but there is another way to construct the Serre spectral sequence by filtering via the postnikov tower, no?

I've never seen this done for a fibration, but that works for the AHSS with fibre a point

there is, but I never learned what that means beyond the idea sounding nice

something something duality of cellular and cocellular

I know I've looked at one of his papers on the subject before, but "looked" did not turn into "read"

This article looks kinda bad

its more like an announcement of results or something more than an actual paper

probably a junk journal, no offense to the authors but full offense to the "editorial board"

@Jakobian how about this one from the same journal ijpam.eu/contents/2013-82-5/4/4.pdf

@leslietownes delightful

@leslietownes Japanese have an interesting perspective to life

found a pdf of “Littlewood’s Miscellany” online, and it’s a really interesting read. I saw the book referenced in the Wikipedia article on “Ross-Littlewood paradox”.

A quote from the introduction: “A good mathematical joke is better, and better mathematics, than a dozen mediocre papers”

In Rudin's Theorem 2.13, he says in equation (18) that the elements of $B_n$ are of the form $(b,a)$ with $b\in B_{n-1},a\in A$, but $b\in B_{n-1}$ is a $(n-1)$-tuple, so the elements in $B_n$ become a $2$-tuple with the first component being an $(n-1)$-tuple, which I guess by the associativity of Cartesian products is the same. Is this a corret interpretation?

well you know what he means :)

whether it is formally correct or whether he's being sloppy with notation there depends on how he concretely defines $n$-tuples, if at all

but up to associativity of cartesian products it's all the same, yes

ok, yes I think he is quietly using the associativity of the Cartesian product there

sshhhhhhh could he use it more quietly please

Let $f:\mathbb{R^2} \rightarrow \mathbb{R}$ with continious partial derivatives of any order at any point. Suppose $p \in \mathbb{R^2} s.t there exists $\delta > 0$ which for every $|t| < \delta$ we have $f(p+te_i)-f(p) \geq 0$ for $i=1,2$ the standard basis vectors. Will this imply that $p$ is a local minimum point of $f$?

no, i don't think it will. there are probably polynomial examples.

I'm amazed you parsed that

or perhaps you just weren't trying to render TeX in the first place

oh, is that messed up? i don't render chatjax

figures, yes it's messed up

try something like x^2 + y^2 - [big number] xy. if you cut the graph with planes like x = 0 or y = 0 you get upward pointing parabolas but if you cut the graph with x = y you get something else.

with p being the origin

If $X\times Y$ are locally contractible metric, is $X$ locally contractble?

how to mathjax large brackets again?

\big[? add gs to increase size

or \left[ and \right]

thx

@monoidaltransform products of contractible spaces are contractible, and you know a convenient neighborhood basis at each point...

if the product of two spaces is an ANR or ENR, is it true that each factor is also an ANR?

I have opened up my old lecture notes from set theory, and the author in there proves that $(0,1)$ is uncountable with the help of decimal expansions not ending with infinite $9$s (aka the diagonal argument). We assume $(0,1)$ to be countable and we write $a_0=0,a_{00}a_{01}\ldots$ and $a_2=0,a_{10}a_{11}\ldots$ and so on. Then we define the real number $b=0,b_0b_1\ldots$ where $b_k=1$ if $a_{kk}\neq 1$ and $b_k=2$ if $a_{kk}=1$.

I'm wondering, does this argument also work if we use that every real in $(0,1)$ has a unique binary expansion not ending in infinite $1$s, say? And we define $b_k=0$ if $a_{kk}\neq 0$ and $b_k=1$ if $a_{kk}\neq 1$? I'm just curious if I can modify it like that.

@psie No, and think about why not.

@monoidaltransform if the factors are non-empty, yeah

@BenSteffan hmm, ok, I must have made a mistake in my definitions then, or? The way the author defined $b$, especially the choice of $2$ in $b_k=2$ if $a_{kk}=1$ seemed like it could just be swapped for say $3$.

With base 2, we only have two digits to choose from, so it gets a little tight.

a little tight indeed

Answer the following: Is your construction well-defined? I.e. does it always yield a number in $(0, 1)$?

its also worth considering, even if this argument doesn't literally work like a 'search and replace', could one make something similar work by adapting "the idea" without being literal

there's certainly nothing all that important about base 10 being base 10

the 'diagonal argument' is a versatile thing that nevertheless has annoyingness built into it by virtue of it framing things in terms of sequences of sequences

@BenSteffan well, if we don't allow binary expansions ending with infinite $1$s, and we have $b=0.b_0b_1\ldots b_k\ldots$, don't we always land in $(0,1)$? From what I'm reading on the web, the diagonal argument applied to showing $(0,1)$ is uncountable should work in any base I think (e.g. the first sentence in this answer).

No. Rethink this.

well, this is disappointing. Rudin says that the set of binary sequences are uncountable implies that the reals are uncountable, and I thought the way one proves the reals are uncountable in this setting is through showing a subset (say $(0,1)$) is uncountable using a base-$b$ expansion argument. Maybe I need to change the interval...

I mean, you can make that work

I don't know what else he would mean. I guess one could use the Cantor set as well, but this hasn't been introduced and is maybe overkill.

there are only countably many reals that have two decimal expansions

rudin is 1000% assuming that his reader is capable of vibing with things like this

not like rudin is any kind of gold standard but if you are reading rudin you had better get used to this

he is very much That Kind of Textbook Author

$$\{0,1\}^\mathbb{N} \longrightarrow [0,2], (a_n)_n \longmapsto \sum_{n=0}^{\infty} a_n 2^{-n}.$$I'm out :)

@psie that map is not injective. However, if you remove all elements of the domain that end in $...0111111....$ then it is injective.

@psie that's what I was thinking

@copper.hat yes :) remove the sequences ending with infinitely many $1$s, and in doing so, we need to change the domain to $[0,2)$ rather than $[0,2]$ (I don't know what I was thinking with the interval $[0,2]$)

because if we remove all the sequences ending with infinitely many $1$s, there's no way to get $2$ if we start the sum at $k=0$

@psie I meant codomain, not domain

@Thorgott Ah I missed some notations sorry

@Ben do you know where in HA Lurie actually explains how to define the cartesian monoidal structure on a category with finite products?

Section 2.4.1

he does it in a rather indirect way

the key proposition is 2.4.1.5

thanks

I'll save understanding for later, for now I just wanna put the results together properly