Mathematics - 2023-07-17

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Ted Shifrin

00:01

You’re just a yak man, @Xander.

Xander Henderson

@TedShifrin Yakkity yak.

(Don't talk back.)

Ted Shifrin

00:28

Back into your rucksack.

shintuku

00:54

$$\sum_I f(r_i)\alpha_{1\cdots n}^{i_{1\cdots n}} + \sum_J f(r_j)\alpha_{1 \cdots n}^{j_{1\cdots n}}= \sum_{(I,\mathbf 0) \cup (\mathbf 0, J)} f(r_t)\alpha_{1\cdots n}^{t_{1 \cdots n}}$$

right?

$I,J$ are multiindexes, $(I,\mathbf 0)$ is the multi index like $I$ which has zero entries afterwards, and $(\mathbf 0,J)$ is the multi index like $J$ which has zero entries before its first nonzero entry

so what I mean is that $(I,\mathbf 0) \cup (\mathbf 0, J)$ counts through all of $I$, then counts through all of $J$, but doesn't do both at once

i.e., in $(I, \mathbf 0) \cup (\mathbf 0, J)$, there is no entry $(x_1, \cdots, x_n, y_1, \cdots, y_n)$, only $(x_1, \cdots, x_n, 0, \cdots, 0)$ or $(0, \cdots, 0, y_1, \cdots, y_n)$

maybe there's some more natural notation to say this?

shintuku

01:16

forget the above, made a mistake in setting up the multi indexes

proper setup would have been $\sum_{(I, \mathbf 0)} + \sum_{(\mathbf 0,I)} = \sum_{(I, \mathbf 0)\cup(\mathbf 0, I)}$

shintuku

01:37

whole idea is inconsistent, starting from zero heheh

冥王 Hades

I had cheese added to my ramen. It tastes amazing, and the texture is so much better. Did I discover something new

Ted Shifrin

01:52

Yuck. What kind of cheese?

Ramen with soyish broth is not like Italian pasta with parmigiano reggiano.

冥王 Hades

It didn’t have any broth it was mostly dry

Ted Shifrin

02:09

Not what I’m used to for ramen. What cheese?

leslie townes

hades: i've done that with parmesan

Ted Shifrin

And sushi with pecorino romano.

leslie townes

02:37

make your favorite brand of chicken ramen and throw a handful of white cheddar cheez-its in there.

it's an old ted shifrin family recipe.

Ted Shifrin

03:33

Barf.

7 hours later…

冥王 Hades

10:22

@TedShifrin just shredded mozzarella

冥王 Hades

10:43

user image

Soumik Mukherjee

11:26

Both will be under a lot of pressure while fighting.

sunny

16 hours ago, by Ted Shifrin

@sunny That is a total mess. You need to say this: Suppose $f_n$ are continuous and $f_n\to f$. If $f_n\to f$ uniformly, then $f$ is continuous. This is the statement which you want the contrapositive of — the opening hypotheses remain outside.

Hey @TedShifrin, could you just clarify if by your first occurrence of $f_n\to f$, you mean $f_n\to f$ pointwise, right?

3 hours later…

geocalc33

14:07

I

would like someone to look at my purported proof

it's always good to get a second set of eyes on anything

if you're available to look it over respond with "I'm ready to look it over"

nevermind

I found a crucial flaw in the argument.

Jakobian

15:10

I'm afraid I wouldn't know what you'd be talking about

:)

geocalc33

15:29

I think you would actually. It's a very straightforward "proof." But it's wrong of course.

just can't figure out what's wrong

Jakobian

you sure? I don't have any experience with PDE or manifolds. I avoid all things differential

Astyx

Anyone know anything about elliptic surfaces?

I'm trying to understand the Dynkin diagrams associated to singular fibres

Specifically I do not understand why some irreducible components get a factor of 2/3/4/5/6 in front of them in the case of fibres of type _*

Shaun

16:15

Hi :) Please would someone give some feedback on my question below? It was downvoted just now and I'd like some idea of why and of how to improve it.

0

Q: Verifying my proof that $Y\subseteq B_X(n)$ implies $B_Y(m)\subseteq B_X(mn)$.

