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00:08
Zero phase shift.
so two wave expressions with different amplitude
but no phase shift?
00:27
Yup. That would be my interpretation.
 
2 hours later…
02:37
Hi
If $M_1$ and $M_2$ have non negative sectional curvature
then $M_1\times M_2$ with the product metric has non negative sectional curvature as well, right?
I don't understand, can't we just pick an orthonormal basis in our product tangent space and then we would get $sec(u,v)=\langle R(u,v)v , u \rangle$ which is always non nehative?
Just do the three cases.
what three cases?
$sec(u,v)= \frac{\langle R(u,v)v,u\rangle}{|u|^2|v|^2- \langle u,v \rangle^2}$ for $u,v$ linearly independent tangent vectors. Can't we always assume $u,v$ are orthonormal?
So what?
So that would mean the denominator is just $1$ and $sec(u,v)\geq 0$
Why?
Just because you say so?
02:51
for which bit? Denominator $1$ because $|u|^2=|v|^2=1$ and $\langle u,v\rangle =0$ and inner product always non negative, no?
Huh? Why?
which bit?
Where are you using all the hypotheses?
that's my issue, i'm trying to see why what i'm doing doesn't make sense
You haven’t used anything yet.
02:54
yes, that I know. There's something I don't understand or confused by, but not sure what exactly
why doesn't my argument show that $sec(u,v)\geq 0$ for any two linearly indepoendent vectors $u,v$?
isn't sectional curvature independent of the vectors you choose?
What?
Do you even know what sectional curvature is?
Only for spaces of constant curvature does it not depend on things.
if $M$ is a Riemannian manifold and $p\in M$ then for any two linearly independent vectors $u,v\in T_pM$, $sec(u,v)=\frac{\langle R(u,v)v,u\rangle}{|u|^2|v|^2-\langle u,v\rangle^2}$ and $sec(u,v)$ is independent of the vectors you choose in $span(u,v)$
How would you write the following Knights and Knaves-type sentence as a logical expression? "Homer claims that he would say Tom is a knave." Homer specifically hasn't said "Tom is a knave" and instead has said "I would say that Tom is a knave", which may or may not be a lie. It's driving me bonkers.
Originally, I had $H \implies (H \implies \neg T) = \neg H \vee \neg T$, but intuitively, that doesn't seem to encapsulate as much information. I may be wrong, but I think the Homer's claim gives us enough information to conclude that $\neg T$.
03:45
one exercise says: find all ring homomorphisms from R to R.
I think that there are only two: 1) the zero map, 2) the identity.
I got no idea what $d(v\cdot v)$ means? Applying product rule for dot product.
because R has only two ideals -{0} and R, so by FIT, homomorphic image of R is isomorphic to either R/{0} or R/R. In the former case, the homomorphism is an automorphism so must be an identity.
In the latter case, we get a zero map (a map that maps all elements in R to element 0).
$\int_0^t d(\frac{mv^2}{2})$ is next level mathematics. So is it true using Riemann sum?
@MethNoob it's differential.
like $d(x^2)=2x dx$ etc.
03:53
@Koro so they replaced dot product with product?
@MethNoob do you know how to differentiate a dot product?
@Koro no but now I see I proved it.
Dot product also follows product rule.
yes :).
Thanks I was so confused.
you're welcome.
04:03
koro: why is the homomorphism an automorphism in the the former case? (being one-to-one, it's an isomorphism onto its range, but why is the homomorphism surjective?)
It is confusing, because there are many notational shortcuts that are typicallly not mentioned.
koro: side note, note also that in many books the zero map would not be a ring homomorphism (because rings are required to have 1 and ring homomorphisms are required to take 1 to 1)
koro: note that all you seem to be using in "R has only two ideals" is that R is a field, but that other fields (e.g. C) do have nonzero, non-identity endomorphisms
complex conjugation being a good example of one
@leslietownes because if $\phi: R\to R$ is a homomorphism with kernel {0} then R/{0} is isomorphic to $\phi(R)$. Since R/{0}$\sim R$, I used equivalence of isomorphism.
@leslietownes In my case, rings can be without unity :).
There are symbols other than R...
copper, what do you mean?
@leslietownes oops, yes.
complex conjugation is clearly not an identity.
thanks a lot Leslie. I have got it now.
