00:00 - 23:0023:00 - 00:00

12:38 AM
Why is the zeroth homotopy set the same as all path components of the space? Since it is defined with a basepoint, wouldn't it be the set of all points that can be path-connected to the basepoint? Thanks!

@pritchard Reread the definition. Isn’t there equivalence classes in there somewhere?

What sucks about equivalence classes $[x]$ is that as soon as their defined the author then on forever writes $x$ instead.

@TedShifrin My definition is $\pi_0(X,x_0)=[S^0,1;X,x_0]$ which is the set of homotopy classes of based maps between the sets, so since $1$ always maps to $x_0$, doesn't it mean what I said above?

12:57 AM
What is a homotopy class?
@I'manalienImaneaglealien i sure hope not. The author should be fired.

pilloried.

Tortured with glass under his nails.

made to eat the entirety of a well done steak.
i may have gone too far.

@TedShifrin The set of all maps homotopic to one another. In this case with relative homotopy, where 1 is mapped to $x_0$.

1:00 AM
Ted Shifrin, can you briefly answer "why study Algebraic Topology"?
I have a book on my digital shelf but am hesitant to start it up

my answer was, because you need something other than analysis on your qualifying exam syllabus.
but it's cool and it works. it's a lot of fun. treat it as a black box if you must.

Well, it's neat because of it's relation to 3/4D spacetime

So all the points in each path component are equivalent.

Yes but wouldn't there only be one equivalence class then, being the set of all points path connected to the basepoint?

You don’t do basepoint unless the space is path-connected.

1:06 AM
Do you mean my definition is not correct?

I’m not used to that definition. Keep your base point. Where does the other point of $S^0$ map?
Then ask when two such maps are homotopic.

All right, you've convinced me. I will study AT now ;)
I'm starting out with Vick's "homology theory" book

Do you know differential forms? Start there with deRham cohomology.

I know a little bit, but I should review from Tu's undergrad text
on manifolds

Asking when closed forms are exact ties in much more to physics, Maxwell, etc.

1:16 AM
Nice. Physics would be great to know

bott and tu is a good book.

more of
I like Szekeres' "a course in modern mathematical physics.. groups, Hilbert space and differential geometry"

that's a good book too. i never owned it but i did check it out of the library.

I will try not to invent any nuclear weapons - only flying cars.

why not compromise, and invent some flying nuclear weapons.

1:19 AM
I like how AT was the precursor that started up homological algebra
and that HA then applies to all other fields (usually)

AT is the virus and HA is the symptoms.
2

Disease "vectors"
I'm wondering how long before I will be dropping some cohomological logic bombs on MSE
I promise to compute all the homology groups that I can !!! :>
With great power comes great responsibility

2:13 AM
@I'manalienImaneaglealien. Europe still have unique taste

1 hour later…
3:23 AM
@robjohn thanks a lot professor Rob :) I never thought about that recursive sequence problem in the way you have considered. Fantastic solution and generalization :)

3:54 AM
tfw one author makes a (mathematical) claim that you don't think is true, and then you find another author do the same
so either i'm missing something obvious or two different people came to the same error without any obvious reason

semiclassical, one time i had a paper rejected, citing a paper whose result was false, if you followed the reasoning or considered even a single example in my paper.

we had to write a letter to the editor saying, uh, this reason they cited for rejecting our paper is wrong, see example 1.2, and this is even more reason to publish the paper.
they published it.

nice
for context, the second person asserting it did so on MathOverflow here: mathoverflow.net/questions/128772/a-regular-polytope/128809

i didn't see a retraction in the journals that had published the false results.

3:57 AM
for the life of me I can't see how these polytopes can possibly be regular

this is out of my expertise. but this stuff does happen.