The Details: Let $G$ be a group and let $X\subseteq G$. Let $n\in\Bbb N$. Recall that for $H\le G$, $${\rm conj}_H(X)=\{ hxh^{-1}\mid x\in X, h\in H\}.$$ Define $$B_X(n)=\bigcup_{k=0}^n\underbrace{{\rm conj}_G(X)^{\pm 1}\dots{\rm conj}_G(X)^{\pm 1}}_{k\text{ times.}} $$ Note that the idea here is...

group-theory solution-verification

PNDas

In an infinite dimensional Hilbert space can 0 be an eigenvalue of a compact operator?

Shaun

It's the second downvote of the question. I don't get either of them.

I think I did all I could to make it a high quality question. If you have any suggestions on how to make it a better one, please let me know.

Jakobian

@PNDas why not? The operator $T = 0$ is like that

PNDas

Ah non-zero compact operator. Sorry

Jakobian

It doesn't really matter if it's zero or not, what matters is that this is a projection

just consider any projection $l^2\to l^2$ of finite rank

something like $(x_1, x_2, ....)\mapsto (x_1, x_2, ..., x_m, 0, 0, ...)$

PNDas

16:34

A statement says: "Consequently, the theory of compact self adjoint operators implies that all the eigenvalues of S are real, positive and..."

They have proved that $S$ is compact, self adjoint and $\langle Sf,f\rangle\geq0$

I don't understand how they get "positive".

I thought self-adjoint gives real. The inequality says that eigenvalues are non-negative.

So compact must be giving the positiveness.

Jakobian

$S = 0$ satisfied all of those

perhaps they are using the French definition of positive

as in non-negative

yeah I think positive here means $\geq 0$

PNDas

But the author wrote: $0<\lambda_1\leq\lambda_2\leq\cdots$

Jakobian

did they wrote that? I didn't see it anywhere

PNDas

I got it. The author wrote $\langle Sf,f\rangle\geq0$ but in that theorem, we actually get $\langle Sf,f\rangle>0$

@Jakobian The author of the book I was reading.

Jakobian

I see. It's hard to help remotely like that

PNDas

16:46

But thank you for the help.

sunny

17:21

Is a function from $\mathbb{N}$ to $\mathcal{F}$, where $\mathcal{F}$ is the set of all continuous, real-valued functions, a well-defined function? In other words, a function that maps a natural number to a continuous-real valued function.

The reason for the question is we can denote a sequence of real numbers $(x_1,x_2,\ldots, x_n,\ldots)$ as a function $f:\mathbb{N}\to \mathbb{R}$ belonging to the set $\mathbb{R}^\mathbb{N}$ (this last notation is simply the generalized Cartesian product, i.e. $\mathbb{R}^\mathbb{N}=\{f:\mathbb{N}\to\mathbb{R}\}$). Now, suppose I have a sequence of continuous, real-valued functions $(f_1,f_2,\ldots,f_n,\ldots)$. How do we denote the set of all those sequences? Is it $\mathcal{F}^\mathbb{N}$?

shintuku

@sunny yeah why not

sunny

@shintuku cool :) I was a little unsure

shintuku

well defined here depends on the explicit definition however

and also it will just not be a bijection of any sort

e.g. $n \mapsto n$, $n \mapsto x+n$ are two such functions @sunny

sunny

ok

shintuku

but uh it's probably a more serious consideration to check whether the set of all sequences of real functions is actually a set

not a trivial question, it sounds like a pretty big set

you can probably ignore it by posing your question adequately, if you're working with pointwise/uniform convergences

sunny

17:38

hmm, ok. I should have probably added a domain of all the functions, like $\mathcal{F}(A)$, where $A$ is the domain of the cont., real-valued functions

shintuku

@shintuku rephrasing: you most likely can and should ignore it by posing your question appropriately

Jakobian

18:13

@sunny wdym by continuous real-valued functions?

But to answer your question, $X^\mathbb{N}$ is the set of all sequences of elements of $X$ i.e. $(x_1, x_2, ...)$ where $x_i\in X$. Formally such sequence is a function $f:\mathbb{N}\to X$

there is no disctinction between a sequence $(x_1, x_2, ...)$ and a function $f:\mathbb{N}\to X$ defined by $f(i) = x_i$. They are literally the same

For finite tuples when introducing set theory, we define a pair $(a, b)$ for example somewhat differently, but ultimately once we have functions we can again say that those are just functions $f:\{0, 1\}\to X$

shintuku

@Jakobian let $C \subseteq \mathbb R$. suppose by continuous real valued function we mean a function $f:C \to \mathbb R$ s.t. for all $c \in \mathbb R$ we have $\lim_{x\to c}f(x) = f(c)$, is the set of all sequences of these functions a set?