I think that's why the solution to this problem said -only 1 homomorphism (the identity) and in doing so they assumed 'R onto R'.
04:24
koro: even if it's assumed to be onto R, why was it the identity?
@Koro I thought you were using two different Rs, but I see you did one with MathJax and one without. Ignore me.
Because then the question is: 'find all homomorphisms from R onto R'. So let f be one such homomorphism, then f(R)=R (because f is onto) and by FIT R=f(R)~R/{0}.
In this case, $f$ satisfies Cauchy Function equation so $f(r)= r f(1),f(1)\ne 0$ for all rational r.
it can be shown that f is continuous at $0$.
that "it can be shown" is the entire exercise, in my view
see the links in math.stackexchange.com/questions/3915406/… (q. yuan's answer) for some backstory on why i might think that
For any x>0 in R, $f(x)=f(\sqrt x)^2>0$ (not writing $\ge 0$ as f is automorphism.)
e.g. if f is only assumed to be a [unital] ring homomorphism from $\mathbb{C}$ to itself, then $f(\mathbb{Q}) = \mathbb{Q}$ but it is not possible to prove that $f(\mathbb{R}) = \mathbb{R}$
04:31
So for any b>a, f(b)>f(a), that is f is increasing.
yes. you have identified a property of R that makes this work
f fixes Q (it follows from homomorphism properties).
now, we can use either Cauchy functional equation argument or just contradiction: if for any x , f(x) is not x then by density of rationals..
wlog, if x<f(x) then there is a rational q such that x<q<f(x). By monotonicity of f and the fact that f fixes all elements of Q, f(x)<q. It follows that f(x)<f(x), contradiction.
we must have f(x)=x for all x in R.
leslie, I understand that. Showing the identity part was actually part of an earlier exercise so I knew that result :).
04:54
Suppose $X \subset \mathbb{R}^{n}$ is a compact set. Suppose $U_1, U_2, \dots \subset \mathbb{R}^n$ are open sets whose union contains $X$. Prove that for some $N \in \mathbb{N}$ we have $X \subset U_1, U_2, \dots U_N$. (Hint: if not for each $k$, choose $\mathbf{x}_k \in X$ so that $\mathbf{x}_k \notin U_1, U_2, \dots U_N$.
So I have to show that every compact set can be contained in a finite subcover. I've been drawing some pics been fiddling with stuff for a good 40mins but can't come up with anything.......ayuda por favor?😢
oh...didn't even see the question architect is in the room....
What definition of compact set are you using @dc3rd?
closed and bounded
closed as in all convergent sequences converge to points in the set
I think you should create sequence $x_n$ in X, and show that it has a limit point.
I see that idea...but I'm struggling to see the contradiction....
let me think for a second...
use openness of those sets
also i presume you note the typo above in the hint (N appears where k may have been intended)
05:03
i noted the typo too :).
@leslietownes reading minds, now?
yes, my mistake......I'm dealing with a neighbour in the next apartment having an acid trip of some sort.........smh.....
what's an acid trip?
2
look, you didn't want any of my acid, so you can't go on my trip and you just have to listen
sorRY
05:05
A walk to get orange juice from the 7-11
best to see that on youtube Koro, than experience it in real life
or may be some acid leakage somewhere
it is 12 am here robjohn.....and not the first one of the night
@robjohn haha
she had quieted down earleir when the cops came, but she's decided to act up again.......maybe some maximum value theorem and topoligy is what would soothe her....
05:07
@dc3rd: there is a textbook also called 'Topology without tears'.
yup I'm very aware of it Koro.
@Koro only continuous surfaces?
no torn surfaces
I have not studied that book yet so I don't know. I'm trying to work my way through Gamelin topology book.
you won't need it if you are doing Gamelin....topology without tears is a very introductory book
05:10
. o O ( tear a piece of paper in two, cry a sea of tears )
I got stuck at understanding why ball centered at origin in R^n equipped with d(x,y)=min{|$x_2-x_1|,|y_2-y_1|$} is a square.
That's where I left it the last time.
squares aligned with the axes
d(x,y)=|x-y| seems very intuitive and balls are 'circles'.
@Koro diamonds
I mean $x,y$ in $R^n$
:(
05:14
oops, I was thinking of $|x|+|y|$
@robjohn oh, i didn't understand the joke.
:D
without tears in the title of the book is probably for 'without crying'
homographs
words that are spelled the same
