@Jakobian is this even pronounceable? I sound like a snake hissing trying to pronounce it.

these almost-elegant results feel offensive to mathematical taste: math.stackexchange.com/q/5017951/137524

(+1)

@SoumikMukherjee you should sound a bit like a snake

Since I want to keep my streak of "go to sleep early wake up early" continue, I guess I'll wake up

Happy New Year

@Jakobian Ive been following a sleep late wake up late streak

sleep early wake up late

@SoumikMukherjee that's a sign of health problems

as far as I know

I was joking:) I sleep late wake up late, I should change that to sleep early wake up early

hard to break the habits

yes

My goal is to wake up at 4am, where as currently I go to sleep at 4am

I try to wake up at 6 am

the time between 10 pm and 6 am is time where there's really nobody awake

@Jakobian do you use alarm or wake up by yourself?

alarm of course

its indispensable

I see, I know some people who can wake up by themselves at 4am, I wonder how

@SoumikMukherjee drink a lot of water before going to sleep so you have to wake up to go to the bathroom

I tried that too but I went to sleep afterwards:(

usually it helps to have something to do that early. Say, a deadline at 8 am xD

The key thing is mindset I guess

Does the question mean that $T$ is a topology on $X$?

I dont rly understand the question

As far as I understand, $X$ is a member of $\mathcal{T}$ which is a topology on some other set..?

then what is the topology on $X$ and $X'$, many such can be defined without reference to $T$

@nickbros123 weird phrasing

@nickbros123 I mean it's obvious what this question is supposed to be

@nickbros123 well theres this result that $Y\subseteq X$ is open in $X$ (or is in $T_X$) if and only if every set in $T_Y$ (the subspace topology) is also open in $X$, so there is a restriction

but no, I think the person who wrote that doesn't speak English or is at a beginner stage of learning

unfortunately this person is dead

hes also James Munkres

well basically it means that $X, X'$ are equal as sets, and same for $Y, Y'$

or, that's what it ought to mean

ohh I get it now

I know of this lemma;

> Lemma A family $\mathcal B$ of open subsets of a metric space $X$ is a base of open sets iff for each $x\in X$ and each open neighborhood $U$ of $x$, there exists $V\in\mathcal B$ such that $x\in V$ and $V\subset U$.

@psie first lets agree on what topology you put on $\mathbb{R}^n$

I mean that to be precise I'd like you to give me definition of its topology, of course it'll be Euclidean one

@Jakobian yes. If I knew how open balls "looked like" in $\mathbb R^n$, that is, if they could be represented as Cartesian products of certain sets, then I think I could show this myself. After all, $(a_1, b_1)\times \cdots \times (a_n, b_n)$ with $a_i,b_i$ rational is simply the Cartesian product of open balls with rational centres, but I don't think this Cartesian product is a ball itself, or?

@psie you didn't answer my question

do you know what topology is? Topology on a set/topology of a metric space

@Jakobian well I'm thinking, is there a metric we can put on $\mathbb R^n$ that simplifies showing what I want?

@psie you can't answer a question with a question

it's not a back-and-forth, I'm not trying to argue with you to come to some kind of conclusion, I am trying to agree on definitions

explain your definitions - you still haven't done that

@Jakobian but you already said it'll be Euclidean one, so why do you want me to answer a question you have already answered? Feels kind of weird.

do you think there's just one way to define Euclidean topology on $\mathbb{R}$

but sure, let's define it to be topology generated by sets of the form $(a_1, b_1)\times ...\times (a_n, b_n)$ where $a_k, b_k$ are rational in this case

@Jakobian I feel like it can't be a definition though. How do you know that $(a_1, b_1)\times \cdots\times (a_n, b_n)$ with $a_k,b_k$ rational generates the Euclidean topology? If we were to define the Euclidean topology, I'd say it is the topology induced by the Euclidean metric or any of the $p$-norms.