wait i think I got the quantifier for $C$ at the wrong place

2 sec

Jakobian

That's a somewhat sketchy definition

冥王 Hades

I love chocolate

shintuku

the function $f$ satisfies the proposition $P$ if there exists a set $C \subseteq \mathbb R$ such that $f:C \to \mathbb R$ and for all $c \in C$ we have $\lim_{x \to c}f(x) = f(c)$. is the set of all sequences of functions satisfying $P$ a set? at Jakobian

missing a condition on $C$, two sec

Jakobian

I hate chocolate. But I ate some vegan ready-made food and it was good. Rice and carrot, with some cicer and a bit of sauce of some kind.

@shintuku is this discussion necessary

I'm sure sunny just meant continuous functions on some fixed set like $\mathbb{R}$ or $[a, b]$ or other kind of interval

shintuku

18:25

no it's super not necessary that's why I suggested they avoid this heheh

Jakobian

well, if we understand it as all real-valued continuous functions $f:X\to \mathbb{R}$ then those don't form a set

if we allow $X$ to be any topological space

shintuku

yeah, i'm thinking of a decent condition on $X \subseteq \mathbb R$

well say for a simpler problem we're talking about $C[0,1]$

is the set of all possible sequences of functions in $C[0,1]$ a set?

Jakobian

of course

I'm just avoidant of notation $\lim_{x\to c} f(x)$ when $f$ is defined on an arbitrary subset of $\mathbb{R}$

shintuku

right that makes sense

Jakobian

Note that $\lim_{x\to c} f(x)$ is defined by demanding that $0 < |y-c| < \delta\implies |f(y)-L|<\varepsilon$ and for arbitrary sets $C$ we can just add that $y\in C$

now say $c$ is an isolated point of $C$

shintuku

18:33

yeah, need some conditions to get rid of pathological domains

Jakobian

then this implication will always hold no matter what $L$ is

it's not a problem with the domain, but using limits in the first place

well, limits of the form $\lim_{x\to c} f(x)$

If you were to use sequences instead nothing like this would be an issue

or just use the standard epsilon-delta definition of continuity

$|y-c|< \delta \implies |f(y)-f(c)| < \varepsilon$

@Jakobian this is a nice definition to have for calculus where you won't really consider a continuous function which won't be defined on some kind of interval etc. but maybe not the best in general

Shaun

2 hours ago, by Shaun

Hi :) Please would someone give some feedback on my question below? It was downvoted just now and I'd like some idea of why and of how to improve it.

0

Q: Verifying my proof that $Y\subseteq B_X(n)$ implies $B_Y(m)\subseteq B_X(mn)$.

The Details: Let $G$ be a group and let $X\subseteq G$. Let $n\in\Bbb N$. Recall that for $H\le G$, $${\rm conj}_H(X)=\{ hxh^{-1}\mid x\in X, h\in H\}.$$ Define $$B_X(n)=\bigcup_{k=0}^n\underbrace{{\rm conj}_G(X)^{\pm 1}\dots{\rm conj}_G(X)^{\pm 1}}_{k\text{ times.}} $$ Note that the idea here is...

group-theory solution-verification

Jakobian

didn't you already get feedback?

> Please use the space in your post wisely. It is really tiresome to have to scroll as if one is reading those recipe webpages that have unnecessary padding before one gets to the actual content. ;)

> Sorry, but again your theorem is an obvious statement with correct notations. I agree with Pedro 100%.

shintuku

at Jakobian, i see. thanks for the comments

Jakobian

> @Shaun It does, but it is not the answer you want: what you need is to use better notation which will allow you to bypass the step where you ask us if the proof is right, since no hand waving will happen.

They seem to be criticizing your style of post, and claim it's obvious if you simplify your notation and stop hand-waving

I can see why would someone downvote your question

especially someone coming in and reading those comments

I'm of course, giving an opinion as an outsider

well, now that I had some delicious food, I probably should make some coffee

also you've got to admit that ultimately it's a pretty elementary calculation

冥王 Hades

18:54

user image

Fried rice^^

geocalc33

19:04

anyone good with..metrics?

$(\Bbb R^2_{\gt 0},g)$ with $g=2 \sqrt{\frac{r}{s}}K_1(2\sqrt{r s})drds$

I'm looking to understand this space a little bit.