I called them homonyms.
@Koro those sound the same - oh, homophones
and also spell the same.
05:18
Hi folks. I have a simple question about Second Order Ordinary Differential Equations
Say we have $y''+P(x)y'+Q(x)y=0$
With initial conditions $y(x_0)=y_0$ and $y'(x_0)=y'_0$
@Koro yeah, homonyms can be spelled or pronounced the same
i'm dying of suspense
both homophones and homographs are homonyms
@copper.hat I refuse to go there
Where then do the equations $c_1 y_1 + c_2 y_2 = y(x_0)$ and $c_1 y'_1 + c_2 y'_2 = y'(x_0)$ come from?
I'm sure I'm missing something obvious.
05:21
During school, they were sometimes asked in our exams :).
antonyms too.
If anyone could explain, that would be appreciated
@rb3652 Presumably $y_1,y_2$ are two independent solutions of the ODE?
first of all let's assume P and Q to be continuous.
@copper.hat Indeed. I should have specified.
@rb3652 the linear combination of any two solutions is also a solution
05:24
@robjohn I understand that if $y_1,y_2$ are linearly independent solutions, then so is $c_1 y_1 + c_2 y_2$?
@rb3652 so the general solution is a linear combination of particular solutions.
There is exactly one solution that passes through $y(x_0),y'(x_0)$ at time $x_0$.
The entire space of solutions is given by the linear span of $y_1,y_2$.
@rb3652: what cucap said
05:25
cucap?
copper = Cu, hat = cap
So, $y_1(x_0), y_1'(x_0)$ will have some values and same for $y_2$.
:60581301 We need to iron out our difficulties
so to find the linear coefficients we solve $y(x_0) = c_1 y_1(x_0) + c_2 y_2(x_0)$ and the same for the derivative.
So, let me get this straight: the solutions to $y'' + P(x) y' + Q(x) y = 0$ are equations. In particular, if you can find two linearly independent solutions $y_1, y_2$, then $c_1 y_1 + c_2 y_2$ is also a solution.
05:27
@Ted: I deleted my message thinking it was way out of context. copper to ferrum.
Koro is an alchemist
fehat?
fecap
@robjohn Evidently
@TedShifrin I don't think alchemy is on topic here.
05:29
@rb3652 the ODE by itself (without specifying initial conditions) describes a whole linear space (2 dim) of solutions $y(x)$.
We are given two initial conditions, since this is a second-order ODE, with $y(x_0) = y_0$ and $y'(x_0) = y'_0$, where $y_0$ is just a number and so is $y'_0$. Since a linear combination of the solutions is also a solution, we have $c_1 y_1 + c_2 y_2 = y(x_0)$. And then differentiating this, we get $c_1 y'_1 + c_2 y'_2 = y'(x_0)$. Is that right?
@robjohn few things are
@copper.hat I see, because given two solutions, their coefficients $c_1,c_2$ describe $R^2$, right?
When you select a specific set of initial conditions this selects exactly one of the elements of the 2 dim space.
how can I be deranged with only one personality?
05:30
And so, your initial conditions must fall within the column space $col{[c_1 c_2]}$, I believe.
So, the $y_1,y_2$ are actually a basis for the entire solution space.
$\mathcal{D}(1)=0$
@rb3652 makes no sense
@copper.hat Thanks for clarifying. Is my explanation above correct?
@TedShifrin Ah, as usual.
Where am I off?
@rb3652 this is hard to do in a chat room.
05:32
Hm
We’re in an infinite-dimensional space of functions, so matrices and column spaces need care.
you are trying to find the $c_1,c_2$ so that $c_1y_1 + c_2 y_2$ is the solution you are looking for.
@TedShifrin But I had the understanding that one had all the solutions to a second-order ODE if one had two linearly independent solutions, and could thus solve all initial value problems.
the key point is that the initial conditions completely specify the entire solution.
05:33
@copper.hat Yes, I understand this
Yes, nothing to do with column space.
so, if you can make them match at, say, $x_0$ then they match everywhere
Hm, OK. I see
Well, thank you @TedShifrin @copper.hat for the help. I will solve some problems now.
so you have reduced a problem of matching the whole curve to a single $x_0$.
And writing a matrix with the $c$s as columns ….
05:35
@rb3652 hey! I stayed out of it to reduce confusion ;-)
another key thing that makes this work is that $(y_1(x_0),y_1'(x_0)), (y_2(x_0),y_2'(x_0))$ are linearly independent.
@robjohn I should follow your advice.
@TedShifrin well, you got an honorable mention ;-p
@rb3652 a nice example is $y''+y = 0$, with $y_1=\cos, y_2=\sin$.
 