(the place where I saw it originally only asserted it was regular for the (2,2) case.)

it happened at a very inconvenient time in my life, i was trying to run up my publication count and maybe get an academic career going. months matter in this, in terms of what you put on your CV.

yeah. it's just jarring when you see something false(?) confidently asserted in passing

it was, pardon my french, excremental to have someone stomping on my result when they didn't have a leg to stand on.
so i said f- it and lived off my girlfriend for a year. and now i am able to support the two of us and our child doing something else.
i don't mean to lure people out of academia, there is a lot of unfair stuff that happens in my business. it's just different stuff and not so much tied up with people's personal identities.

4:02 AM
yeah, i hear you

Lure? Are we going fishing?

nah. i'm just waxing frustrated over not knowing if i'm wrong or other reputable people are :P

Other? :)

specifically 1) an author I know who claimed a certain polytope is regular, and 2) Robert Israel doing the same in a MathOverflow question

Your use of “other” suggested that you too are reputable.

4:05 AM
ah

Israel seems to be dependable in my experience.

agreed
so i'm in a stuck position of not knowing how it can be true, but also having good reason to believe the people who say it's true
i've left a comment so hopefully he'll get back to me

robert israel is a very high-quality participant in this discourse.
better than we deserve, on average. which is not to say that he can't be wrong, but i haven't seen him be wrong.

right

Robert Israel has been a high quality contributor since I knew him on sci.math

4:10 AM
Well, in fairness, have you seen the instances where I’ve been wrong?

yeah, he goes back to sci.math. i was dumb on sci.math

I know not sci.math.

It was a usenet math group.
I was there until I started here.

is usenet still around

It moved to Google groups and I don't know if it still exists
@Semiclassical I forget when sci.math moved to Google, but I left there about 10 years ago.

4:15 AM
which, given google's track record, doesn't really surprise me

my monocle drops in alarm.

@Semiclassical yeah, I have a copy of the last post I made on sci.math, and I can't find it searching on Google.

newsboards getting deleted is something that irritates the hell out of me

with all the junk that is on Google, they get rid of the content of decent newsgroups. The largest post I made on sci.math is probably dwarfed by one image posted here.
sci.math being all text
don't get me wrong, there were cranks there, too.

4:32 AM
yeah, but it was a community. wheat and chaff

indeed. Robert Israel was a very serious contributor there.

2 hours later…
6:23 AM
I understand what the question wants me to do. However, I would like to find out if there a clever way or a pattern that can be observed to find the sum the of $\sum \alpha^n$? I brute forced the $\sum \alpha^2$ and $\sum \alpha^3$. But I think the question wants me to observe a pattern or something, but I don't see it. By $\sum \alpha^3$, I mean $\alpha^3 + \beta^3 + \gamma^3 = (\alpha+\beta+\gamma)^3-3(\alpha+\beta+\gamma)(\alpha\beta+\beta\gamma+\alpha\gamma)+3\alpha\beta\gamma$

7:13 AM
Nevermind, figured it out. Seems instead of expanding out $(\alpha + \beta + \gamma)^n$ to get an identity, you were supposed to make use of the cubic equation. I.e. $\alpha^3 = 2\alpha^2 - k$ and from there it's easy to find $\alpha^3+\beta^3+ \gamma^3$, $\alpha^4+\beta^4+ \gamma^4$ and $\alpha^5+\beta^5+ \gamma^5$

5 hours later…
12:15 PM
1

The following is an Example 6.7.8 in Spanier AT If $C^*$ is the relative Alexander presheaf of $(X,A)$ (with some coefficient module $G$), the kernel of $\alpha:C^*\to\hat{C}^*$ is $C^*_0$ (the locally zero functions). To show that $\alpha$ satisfies condition (b) (which is gluing property), let...