$r,s>0$ as well. As for $K_1$

https://mathworld.wolfram.com/ModifiedBesselFunctionoftheSecondKind.html

Shaun

Thank you for the feedback, @Jakobian.

Jakobian

@冥王Hades amazing

geocalc33

19:30

actually I meant to write the product metric $$ dt^2=(2K_1(2\sqrt{r})/\sqrt{r})dr^2+(2K_1(2\sqrt{s})/\sqrt{s})ds^2 $$

$dt^2=g(r)dr^2+g(s)ds^2$

this tells me that this space has 0 curvature

iirc

$$dt^2=\frac{2K_1(2\sqrt{r})~dr^2+2K_1(2\sqrt{s})~ds^2}{\sqrt{s}}$$

I like this one though.

Jakobian

19:49

Correct me if I'm wrong, but any closed discrete subspace of $[0, \omega_\alpha)$ is finite, right

Since if it were infinite then letting $\alpha_1 < \alpha_2 < ...$ we'd have $\sup_n \alpha_n\in D$

well I guess I do need $\omega_\alpha$ to have uncountable cofinality

冥王 Hades

I got a fine yesterday for honking unnecessarily. I was stuck behind an awfully slow Toyota Prius and, in frustration, I honked telling him to move it.

I was fined

Jakobian

Even if $D$ is infinite, we can demand that $\alpha_1 = \min D$, $\alpha_n = \min D\setminus \{\alpha_1, ..., \alpha_{n-1}\}$

Astyx

chat.stackexchange.com/transcript/message/63993266#63993266 got it

Jakobian

Then if $D\neq \{\alpha_1, ...\}$ we'd obtain that $\sup_n \alpha_n < \omega_\alpha$ which again would contradict that $D$ is closed and discrete

so $D$ needs to be at most countable

so the extent of ordinal spaces $[0, \omega_\alpha)$ is $\aleph_0$

So it's not only that $D$ is countable but also that it's order isomorphic to $\omega_0$

kind of an interesting property

Actually no this isn't really interesting, $\omega_0$ is the only infinite discrete ordinal in the first place

actually maybe it is interesting because here $D$ inherits the topology from being an ordered space

@Jakobian it must be club in that case of course. So the only discrete closed sets are countable clubs and finite sets. The former doesn't exist when $\omega_\alpha$ is of uncountable cofinality

well instead of $\omega_\alpha$ I can just put any ordinal $\alpha$

user10478

20:20

If the limit as a circle's radius approaches infinity is a straight line, what is the line's slope?

Xander Henderson

@user10478 Depends on how the limit is taken.

leslie townes

dunno, but note that "slope" is coordinate dependent. even the same actual, non-limit-of-a-circle, standard-issue line will have a different slope in different coordinates.

Xander Henderson

Oh, that too.

I had assumed that we were just working in a given Cartesian plane.

leslie townes

yeah, where the question would still be, i guess, which circles and what point.

Xander Henderson

Exactly.

user10478

20:23

Would it be more accurate to say an infinite radius circle is a family of lines rather than one line, given by y = Cx where C is an arbitrary constant in a differential equations sense?

Yeah I'm thinking Caresian plane.

Xander Henderson

None of what you've said makes sense to me.

leslie townes

it would be even more accurate not to say "infinite radius circle" at all.

if you have a family of circles with a common tangent line, that line has a well-defined slope, but maybe you wouldn't call it the slope "of an infinite family of circles"

Ted Shifrin

@冥王Hades Damn right! Good!

leslie townes

or maybe you'd say something like, "'infinite radius circle' is just my funny term for 'line' and here perhaps are other aspects of this funny naming system that i define while leaving other aspects not defined"

冥王 Hades

@TedShifrin You don’t understand. I feel the need, the need for speed.

Xander Henderson

20:25

I also question the initial statement: "The limit as a circle's radius approaches infinity is a straight line." It is possible to construct a sequence of circle's who's limit is, in some sense, a straight line. But that is a very different statement.

user10478

If I said "infinite radius circle in a limiting sense" would that help?

Ted Shifrin

@冥王Hades They should throw the book at you!

@user10478 No.