1 hour later…
06:53
4
Q: Understanding discretized FIRE; how to get from Equation 1 to the recommended algorithm in Bitzek, Koskinen, Gähler, Moseler and Gumbsch (2006)?

uhohIntroduction/Preamble @SusiLehtola's answer to Basics of numerical energy minimization techniques used in molecular dynamics? mentions conjugate gradients, BFGS for energy minimization and Metropolis Monte Carlo (and the) the FIRE algorithm for dynamical simulations. The FIRE algorithm was in...

 
3 hours later…
10:08
quite confusing
I don't know how to expand this if N=2
Quite confused
for i=/=j is making me even more confused
everybody is offline ;_;
Weird sum notation
10:24
@MethNoob first expand the sigma that's inside.
@Koro yeah I did that
Sigma that inside is $\frac{q_iq_1}{r_{i1}}+\frac{q_iq_2}{r_{i2}}$
For $N=2$, you'll get: $\frac{q_iq_1}{r_{i1}}+\frac{q_iq_2}{r_{i2}}$ for the inside sigma. So $\sum_i \frac{q_iq_1}{r_{i1}}+\frac{q_iq_2}{r_{i2}}=\frac{q_2q_1}{r_{21}}+\frac{q_2q_1}{r_{12}}$
$i\ne j$ prevents terms like $x_{11}, x_{22}$ etc.
Ok I will take time to digest it.
first try $\sum_i^N\sum_j^N x_i x_j, i\ne j$ for N=2.
@Koro Derivative part of Zorich Analysis is really good.
10:32
so it will eliminate the term $\frac{q_1q_1}{r_{11}}$ first time and then $\frac{q_2q_2}{r_{22}}$ at second time right?
@love_sodam glad that you liked it :). Did you also notice how existence of $\lim_n (1+\frac 1n)^n$ is shown there?
@MethNoob yes, something like that.
@Koro Oh I see how it works. Thanks for saving my life again B-]
:-)
@Koro Using Bernoulli's inequality you mean. I actually read limit and continuity part selectively. Too many definitions.
@love_sodam yes :).
can anyone please tell me the difference amongst Lang's introduction to linear algebra, Lang's university algebra and Lang's linear algebra?
 
4 hours later…
14:33
@Koro is it similar to this answer? That is, using Bernoulli to show that $\left(1+\frac1n\right)^n$ is increasing and $\left(1+\frac1n\right)^{n+1}$ is decreasing.
15:30
@robjohn the later term is shown decreasing and since its bounded below its limit exists, then by limit rules $\lim (1+1/n)^n = \frac {\lim (1+1/n)^{n+1}}{\lim (1+1/n)}$ exists :).
 
2 hours later…
17:37
0
Q: How to determine equation of a normal to vector parametric curve.

unit 1991If curve is given with $r=r(t)$ vector parametric equation, then equation of tangent line at point $M_0$ corresponding to $t_0$ parameter value is given by $R(\tau)=r(t_0)+$ $\tau$ $r^\prime(t_0)$ where $\tau$ is parameter. Now I understand how we are getting this equation but have question conc...

From what this follows?
18:18
If we have isomorphic fields $\mathbb{Q} \cong F$ then $F$ has an arithmetic mirroring the rational arithmetic. Can you extend $F$ to a field $G$ in such a way that $G$ is not isomorphic to $\Bbb R,$ $\Bbb C$, nor the rational complex numbers?
$\mathbb{Q}_p$.
so you would have $G \cong \mathbb{Q}_p$?
but $G$ wouldn't necessarily be a number field
because no number field is isomorphic to $\mathbb{Q}_p$
Sure. But you didn't specify that.
 