Google doesn't know the terminology 'relative alexander presheaf'

3 hours later…
2:51 PM
in ZF, is $\{A\} = \{x: x = A \lor x = \varnothing \}$ an exhaustive definition of $\{A\}$? I'm trying to figure out if $x \in \{A\} \iff x = A$ is a tautology

3:39 PM
$\{\emptyset,A\}$ is not the same thing as $\{A\}$

oh right

well, it might be, if A is emptyset :)

but if $\varnothing$ is a subset, don't $A$ and $\varnothing$ share at least one element, $\varnothing$?

no?
emptyset doesn't share elements. that is not in the nature of emptyset

3:44 PM
haha, alright thanks

4:04 PM
Like your daughter? Doesn’t play well with others?

very much so.
today the cat was attacking the reflection of sunlight from my wife's watch on the wall.
my daughter saw it and joined in.

how are the logic laws called? (something with M?)
moore or so

there's a couple called de Morgan's laws

ah yes

de morgan wrote a very funny book, 'a budget of paradoxes,' compiled from his reviews of the work of what we might call 'cranks,' and correspondence with same. it's on the project gutenberg website and in a dover hardcover from 1910 or so. he doesn't just poop on them. when he sees original ideas he praises them for what they are worth.
his father in law was obsessed with the idea that negative numbers did not exist. that comes up a few times.

4:11 PM
how did his father substract?

i think he would only do it some of the time.

@SAJW euclid-style technique I guess

5-3, [drake thumbs up], 5-7 [drake turns away]

geometrical substraction

i used to have a habit of putting bills of large denomination that would be hard to pass at normal stores in my copy of that book. then i forgot about the habit. a few years went by and i opened up the book and $400 fell out. 4:14 PM that's quite a hourly better than any interest a bank could give :D If I had known, I would have borrowed the book! for a few years my now in-laws had a habit of giving me$100 on my birthday. they stopped that once i married their daughter.
haha

They were bribing you to disappear?

probably? it didn't work. i can only appear.

I have to learn geography, history and politics (of my country), not in detail, but for example from which year to which was ww1 or ww2

4:16 PM

those are fairly easy questions to answer, although they do vary from country to country, if you measure duration by your country's involvement.

or the fall of the berlin wall

is that even history? i remember that, so it wasn't history.

geography is the biggest hassle, europe is too big haha

it is productive to consider one at some point becomes history

4:19 PM
but will books be written, that's the question.

i remember when the soviet union began to disintegrate. we were camping in the middle of nowhere and my dad gave me a quarter to go down and buy a newspaper. it was the failed coup attempt in the USSR.
i took one look at it as i brought it back and said "dad, you might want to read this."

the august coup :O

Furthermore, I should hone my reading abilities.

NVM got it!

The best technique I came up with to understand a sentence(correctly) is to translate it. But you can't do that always, so it still sucks.
maybe the sailing from Christoph Columbus is important too.

4:26 PM
i think that's the last time i remember getting news from an actual newspaper.
SAJW, there's a rhyme, in 1492, columbus sailed the ocean blue. that's one date for you.

4:46 PM
you can also remember that it's the year the Reconquista ends, which explains why morocco isn't part of the iberian peninsula

5:27 PM
Hello, everyone. I'm convinced I've finally found something that I can use for computing quotients in general, however, I just have one question now: what is the formal name of the algorithm or process by which you find the largest element common among a number of discrete sets?
Took me long enough to realize that you can compute $\frac{x}{y}$ in binary by computing the quotient of the individual powers of two, then finding the largest integer that is less than or equal to all the quotients.
err I suppose in this case, less than or equal to the smallest quotient.
Eh, something like that, actually.
A simple example: $\frac{7}{5}$. $5 = 2^2 + 2^0$. $\frac{7}{4} = 1$; $\frac{7}{1} = 7$. The largest value common to the quotients 1 and 7 is 1, therefore the quotient of 7 and 5 is 1.

