefully going through each step of the proof, and that took a while.) Would it be okay to ask a few questions about it? | |
| Feb 26, 2023 at 21:14 | history | edited | James E Hanson | CC BY-SA 4.0 |
deleted 37 characters in body
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| Feb 26, 2023 at 21:04 | history | answered | James E Hanson | CC BY-SA 4.0 |