Xander Henderson

@user10478 How are you taking the limit? (which was my original comment)

leslie townes

user: you might be getting at the idea that there are lots of points on any given circle, and depending on which ones you consider in a family of circles, you could get different "limiting behavior" (putting in quotes as we have not given this a definition yet) of the family.

user10478

So in my mind, I'm thinking of a geometric animation, where one point on the circle remains fixed, and everything else expands outward.

Ted Shifrin

20:28

There is no geometric limit. The limiting curvature is $0$, so that says line. But …

leslie townes

it's not particularly unusual, or circle-specific, for a limit to depend on the thing that you're taking the limit of. e.g. you could consider the x-axis as some kind of limit of the family of circles that have centers (0,R) and radius R. you could consider the y-axis as some kind of limit of the family of circles that have cneters (R,0) and radius R.

冥王 Hades

@TedShifrin They barely reach my shoulders Ted.

leslie townes

so, different limiting lines, and different slopes, associated with different families of circles. maybe all that suggests is that how you think of a line as a limit of circles depends on which circles you start with.

Ted Shifrin

If the circles are all tangent to a fixed kine, then … you win.

user10478

So depending on which point on the circle you hold in place, you end up with a different line. Is that the idea?

Ted Shifrin

20:29

You need more than a point tixed.

user10478

How many points remain fixed? It seems to me that if you fix more than one point and try to expand a circle you necessarily deform it.

Ted Shifrin

Read my comment about same tangent line at the point.

user10478

Okay, I think I get the problem, if I want to truly say I'm in the Cartesian Plane I have to treat multiple shifted lines as different. So the point on the circle I fix when I expand it lets me set the slope of the line, and the position of the circle in the plane lets me set the y-intercept of the line.

But I still only fix 1 point I think?

冥王 Hades

20:47

user image

Jakobian

I dislike this

wrong attitude

冥王 Hades

21:45

@Jakobian it’s not what you think it means. It’s a jab at Intel, the company, for recently manufacturing dead-on-arrival CPUs

GratefulDisciple

@冥王Hades The crime is not putting meat in it, like the Italian Arancini al ragù, which I would prefer any day over the Japanese Yaki Onigiri.

geocalc33

22:17

wow - been 6 years already

Xander Henderson

@冥王Hades I think that something you have maybe not understood is that a large number of folk here are extremely annoyed by your childish memes, and would prefer that you just stop. This is a semi-professional, semi-focused space. Some of your communication habits might be more welcome at... I don't know... 4chan?

冥王 Hades

@XanderHenderson A simple “please stop the memes, it’s a bit annoying” would’ve done the trick.

But you’re Xander.

shintuku

4chan is so 2000s

Xander Henderson

@冥王Hades I've previously asked you to knock it off. It doesn't seem to stick.

shintuku

22:32

a more modern reference might have been discord

冥王 Hades

@XanderHenderson I don’t remember but had you asked again I would’ve stopped even if temporarily

shintuku

2/10, unaesthetic communication, dated references

geocalc33

So...no memes?

shintuku

no 4chan, geocalc. no 4chan

冥王 Hades

You know I can’t be mad at him, I used to be like this too as a mod in discord. Berating whoever I like. It’d be hypocritical of me to criticize something even I’ve done

S Brian

22:34

@shintuku lol

it's difficult to be a person of tolerance, because when one tries to act intently in kindness, negative feedback stings a lot more

geocalc33

I mean, 0 memes in the chat would be very refreshing honestly

7

S Brian

and then, after that memes, so we are refreshed from having no memes?

an infinite loop of memes and no memes

shintuku

no memes? what's next? no jokes? and then what? no capital letters? bye bye punctuation? gee, i wonder when was the last time this happened

S Brian

Sehr schrecklich.

We will reduce to nothing, a collapse, and then from nothing, we would rebuild the Mathstackexchange chat till we reach a point where people start posting memes again

CroCo

23:02

Hi everyone,
I'm reading Elementary Linear Programming with Applications by B. Kolman and R. Beck. The following definition caught my attention.

Definition. Let $\boldsymbol{x}_1$ and $\boldsymbol{x}_2$ be two distinct points (or vectors) in $\mathbb{R}^n$. The line determined by $\boldsymbol{x}_1$ and $\boldsymbol{x}_2$ is the set of points
$$
\{ \boldsymbol{x} \in \mathbb{R}^n | \ \boldsymbol{x} = \lambda \boldsymbol{x}_1 + (1-\lambda) \boldsymbol{x}_2, \quad \lambda \text{ real} \}
$$

In this case, I think the vectors should be linearly independent, but I don't understand what distinc…

Ted Shifrin

23:16

No.