If $n = a\cdot 2^b$ ($a$ is odd) is given natural number and $f:\Bbb N\to\Bbb Z$ is a function given by $f(n) = b$ (i.e. counts the number of $2$ in the unique factorization in to a prime number). Then $\sum_{k=1}^n f(k)\leq n-1$.
I want to prove this by induction. Base case is clear and I consider the case $n$ (and suppose the statement is true up to $n-1$, $n\geq 2$).
$\sum_{k=1}^n f(k) = \sum_{k=1}^{n-1}f(k) + f(n)\leq n-2+f(n)$ by induction.
I need to conclude $f(n)\leq 1$ which is false in general.
I need to approach in different way. Could you help?
Minor observation is $f(n\cdot m) = f(n)+f(m)$

Minor observation about that observation: that property applies to logarithms which implies f is a logarithm.
This is supported by the fact that we're working in the context of exponential functions.

5:42 PM
@AMDG if $f$ is continuous, then it is a multiple of the logarithm (which one might call a logarithm to a different base)

Right, and it also fits the description for the mapping from N to Z.

Ok so I need to prove $f(n!)\leq n-1$
I don't want to use that analysis fact

It's easy to prove graphically. And I believe that's the most my brain can do at the moment.
So I'm afraid this is the limit of my help, if it can even be called that...
Though if I had to guess, you could probably compare the rate of growth for each $f(n)$ and $n - 1$ to determine that analytically.

@AMDG however, since these are defined only on integers, it does not need to be any logarithm. We could even define $f(p)=1$ for all primes $p$ and get a function that is not a logarithm.

Yeah, hence why at the last moment I wrote "implies" instead of "means" because of the uncertainty.

5:49 PM
Ah, I see they have defined $f(2)=1$ and $f(p)=0$ for all other primes

I think it's a simple induction problem

Sadly, I only have like two brain cells functioning right now, and I must dedicate them all to efficient computation of quotients.

if the sun is shining one should weigh less, I conclude
the mass however stays roughly the same of course

Well that certainly is a heuristic.
The weight will be less on one side of the earth and greater on the other side.

@PeterJohn Look at Legendre's Formula
That will show that $\nu_2(n!)\le n-1$

5:54 PM
@robjohn How?

@PeterJohn Look at the definition of $\nu_p$

@robjohn I mean I read it. From that formula, how can I conclude $\leq n-1$?

Do you know that $\sum\limits_{k=1}^\infty\frac1{2^k}=1$

Is $\lg(x)$ the convention for $\log_2(x)$?

@robjohn Yes but it only shows $\leq n$

5:57 PM
Think a bit and you will see that it has to be $\lt n$
we are only ever taking a finite sum of $\sum\limits_{k=1}^N\frac1{2^k}\lt1$
and $\lfloor x\rfloor\le x$
put all of this together

@robjohn Thanks

did you get $\le n-1$?
$$\sum_{k=1}^\infty\left\lfloor\frac{n}{2^k}\right\rfloor\lt n$$

From $\lfloor x\rfloor+\lfloor y\rfloor\leq \lfloor x+y\rfloor$, $<n$.
@robjohn But I believe there is an easier way not using that formula

why is the night shorter than the day >:|

@SAJW because it is spring or summer?

6:05 PM
nobody can sleep long enough

Actually this problem was introduced as an induction problem

That has to do with the natural properties of the sun as a light source or lamp and the earth represented as a sphere.

@SAJW Not if you live closer to the poles!

'course, if you define it strictly by the hours of the day, then there are 12 hours for the night and 12 hours for the day.

@TedShifrin to the south pole at this time. It's even more exaggerated near the north pole.
@SAJW Near enough to the north pole, there isn't even any night, now.

6:17 PM
ireland has about 6 hours of night at the moment.
of course, the opposite story in winter...
i like both

Here in Florida, the amount of sunshine doesn't change too much being so close to the equator.

6:35 PM
I imagine the insects and alligators love it ;-)

i'm surprised at the level of seasonal variation around here. it is less than what i grew up with but people who say LA doesn't have seasons are overlooking a lot of obvious signs.
it doesn't have months of crummy weather, is what they're saying.