@psie because it generates it by definition of Euclidean topology

that's the point, if you don't define things then I'll just take the most convenient for me definition which allows me to say basically nothing, but also proves what you want

okay, so topology induced by Euclidean metric?

yes

you might know that sup metric and Euclidean metric are equivalent

and that implies they generate the same topology, so we can take sup metric instead

now show such products of open sets are open first

that is, show that if $U_k\subseteq\mathbb{R}$ are open, then $U_1\times ...\times U_n$ is open in $\mathbb{R}^n$

@Jakobian but this is true for $\mathbb R^n$ equipped with Euclidean metric too, or?

?

didn't you hear me saying that sup metric and Euclidean metric generate the same topology

it's here above your message

@Jakobian is that reasonable? in a nice enough space, every open is cozero, but disjoint opens certainly wouldn't be expected to have disjoint closures?

@Thorgott it's one of the weakenings of the concept of extremally disconnected space

so while I'm not sure what you mean by "reasonable", it would be expected to hold in a space with properties that break connectedness a lot

for instance, the Stone-Cech compactification of the integers might not be extremally disconnected, but its still an $F$-space

the property of being an $F'$-space is the weakest property I know which generalizes extremally disconnected spaces

ah ok, it does sound reasonable like that

I was just thinking that in a space where zero = closed (e.g. metric space), this condition doesn't hold

unless it's discrete, I suppose

no wait, that isn't even true

@Thorgott yeah, it should imply that if a sequence converges to a point, then its eventually constant

so that its sequential coreflection is discrete

topology.pi-base.org/properties/P000167

If $x_n\to x$ and $x_n$ are distinct, take disjoint cozero sets $U, V$ such that $\{x_{2n}: n\in \mathbb{N}\}\subseteq U, \{x_{2n+1}:n\in\mathbb{N}\}\subseteq V$, then $x\in\overline{U}\cap\overline{V}$

of course assuming Tychonoff

So an $F'$-space is sequentially discrete

so if its sequential, it's already discrete

there do exist $T_6$ extremally disconnected spaces which are not discrete though

topology.pi-base.org/spaces/S000111

and perfectly normal spaces are precisely those in which every closed set is a zero set/open set is a cozero set

so even under this assumption, it's not enough to obtain that your space is discrete

so, it's more about being sequential than separation axioms

recently I was also thinking about regular open sets

If $X$ is a topological space, then $\text{RO}(X)$ is a complete Boolean algebra, and if $B$ is a Boolean algebra, then it has its corresponding Stone space $S(B)$, and from Stone space $X$ you obtain Boolean algebra $B$ of clopen sets.

or, I guess $X$ can be arbitrary space, then you can obtain Boolean algebra of clopen sets

so we are dealing with two functors here, one from topological spaces to complete Boolean algebras, and another to Boolean algebras

I think it's a little surprising that you are losing more information by considering $\text{RO}(X)$?

well, Stone spaces don't concern me, its the Stonean spaces

I think I was talking about Stonean spaces before? I don't remember the context

oh I was trying to create some kind of functor but it was all incomprehensible

I really shouldn't post half-assed questions

nothing but whole ass in your questions, please

the converse is always true

w

whole ass question

Is there an "easy" way to see a manifold with constant sectional curvature is an Einstein manifold? (without using Weyl tensor)

In a text I'm reading, the author defines Lebesgue outer measure $\lambda^\ast$ on $\mathbb R^n$ in terms of countable coverings of closed rectangles $R=[a_1,b_1]\times\cdots\times[a_n,b_n]$ whose volume $\mu(R)$ is $(b_1-a_1)\cdots(b_n-a_n)$. So $$\lambda^\ast(E)=\inf\left\{\sum_1^\infty\mu(R_i):E\subset\bigcup_1^\infty R_i, R_i\in\mathcal R(\mathbb R^n)\right\},$$where $\mathcal R(\mathbb R^n)$ is the collection of all $n$-dimensional closed rectangles.