1 hour later…
19:52
To what extent is it appropriate to think of the recurrence relation behind Fibonacci Numbers, and the conditions under which the recurrence relation applies (an in the rules for the rabbit mating game), as explaining the numbers' (would-be) surprising appearances? When one studies the appearances not as numbers, but as conditions under which the numbers emerge, does identifying where to expect the numbers become obvious?
20:13
I don't know what you are asking really.
20:24
I guess another way to ask is, to what extent does math elucidate the mystery of the prevalence of Fibonacci Numbers, versus to what extent is the mystery empirical in nature?
How do you quantify the prevalence?
I imagine this is more to do with biological dynamic than mathematics?
Not sure about quantifying the prevalence. It just seems to be a common talking point that the numbers are interesting because they appear in so many surprising contexts.
I think people just notice is because of familiarity. Numerology.
Ahhh, maybe a combination of that and trying to make math sound exciting to lay people. So you would say mathematicians aren't really surprised at all to find Fib numbers all over the place, because the conditions that produce them are so easy to trigger?
21:06
some of the spotting of fibonacci 'in the wild' is pure numerology based on familiarity, as copper hints. sometimes a great case of en.wikipedia.org/wiki/Strong_law_of_small_numbers
2
@leslietownes I shot several $13$s and a $21$ on a recent Phibonotchy Safari.
@robjohn Did they growl?
no, but they repurred, however
Lately I am seeing a lot of negative numbers in my investments.
21:21
well, just hope for a negative negative.
double negative alert
Yeah right as they say.
The double positive.
we weathered Tromp surprisingly, but Vlad is too much.
I suspect they are related.
There are many similarities, but not being in the US gives Vlad more freedom
21:24
Well, he has implicit support from some big neighbours.
@copper.hat I think they are secretly married
A Stormy relationship I suppose
seems like we were within $\epsilon^2$ of his taking down democracy as it was
@TedShifrin that is true and scary
plenty of co-conspirators still in the game
21:25
My concern is more atomic.
your atoms are governed by big molecules
Here is an odd one, math.stackexchange.com/q/4397497/27978 but the OP says they are not familiar with Riemann integration. How does one do distribution theory without integration?
I think Vlad is reacting to more NATOral forces
Certainly there are few innocent national bystanders in all of this.
Suggesting that Georgia & Ukraine join NATO was an incredibly stupid move.
Lebesgue integration, in fact …
21:28
Indeed, but they could escape with Riemann for the moment.
what course is this person taking without basic integral calculus?
I didn't inquire.
@copper.hat not at all — 14max 13 mins ago
really?!
M17
M17
Pq=n
P, q are prime numbers
P is not equal to q
If p and q are huge prime numbers,
And we only know n in this equation

