Timeline for Mapping space from a quotient space
Current License: CC BY-SA 3.0
30 events
| when toggle format | what | by | license | comment | |
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| Jun 15, 2020 at 7:27 | history | edited | CommunityBot |
Commonmark migration
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| Jul 16, 2017 at 3:01 | history | edited | Wlod AA | CC BY-SA 3.0 |
Shorter version of the note about the pre-version.
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| Jul 15, 2017 at 23:42 | comment | added | Wlod AA | @Victor, thank you for your patience and understanding. | |
| Jul 15, 2017 at 21:34 | comment | added | Victor | Good job! A very nice example and an elegant proof! | |
| Jul 14, 2017 at 3:34 | history | edited | Wlod AA | CC BY-SA 3.0 |
English + an extra explanation (one word "compact").
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| Jul 14, 2017 at 3:29 | history | edited | Wlod AA | CC BY-SA 3.0 |
Last typos, no more.
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| Jul 14, 2017 at 3:16 | comment | added | Wlod AA | I have fixed my $X$ already. Give me a couple minutes extra for LaTeX. | |
| Jul 14, 2017 at 2:56 | comment | added | Victor | It looks like the example is good, otherwise Taras would not accept it :) In case you won't succeed to prove, please keep it as an open question. It's an interesting example. | |
| Jul 14, 2017 at 2:20 | history | edited | Wlod AA | CC BY-SA 3.0 |
English
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| Jul 14, 2017 at 2:14 | history | edited | Wlod AA | CC BY-SA 3.0 |
An accent of "will remove".
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| Jul 14, 2017 at 2:09 | comment | added | Wlod AA | @Victor, you are right, I have a HOLE in my proof, you've pointed to it (if $X$ were wrong too then it would be an ERROR). Thank you very much, and for your patience. I'll edit and keep for a moment my "answer" under construction. I feel that there is a 50-50 chance that my $X$ is correct. But if not, than after a couple of days I will remove my "answer" completely. Thank you again. | |
| Jul 13, 2017 at 20:05 | comment | added | Victor | But why $(\{a\}\times \mathbb{R}) \cap Y$ is finite? As I said earlier irrational numbers contain even uncountable compacts. | |
| Jul 13, 2017 at 16:42 | comment | added | Wlod AA | (Formally), you WERE right. It was another nasty typo. Now it's FIXED -- now it's $Y$ (twice, on the right hand side). A good pedagogical computer should give me a tutorial in avoiding typos. | |
| Jul 13, 2017 at 16:39 | history | edited | Wlod AA | CC BY-SA 3.0 |
A persistent typo. Now finally fixed.
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| Jul 13, 2017 at 15:33 | comment | added | Victor | But isn't the right-hand side of your big union formula just $X$ itself? | |
| Jul 13, 2017 at 4:58 | comment | added | Wlod AA | By the way, absolute value stands here for cardinality (as is common in set theory, e.g. $|\mathbb Z| = \aleph_0$. | |
| Jul 13, 2017 at 4:55 | comment | added | Wlod AA | It's not a typo. When I mean a multiplication than I say so: $\cdot\ $ (I write "cdot"). Most of the time I simply treat a blank space as a separator, there is no need of any "," most of the time. To me unnecessary commas "," are a nuisance, are eyesores. (Please, let me write this way). | |
| Jul 13, 2017 at 0:17 | comment | added | Victor | I am sorry, I still didn't get your example. I think it still has another typo: $\{x,y\}\cap Q=1$, not the product of $x$ and $y$? Also, please check your formula for Y proving that it is a countable union of finite sets - it does not seem right. Also irrational numbers contain uncountable compact sets, take for example a complement to a union of neighborhoods over every rational. This union can be of any small measure as possible. | |
| Jul 11, 2017 at 19:06 | comment | added | Wlod AA | BTW, I had a trivial but nasty typo, now fixed. | |
| Jul 11, 2017 at 19:04 | history | edited | Wlod AA | CC BY-SA 3.0 |
a trivial but nasty typo fixed
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| Jul 11, 2017 at 19:03 | comment | added | Wlod AA | @Victor, they are not in $X$. (BTW, simply the horizontal line would be enough, but it's not in $X$). | |
| Jul 11, 2017 at 18:59 | history | edited | Wlod AA | CC BY-SA 3.0 |
a trivial but nasty typo -- an omission of cartesian exponent (square)
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| Jul 11, 2017 at 18:50 | comment | added | Victor | What happens with the sets $\{(t,0):\, t\in J\}$ and $\{(0,t):\, t\in J\}$? Are they in $X$? If yes, their union is the required $Y$. Or, perhaps, you meant $\{x,y\}$ not $\{xy\}$ in your formula? | |
| Jul 9, 2017 at 16:38 | comment | added | Taras Banakh | and the map is open, not just quotient! | |
| Jul 9, 2017 at 16:34 | comment | added | Wlod AA | My example has its modest advantage, it is metric separable (i.e. a kind of small). | |
| Jul 9, 2017 at 16:17 | history | edited | Wlod AA | CC BY-SA 3.0 |
doubts (about the competition :-)) erased
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| Jul 9, 2017 at 15:58 | comment | added | Taras Banakh | By the way, open maps between completely metrizable spaces are compact-covering (this follows from the 0-dimensional Michael Selection Theorem). So, the space $X$ in the example of @Wlod-AA is not (and cannot be) complete (unlike to the topological sum of all convergent sequences which is completely metrizable). | |
| Jul 9, 2017 at 15:26 | history | edited | Wlod AA | CC BY-SA 3.0 |
a LaTeX "error" finally fixed
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| Jul 9, 2017 at 15:17 | history | edited | Wlod AA | CC BY-SA 3.0 |
LaTeX typo
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| Jul 9, 2017 at 14:57 | history | answered | Wlod AA | CC BY-SA 3.0 |