I wonder, what exactly is $\mu$? A measure, a premeasure, something else? Clearly $\mathcal R(\mathbb R^n)$ is not closed under unions nor intersections I believe, so it's not an algebra or a $\sigma$-algebra. So I'm not sure what $\mu$ is.

allow me to challenge the framing of this question, does it matter?

could it just be a notational choice? does it have to "be" something? what would be the significance of it being something

the definition of mu tells you what it is, it assigns a number to cartesian products of closed intervals, why isn't that enough

@SoumikMukherjee last game I've played has 98% accuracy. I'll let you start the procedure

@leslietownes well, if you develop your theory with closed rectangles, how do you prove that $\mu(R)=\mu(R^\circ)$ where $R^\circ$ is an open rectangle?

well in the context you have given, mu is not defined on open rectangles, although lambda^* is

what difference does it make

@psie The choice of open vs closed rectangles really doesn't matter in this case---the interior of the closed rectangle is an open rectangle with the same volume.

i'm not 100% sure what you mean by "develop your theory with closed rectangles" if one is just using mu as an abbreviation for a natural candidate for the measure of a cartesian product of closed intervals, how is this a significant limitation

what is anyone missing by introducing a notation for this vs. not

If you want a name for $\mu$, the author gives it to you: it is the volume of a rectangle.

What book is this, by the way?

nothing prevents you from extending a definition of mu to some broader setting, and nothing requires you to do that

it's just a stepping stone on the way to something called lebesgue measure which more or less doesn't depend on notational choices like this, but also does extend the idea that the measure of a cartesian product of intervals is a product of their lengths

you're basically asking "you said the length of [a,b] is b-a, what about (a,b] and [a,b) or (a,b)" and it's hard to imagine what "real life" concerns would be addressed by going there

or not going there, it's entirely up to you, you can decide what mu means

@SineoftheTime op

Ok, I'll have to think some more on this.

it might help to take n = 1 here, i think the essential issues are present in the one dimensional case

@leslietownes The two dimensional case might be slightly more interesting, as any open set in the one dimensional case is just a union of open intervals. But yeah, figure out the 1D case first.

the key point as i see it is how, whether, why defining mu only for closed intervals [a,b] would somehow prevent that formula for lambda^* from giving rise to something that is worth being called lebesgue measure

and yeah that is present in n = 1

there are a million reasons why R^1 is not a good model for intuition but it will do in this case

@XanderHenderson in the definition of Lebesgue outer measure, what is the reason behind the fact that it doesn't matter that we use open versus closed rectangles? Is it because we can write a closed rectangle as a countable union of open rectangles and vice versa, and measure theoretic operations are preserved by countable unions?

psie that is a highly formalistic way of looking at it, as i think of it, the idea is to extend the basic notion of "the length of [a,b] [or (a,b) or any choice of endpoints really] is b-a" to a larger collection of shapes, or subsets of R that one might not readily think of as 'shapes'

and you're forced into something like what lambda^* turns out to give you, regardless of how you set it up notationally

if you want to preserve whatever standard ideas you have about measure (which might include countable additivity or monotonicity or whatever)

@leslietownes yeah, I realize that probably at the beginning of defining Lebesgue outer measure, one has to simply accept that the volume of (a,b) is b-a and that of [a,b] is also b-a, and then choose one of these types of intervals to develop more theory

yes, i think a kind of vibe prerequisite for lebesgue measure is being fine with [a,b], (a,b] and [a,b) and (a,b) all being declared to have the same size

but which sets you regard as primitives in setting up some sequence of definitions that ultimately gives you lebesgue measure, maybe doesn't matter so much. it is a choice that only matters at the level of textbook drafting

a useful exercise you should do before worrying about anything in R^n is that if [a,b] is a subset of a union of intervals [a_j, b_j], then sum_j (b_j - a_j) >= b - a

