Timeline for Does ZF prove that topological groups are completely regular?
Current License: CC BY-SA 3.0
14 events
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| Jun 15, 2020 at 7:27 | history | edited | CommunityBot |
Commonmark migration
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Mar 3, 2015 at 17:35 | comment | added | Rahman. M | @ToddTrimble oh,I meant $ZF+DC$ by 2! It was my fault. I understand what happened. Both of us mean one thing!!sorry! | |
| Mar 3, 2015 at 17:32 | comment | added | Todd Trimble | Rahman. M: no, it doesn't answer question 2. Question 2 reads, "Does ZF prove that $\langle G, \mathcal{T} \rangle$ is completely regular?" Since ZF does not prove ZF + DC (i.e., since DC is not provable in ZF), this answer does not address question 2. It only addresses part 2 of question 1 ("Are those right?", with part 2 asserting "ZF + (Dependent Choice) proves that $\langle G, \mathcal{T} \rangle$ is completely regular". | |
| Mar 3, 2015 at 15:19 | comment | added | Andreas Blass | Although $T_1$ is stronger than $T_0$ in general, they're equivalent in topological groups. | |
| Mar 3, 2015 at 13:37 | comment | added | Rahman. M | @AndreasBlass since $T_{1} \longrightarrow T_{0}$, so if my answer uses only DC, then it answers question2. | |
| Mar 3, 2015 at 13:19 | comment | added | Andreas Blass | About" maybe closedness of $\{e\}$ is sufficient": In a topological group, if $\{e\}$ is closed, then, by homogeneity, all singletons are closed, and so the topology satisfies $T_1$. | |
| Mar 3, 2015 at 12:52 | comment | added | Rahman. M | I am not sure about the necessity of $T_{0}$, maybe closed-ness of $\{e\}$ is sufficient. | |
| Mar 3, 2015 at 12:47 | comment | added | Todd Trimble | No, it just answers (part of) question 1. | |
| Mar 3, 2015 at 11:21 | history | edited | Rahman. M | CC BY-SA 3.0 |
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| Mar 3, 2015 at 11:19 | comment | added | Rahman. M | @AsafKaragila Thank you. I don't know much about AC and related things, Is it an answer of question 2. | |
| Mar 3, 2015 at 10:31 | comment | added | Asaf Karagila♦ | Defining a sequence by induction is the quintessential use of dependent choice. | |
| Mar 3, 2015 at 8:09 | history | edited | Rahman. M | CC BY-SA 3.0 |
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| Mar 3, 2015 at 7:39 | history | answered | Rahman. M | CC BY-SA 3.0 |