# User talk:Usagiop

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Cheers! prime mover (talk) 20:09, 7 May 2022 (UTC)

## Preimages

FWIW $f^{-1} \sqbrk U$ as opposed to $\map {f^{-1} } U$ is the preferred notation for preimages. Caliburn (talk) 10:39, 21 May 2022 (UTC)

OK, thanks! Usagiop

added that essential note of clarifxion to page in qn --prime mover (talk) 13:45, 21 May 2022 (UTC)

Please, when implementing an internal link, enter it without underscores. Otherwise it makes maintenance irritatingly more difficult.

Thx for co-ops. --prime mover (talk) 19:38, 21 May 2022 (UTC)

OK. Tanks for a lot of improvements. Usagiop

## General structure of definition pages

We have standardised on $\mathsf{Pr} \infty \mathsf{fWiki}$ to use if and only if on all our definition pages.

Is there a solid mathematical reason why at any stage this may not apply? It would make for some complicated reasoning if for any reason if and only if does not apply in some circumstances.

And while I am about it, "any" is ambiguous. Defining a quantifier as "for any $A$ in $B$" or whatever could be interpreted to mean "there exists" as well as "for all".

You will find this in our house rules: Help:Editing/House Style/Linguistic Style#Any

There may be discrepancies between how you are used to presenting mathematical arguments and how they are presented in $\mathsf{Pr} \infty \mathsf{fWiki}$ standard style. We take pride on internal consistency of presentation. Hence you may find that if you were to change pages so as to look how you like it, they may well be changed back again.

Thanks for your cooperation. --prime mover (talk) 05:07, 23 May 2022 (UTC)

OK, sorry I just did not know about the if and only if convention. I am going to follow the convention. Just to be sure, you are trying to formulate all definitions by if and only if, right?

For example, Definition:Embedding (Topology) should be rewritten as follows?

$f$ is an embedding (of $A$ into $B$) if and only if the restriction $f {\restriction_{A \times f\sqbrk A }}$ of $f$ to its image be a homeomorphism.
No, just those that are shaped the way they are shaped. Stuff that's already there does not need to be changed unless it is wrong.

Also thanks for any. OK, I will try to do it better. Usagiop

## FA

Most of the FA pages are done with normed vector spaces for now. My intent is to come back to generalise (eg. locally convex spaces) later. This'll be done by adding new separate proofs, since the spaces being normed of course simplifies the presentation massively, and so is worth treating separately. Before I came, sometimes the Hilbert condition was unnecessarily included, and there are a few remnants of that I haven't got to yet. Just saying this to avoid the hassle of going through all the FA pages tagging that it can be generalised. I think I will be using Introduction to Banach Spaces and Algebras by Allan and Dales for some of the more general theorems, (the other books I have only deal with NVSs) but I've only just picked it out so I'm not sure what it has yet. You're welcome to offer any input of course and I'd appreciate having people to proofread my FA work. Caliburn (talk) 10:46, 29 May 2022 (UTC)

I may add that here we are trying to make maths accessible to anyone interested. This means that even though a lot of results could be described at the level of topological spaces, it is still worth repeating same results in metric spaces, normed vectors spaces etc. Not everyone understands or wants to read the most general abstract version of the theorem there is. Hence, instead of asking for generalization I invite you to write up equivalent statements in those more general spaces and then create a disambiguation page where you would host all the incarnations of the concept.--Julius (talk) 07:16, 31 May 2022 (UTC)

## Krull's Theorem

Leave it alone for the moment, will you? I told you I'm working on this.

OK, sorry but I did because you wrote feel free to change everything back the way it was. Don't worry, I will leave it. Just please be aware, if you only consider countable chains, the proof is formally wrong. --Usagiop
I was being sarcastic because I was beginning to feel pissed off. --prime mover (talk) 04:57, 30 May 2022 (UTC)

## Signing messages

BTW, you should sign your talk page messages with ~~~~. I'm glad to have another pair of hands on board for analysis. Caliburn (talk) 17:39, 30 May 2022 (UTC)

Thanks. I did not know how to sign. Usagiop (talk) 18:50, 30 May 2022 (UTC)
there's a button on the edit tools panel at the top of your edit pane that looks like a signature. You might want to use that. it keeps everything consistent. --prime mover (talk) 19:19, 30 May 2022 (UTC)
OK--Usagiop (talk) 19:28, 30 May 2022 (UTC)

## Categories

Hey bro, you might want to look at how we use categories here. Definition categories are for definitions. Those are pages whose namespace is Definition. Also, we try to put things in as deep a level of category as the possible, otherwise everything ends up at the top level and it becomes difficult to find. See the recent discussion on the talk page. --prime mover (talk) 05:07, 6 June 2022 (UTC)

OK, thanks. I'm trying to follow it.--Usagiop (talk) 21:47, 6 June 2022 (UTC)

## Save for myself

### Proof of measure is subadditive

For each $n\in\N$, let:

$\ds A'_n := A_n \setminus \bigcup _{i=0} ^{n-1} A_i$

Then, by (SA3) and (SA2'):

$\forall n\in\N : A'_n \in \Sigma$

Furthermore, by construction:

$\bigcup _{n \in \N} A'_n = \bigcup _{n \in \N} A_n$
$\sequence {A'_n} _{n \in \N}$ are pairwise disjoint
$\forall n\in\N : A' _n\subseteq A_n$

Therefore:

 $\ds \map \mu {\bigcup _{n \in \N} A_n}$ $=$ $\ds \map \mu {\bigcup _{n \in \N} A'_n}$ $\ds$ $=$ $\ds \sum _{n \in \N} \map \mu {A' _n}$ (2) of definition of measure $\ds$ $\le$ $\ds \sum _{n \in \N} \map \mu {A_n}$ Measure is Monotone

## $p$-adic Numbers
Thanks for your comments on the $p$-adic Numbers. I hope you are now OK with the statement and proof of Sequence is Cauchy in P-adic Norm iff Cauchy in P-adic Numbers. I have struggled with how to express the fact that the rational numbers are a subset of the $p$-adic numbers given the way the $p$-adic numbers are constructed from the rational numbers. This is similar to constructing Real numbers from Rational Numbers or Integers from the natural numbers or rational numbers from integers. The books I am following just simply assume it.
Yes, everything looks correct, formally. But I feel if you say $\Q$ is (identified as) a subset of $\Q_p$ and the norm on $\Q$ is just (identified as) the restriction of that on $\Q_p$, then the statement of Sequence is Cauchy in P-adic Norm iff Cauchy in P-adic Numbers is trivial and superfluous. I would have expected a more explicit proof based on the explicit isometry $\phi : \Q \to \Q _p$ but now I am not sure where we are going. I think for now, it is fine.--Usagiop (talk) 21:35, 12 June 2022 (UTC)