@psie There is nothing to "accept". It is a definition.

it's not particularly hard as a matter of showing that it's true, it's maybe hard as an expository matter or organizing your proof writing

@XanderHenderson for me it's a mental surrender :)

By definition the length of an interval $(a,b)$ or $(a,b]$ or $[a,b)$ or $[a,b]$ is $b-a$. That is the natural way of defining length. Then the question is "How does one extend this to any subset of $\mathbb{R}$?"

the exercise i have just indicated is a large part of the "obvious" fact that lambda^* assigns the same value to [a,b] as mu does

@onepotatotwopotato isn't this a straightforward calculation

I think I found a generalization of the notion of boundedness: Given a uniform space $X$, a subset $A$ of $X$ is said to be bounded if the closure of $A$ in the completion of $X$ is compact.

@DannyuNDos Is it useful?

Yeah, when I'm looking into uniform spaces.

@DannyuNDos I don't think that's a good definition

it seems to be equivalent to "$A$ is totally bounded as a (sub)uniform space"

a better definition of bounded would be that if $\mathcal{D}$ is the family of uniformly continuous pseudometrics on $X$, then $\sup_{x, y\in A} d(x, y) < \infty$ for all $d\in\mathcal{D}$

In my definition, for example, if $(X, d)$ is a metric space, then $A\subseteq X$ is bounded in the induced uniformity iff it's bounded in the metric space

I've also seen this property pop up when thinking about various generalizations of theorems from metric spaces to uniform spaces

What if your $d$ is the standard bounded metric?

what do you mean

I mean... Under the standard bounded metric, every subset is bounded.

affirmative

Though, yeah, I admit that I forgot the notion of total boundedness.

Let $G$ and $G'$ be two Lie groups. Is it true that $\dim (G \cap G') = \lvert \dim(G) - \dim(G') \lvert$?

where dimension of a Lie group is defined as the dimension of its Lie algebra

@DannyuNDos I think you might be trying to say that if $d$ is Euclidean metric for example, then $d$ is uniformly continuous as a map $X\times X\to \mathbb{R}$ when $X$ is equipped with $d\land 1$ metric

which is true, and I overlooked that

Yeah, exactly.

@SillyGoose what if G=G'

Yes oopsies.

The statement I mean to ask about is if $\dim(G \cap G') = \dim\mathfrak{g} + \dim \mathfrak{g}' - \dim(\mathfrak{g} + \mathfrak{g}')$ (naive application of vector space dimension rule for this case)

since i think $\text{Lie}(G \cap G') = \mathfrak{g} \cap \mathfrak{g}'$

so for this to make sense, you are assuming G,G' are subgroups of a common Lie group H, right?

@DannyuNDos totally bounded implies my definition of bounded, but I think if you have something like $\ell^2$, then probably the bounded sets are those which are bounded in the metric

@Thorgott That is correct

so are your subgroups embedded or immersed?

I do not know what those words mean. I can perhaps try to contextualize how specific the situation I am dealing with is. I am considering two Lie groups $G, G'$ that are sub-Lie groups of $GL_n(\mathbb{C})$.

@SillyGoose I'm asking what you mean by a sub-Lie group

The definition I use I think is $H$ is a sub-Lie group of $G$ is $H$ is topologically closed w.r.t. the subspace topology on $H$ induced by the topology of $G$. @Thorgott

ok, in that case, do you know what the Lie algebra of $H$ looks like?

@Thorgott usually like $\mathfrak h$ :)

@s.harp BOOOO!

Also, I disagree. You may want to call it $H$, but it is spelled $\mathfrak{g}$.

@DannyuNDos something worth looking into is the notion of a bornology, en.wikipedia.org/wiki/Bornology, so you are not reinventing the wheel. but even then not really because the whole idea turns out to be not as useful as literally any other use you could put an equivalent amount of time into