Is there no direct algebraic solution to find p and q, and there is no solution unless the computer analyzes the factors?
@robjohn I mean about what course they are taking. Perhaps it is self study?
21:31
We’ve told you this numerous times, M17.
@M17 Are you a hacker?
@copper.hat I know, but they are not familiar with Riemann integration?
@robjohn At that point I realised my attempts to guide were wasted.
indeed
Lunch would be a better idea, I think...
M17
M17
21:33
@TedShifrin, I was just making sure, because I read something so I asked
@copper.hat my wife and I were just discussing food
I think one needs to ask both context and background explicitly.
@TedShifrin I suppose I always hope for an 'ah-hah' moment with a nudge.
Perhaps MSE should have a qualification process :-).
This is why I complain when people don’t show us their tool-bag with efforts …
M17
M17
@copper.hat, No, this equation caught my attention
21:36
@copper.hat That might not be too popular, but it would cut down on PSQs
@robjohn I wonder if MSE is the bane of educators around the world.
I should stick to convex, non-ML problems.
Not sure what qualification could entail!
I did respond recently to a question with a comment to the effect that I charge by the hour for consulting. Another commenter upbraided me and, I suppose, flagged my comment.
I should not comment at 4am.
I only bitch about giving me time for free when certain people get ornery and expect compliant service.
sku
sku
Hi All, I have a question. Consider a simple real function f(x). In a certain interval, this function is monotonically increasing BUT you can see a slight concave left followed by a slight concave right.. So the slope is not uniform. How does one describe such a function in mathematical terms? Thank you
21:42
say it has an inflection point?
@TedShifrin Yep, I lose it at that point.
sku
sku
@Ted
if you want to be specific, you can say $f’>0$ and give intervals on which $f”$ is pos/neg.
M17
M17
@TedShifrin, You asked a question here and you advised me to learn things about algebra. In the previous period, I learned many things and watched the videos and researched well and understood my mistake in my question I asked.
sku
sku
@TedShifrin thank you. That must be it. Let me think about something and get back to you. Appreciate it.
21:44
sure. :)
M17
M17
yes
I nean i asked*
Mean
If we know p and q, what are the consequences for that?
I read that it is important for encryption, but there is a question: What if a body or institution finds this, will a lot of data be at risk?
21:59
Probably. So it is unlikely in the near future. Just like playing the lottery.
an awful lot of stuff presently depends on an assumption that it is prohibitively difficult, in general, to factor large numbers (without information about how they factor).
knowing the factorization of any specific large number would not change that much.
M17
M17
Are there results other than encryption
So it doesn't change that in coding, but still large number analysis is useful in mathematics?
@leslietownes
encryption
i think encryption (and all of the fields that use it) is the key application. my focus of the above was that it is general methods of factoring, not any set of specific factorizations, that would put those methods at risk.
it's certainly helpful to be able to compute with large numbers in a lot of fields, but i don't know of fields other than applied mathematics where knowing how large numbers factor is a center of attention.
M17
M17
If we knew p and q, I think that would be useful for mathematics? I read it on Wikipedia
Must be true then.
22:10
what do you mean, knowing p and q? context is missing.
for what? why? this is why i made the remark above, specific factorizations of specific integers are not that important even in cryptography, which depends on general algorithms, not specific numbers.
M17
M17
The idea is that p and q are prime numbers, so factor analysis is more difficult than if an odd number is not prime
When n is a very large number, computers have a limit, if there is a direct solution to deal with factors without algorithms I think this is useful
In pure mathematics I think it is useful
I do not understand your sentence.
M17
M17
I mean, we use the computer to analyze factors, but the computer still has a limit, and it can't analyze very large numbers
sku
sku
@TedShifrin so the function I have in mind was already going from a local minima to a local maxima. In that sense it already had one inflection. This other function basically also starts with local minima to local maxima but in the middle has a concave left and concave right adding two more inflection points. Good. Thanks
Yes, that is the basis on which current encryption is based, as mentioned above.
M17
M17
22:21
The computer does the analysis, and the numbers that need to be analyzed are endless. Whatever developments in computers, the problem will remain
Has this problem been proven to be mathematically impossible without any algorithms?
22:41
It cannot be impossible, just random guessing will solve the problem given long enough. As far as I am aware there is no consensus on the complexity of the problem.
M17
M17
I don't mean to guess, I mean to get to the factors directly without testing a set of numbers in a specific area based on the number
@copper.hat
You need to be more specific about what you are asking, because we are sort of going around and around. You can clearly solve the problem by just iterating. So it cannot be impossible. The only issue is complexity. And as far as I know that is not known.
23:01
@geocalc33 hey mon :)
sup
have you ever completed a field isomorphic to the rationals/
Complete in what sense?
Cauchy?
You will end up with a field isomorphic to $\Bbb{R}$. But in your particular case, your viewpoint may be valuable with the maps you're using
I mean completion of a field isomorphic to the rationals that is isomorphic to $\Bbb Q_p$
23:12
I understand some of it, where are you stuck?
I want to understand how to go through and construct it
I would first understand the standard construction of the inverse limit of $\Bbb{Z}/p^i$ in the category of commutative rings
That will give you ideas and relations with the $p$-adic rationals
There is an infinite number of ways to construct it, but one or a few ways we do it "naturally". Check out inverse-limit construction of $p$-adic integers
All the ways which are an object that is such a inverse limit are isomorphic as rings.
Of course an inverse limit is thought of both as an object $X$ together with maps $\psi_i : X \to X_i$ (or is it the other arrow direction) and as the object $X$ itself alone. Depending on your context.
Those maps $\psi_i$ are called "universal cone".
For any other object $Y$ with a cone to the same $X_i$, there must be some commutative relations happening and a unique map between $X$ and $Y$. That's why the $\psi_i$ are called a universal cone
By that I mean commutative diagrams
an infinite number of them are formed
So when you draw out diagrams, just index by $i$, do only a few of the diagrams and let the reader surmise the consequence of everything being infinite in nature
A @geocalc33 would you like a tutorial session about limits/colimits in a category and then relating that back to the $p$-adic special case?
In other words, I recommend studying other math in order to understand more the specific math that is your goal
Limits / colimits are one of the rare cases the mathematician gets to draw node / arrow diagrams for a good reason

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