User talk:4C~enwiki

Welcome

Hello 4C~enwiki, and welcome to Wikipedia. Thank you for your contributions. I hope you like the place and decide to stay. Here are a few good links for newcomers:

I hope you enjoy editing here and being a Wikipedian! By the way, you can sign your name on Talk and vote pages using three tildes, like this: ~~~. Four tildes (~~~~) produces your name and the current date. If you have any questions, see the help pages, add a question to the village pump or ask me on Talk page. Again, welcome! You 20:55, Jun 4, 2005 (UTC)

Kremer prizes

Yes, there was also third prize for MacCready's Bionic Bat :) Vuvar1 T 20:53, 18 Jun 2005 (UTC)

Unfortunately, there is all what I know, and there are only a few information in website: results.Bye! Vuvar1 22:04, 18 Jun 2005 (UTC)

Thank you for your Poland-related contributions

Hello 4C~enwiki! Thank you for your contributions related to Poland. You may be interested in visiting Portal:Poland/Poland-related Wikipedia notice board, joining our discussions and sharing your creations with us.

--_{Piotr Konieczny aka Prokonsul Piotrus | talk} 20:35, 15 August 2006 (UTC)[reply]

drup/smal/klein

Cześć 4C, czy zamierzasz dokonać jeszcze jakiejś edycji na pl:wiki? Wielka szkoda, że udałaś się na emigrację i nie chciałaś budować Wielkiego Porządku w pl ;-) (cóż, ten trend wykracza poza granice wikipedii), ale można było jakoś odnależć się w naszej pl:rzeczywistości (podczas Jedynych Slusznych Porządków Miłosz wyjechał, ale Herbert został - także można sobie jakoś radzić ;-) ). Zobacz na Twojej pl:wiki:stronie wybudowano Ci tablicę pamiątkową. Pozdrawiam, Pan Camel 16:52, 15 October 2006 (UTC)[reply]

Topological characterization of Liuoville numbers

Hi,

I commented out the sentence mentioning that almost all real numbers belong to the set $L$ of Liouville numbers. That's obviously wrong in the commonly accepted measure theoretic sense.

Basides, I cannot understand why $L$ is a G-delta set. Would you mind quoting a reference, or mentioning which is its complement and why it is meagre, please?

TIA ale (talk) 17:38, 28 January 2008 (UTC)[reply]

Thanks for explaining me why

L

is a G-delta set. However, I have difficulty in understanding your proof of meagreness of L^c. You assert that L is dense because its complement is meagre. However, your discussion only involves the relevant intersections being G-delta and the corresponding complements F-sigma. My topological concepts are possibly a bit rusted, so I apologize if I'm missing some obvious point. As far as I can recall, taking infinite intersections of dense sets may result in a non-dense one. ale (talk) 07:56, 2 April 2008 (UTC)[reply]

Yow wrote

If I am not wrong maybe this phrase from the article confuses you: "...and it follows that is a dense G-delta set, since its complement is a meagre set" - if so, it should be rephrased. Maybe "... and it follows its complement is a meagre set". Best, 4^@ 11:04, 3 April 2008 (UTC)[reply]

Yes, that's it! I thought the conclusion was the premise and vice versa. Let me try and reword the article's section below striving to make little steps so that even an idiot like me can understand it :-)

Note that I expunge the center of each small interval, which is different from the original. That way, the $U_{n}$ are nearer to the definition, and it is not necessary to subtract $\mathbb {Q}$ afterwards. However, I would still leave out the sentence saying that from a topological point of view it means that "almost all" real numbers are Liouville numbers (commented out in the article's text), because in the usual measure-related sense of almost all, the opposite is true, and that might thus originate confusion.

Thank you for your patience, ale (talk) 10:53, 5 April 2008 (UTC)[reply]

article's section rewording proposal

For each positive integer n, set

{\begin{aligned}U_{n}&=\bigcup \limits _{q=2}^{\infty }\bigcup \limits _{p=-\infty }^{\infty }\left\{x\in \mathbb {R} :0<\vert x-{\frac {p}{q}}\vert <{\frac {1}{q^{n}}}\right\}\\&=\bigcup \limits _{q=2}^{\infty }\bigcup \limits _{p=-\infty }^{\infty }\left({\frac {p}{q}}-{\frac {1}{q^{n}}},{\frac {p}{q}}+{\frac {1}{q^{n}}}\right)\setminus \left\{{\frac {p}{q}}\right\}\end{aligned}}

.

The set of all Liouville numbers can thus be written as $\;L=\bigcap \limits _{n=1}^{\infty }U_{n}$ .

Each $U_{n}\$ is an open set; as its closure contains all rationals (the {p/q}'s from each punctured interval), it is also a dense subset of real line ${\mathbb {R} }$ . Being the intersection of countably many open dense sets, $L\$ is comeagre, that is to say, a dense G_δ set.

Ok, I've replaced that section with the above text.

I searched "almost all" and Oxtoby on Google books, but that landed on a proof of Lebesgue density theorem, where almost all was being used in its measure-theoretical meaning. Thus I left that sentence out... ale (talk) 15:23, 5 April 2008 (UTC)[reply]

Your account will be renamed

Hello,

The developer team at Wikimedia is making some changes to how accounts work, as part of our on-going efforts to provide new and better tools for our users like cross-wiki notifications. These changes will mean you have the same account name everywhere. This will let us give you new features that will help you edit and discuss better, and allow more flexible user permissions for tools. One of the side-effects of this is that user accounts will now have to be unique across all 900 Wikimedia wikis. See the announcement for more information.

Unfortunately, your account clashes with another account also called 4C. To make sure that both of you can use all Wikimedia projects in future, we have reserved the name 4C~enwiki that only you will have. If you like it, you don't have to do anything. If you do not like it, you can pick out a different name. If you think you might own all of the accounts with this name and this message is in error, please visit Special:MergeAccount to check and attach all of your accounts to prevent them from being renamed.

Your account will still work as before, and you will be credited for all your edits made so far, but you will have to use the new account name when you log in.

Sorry for the inconvenience.

Yours,
Keegan Peterzell
Community Liaison, Wikimedia Foundation

21:31, 19 March 2015 (UTC)

Renamed

This account has been renamed as part of single-user login finalisation. If you own this account you can log in using your previous username and password for more information. If you do not like this account's new name, you can choose your own using this form after logging in: Special:GlobalRenameRequest. -- Keegan (WMF) (talk)

10:06, 22 April 2015 (UTC)

ArbCom elections are now open!

Hi,
You appear to be eligible to vote in the current Arbitration Committee election. The Arbitration Committee is the panel of editors responsible for conducting the Wikipedia arbitration process. It has the authority to enact binding solutions for disputes between editors, primarily related to serious behavioural issues that the community has been unable to resolve. This includes the ability to impose site bans, topic bans, editing restrictions, and other measures needed to maintain our editing environment. The arbitration policy describes the Committee's roles and responsibilities in greater detail. If you wish to participate, you are welcome to review the candidates' statements and submit your choices on the voting page. For the Election committee, MediaWiki message delivery (talk) 13:04, 23 November 2015 (UTC)[reply]

old guards

the old guard welcomes the older guardant!

(in case you have returned after the break).

w1k0 *

03:19, 20 January 2022 (UTC)