Think of the line through the heads of the two vectors.

CroCo

Hi Prof. @TedShifrin

This is my problem actually, the heads.

Ted Shifrin

Why a problem?

CroCo

How can we draw a line between the tips if they point in the opposite direction?

I think there is no line because the vectors are linearly dependent.

right?

but again I don't understand the exact definition of distinct points in this context.

Ted Shifrin

No. Distinct just means they are not the same point.

Xander Henderson

I think that thinking of these objects as "vectors" may be an unnecessary confusion. Just think of them as points.

Ted Shifrin

23:27

You can draw a line through $(1,0)$ and $(-1,0)$.

CroCo

@XanderHenderson exactly my point. Using vectors here confuses me.

Xander Henderson

If you have two distinct points in the plane, then you can construct a line which passes through both points. That line is the set $\{x \in \mathbb{R}^n : x = \lambda P_1 + (1-\lambda) P_2, \lambda \in \mathbb{R}\}$ (among other parameterizations).

Ted Shifrin

Except I don’t know how to add points.

This is worse.

Xander Henderson

@TedShifrin Yeah, I don't really like the "adding points".

Ted Shifrin

You cannot!

In affine geometry there is no sum, only difference.

Xander Henderson

23:30

What I really want to do is define a subtraction of two points to give a vector.

CroCo

@TedShifrin true but is this what is meant between tips? In this case, the line passes via the tail as well.

Xander Henderson

And then consider the set of points $x + \lambda v$.

@CroCo Nothing about your definition suggests that this is impossible.

Ted Shifrin

Still not defined.

Xander Henderson

@TedShifrin I can use other notation if you like, but it is entirely reasonable to define $x + v$ to be the action of $v$ on $x$, i.e. the translation of a point $x$ along $v$.

CroCo

@XanderHenderson If $x_1 = -x_2$,we get $x=2\lambda x_1 + x_2$, right?

Ted Shifrin

23:33

It’s fine as long as the “point” $x$ is a vector.

CroCo

But the definition no longer holds

Xander Henderson

@TedShifrin No. $x$ is a member of some space, which is being acted on by elements of some group.

Jakobian

@XanderHenderson well, I thought this was self-deprecating hence why I commented. But it's true that this is a semi-professional space

Ted Shifrin

Have you never worked with parametric equations of lines using vectors, Croco?

I won’t continue this, Xander.

CroCo

I've read it long time ago, not sure if this will help me understand this definition better.

Ted Shifrin

23:36

Of course. That’s what this is.

CroCo

I'm happy with the points though.

or should I forget the notion of vectors here and pay attention the points.

Jakobian

Worst moment of my life. I unironically thought that the long line is perfectly normal

Ted Shifrin

You need vectors.

It’s perfectly abnormal.

Jakobian

@冥王Hades I agree that Xander was a bit harsh especially since your memes aren't that bothersome and are easy to ignore. But I wouldn't compare his status as a moderator on this site to a mere discord mod.

What Xander said is true though

冥王 Hades

@Jakobian I wasn’t drawing a comparison but I don’t wanna probe this further

CroCo

23:41

@TedShifrin It's interesting how a banal definition can become a hassle.

:<

Ted Shifrin

It’s not a hassle. You have a lack of understanding.

This is at the beginning of every reasonable linear algebra course.

Xander Henderson

@CroCo I think you have it backwards. It is easy to think that an idea is intuitive and obvious, but when you start trying to use that intuition, it turns out that things are actually kind of complicated, and you need more precise definitions.

And that the definitions make it much, much easier to both understand and communicate.

It just takes time to learn the definitions.

CroCo

Is there any thing prevents me from assuming $x_1=-x_2$ in this definition?

Xander Henderson

@CroCo No.

And if $x_1 = -x_2$, you end up with a line through the origin, which is parallel to either of the vectors you started with.

CroCo

My stupid mistake is now clear to me. There is a feeling of humiliation in my heart.

Xander Henderson

23:53

@CroCo Meh. It's part of learning.

I wouldn't feel shame about it.

I doubt that anyone here is judging you.

CroCo

@XanderHenderson Your words are warm and welcome.

thanks.

冥王 Hades

Huh, they ran out of chocolate stuff dorayaki

Wonder why

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