6:51 PM
Ok, so assuming what I have here is a valid algorithm for computing quotients, then the theoretical optimum will be O(n) to compute the power of two quotients, O(lg(n)) to compute the comparisons. On Zen 2, that should translate to 1n cycles for n concurrent shifts, and 6n cycles for n concurrent compares.
In theory that means a theoretical optimum of 7c for a single integer divide compared to 41c for div/idiv.
So it would be only $2 \frac{2}{3}$ times slower than multiply.
We can use an n-bit LUT to get us closer to this theoretical optimum, accounting for cache misses, data location (icache is optimal), cache line, and fetch size (32 bytes iirc).
To remove the cache misses, and in consideration of the 32 bytes that are pulled from main memory every time there is a cache miss, we can make a 5-bit unconditional jump table to keep us in icache and our data to operate on at $\lceil \frac{64}{5}\rceil = 13$ as the practical software optimum implementation.
Then our value $n$ here can be computed from $13$ as $\lceil\frac{13}{8}\rceil = 2$ assuming you can operate on eight bytes at a time per 64-bit SIMD register of your choice.
@robjohn Since you asked for me to mention it, I figure now is a good time to mention it "officially" and ask what your thoughts are. I'll have to respond later, though, as we'll be going to a farm for groceries right now.

7:13 PM
@AMDG most of what I've seen you mentioning are divisions with small quotients. How does it work with things with larger quotients, say $\frac{2^{63}-1}{11}$?

7:43 PM
@robjohn Thanks for the reply. Why is it necessary to specify uniform little-o-ness?
In a message you wrote earlier:
> 8 days ago, by robjohn
> @schn well, I mean that for different $u$, the convergence of $\frac{f(h)}{h^m}$ to $0$ might be altered. The limit may be $0$, but the rate of convergence might change. It depends on the context how and if this affects the result. That is why it is difficult or impossible to create a calculus for little-o.
Take for instance the Taylor expansion. Are you saying that one needs to know the rate of convergence of $\frac{f(u,h)}{{(hu)}^m}$ to $0$ for every $u$?

@schn it all depends on the application to which it is being applied. There is no way to make a generalization that will be true in every case. Some cases require a uniform convergence and some don't.
As I said before, look at actual cases where little-o is being used before trying to generalize things.
it looks simple, and in definition it is, but it can be tricky to use in some situations.

How can one be sure though this problem is one of those applications where a generalization will not work?

7:58 PM
Use the definition. work from there. $f(x)=o(g(x))$ near $0$ if $\lim\limits_{x\to0}\frac{f(x)}{g(x)}=0$
that is the extent of the definition. That there is only one variable, and often there are more in a given problem, leaves a bit to be handled by the author.
it is a tool, and one often needs experience or training to use a tool
that is why I say to work examples.

Thanks for sharing some viewpoints, @robjohn. I'll get back to you :)

8:24 PM
@robjohn I think we all agree this is the chatroom mantra!

Sorry!!! Can someone help me with this question? Thanks.
0

Question: Let $P$ be the stationary transition probability matrix of the Markov Chain $\{X_n:n\ge0\}$ which is irreducible and every state has period $2$. Further, suppose that Markov Chain $\{Y_n:n\ge 0\}$ on the same state space has transition probability matrix $P^2$. Both the chains are assum...

8:50 PM
Hi, I was trying to figure what’s the dimension of 2 by 2 hermitian matrices consider as a vector space over C. If these 2 by 2 hermitian matrices are considered a vector space over the reals I know the dimension is 4 when considered as a vector space I think their dimension should be 3 because we will have one constraint that off diagonal terms are complex conjugates of each other. But when I tried to find a basis for these hermitian matrices considered as a vector space over C, I couldn’t
Can anyone help me figure out a basis for hermitian matrices considered as a vector space over complex numbers.

Hello, any idea why $for ~ i \leftarrow 1~ to~ (n-1)$ takes $3n-1$ operations? I guess assignment to 1 is 1 op, comparing to $i<(n-1)$ takes another 1 op and finally $i++$, but $i++$ is supposed to take 2, so the total should be $4n-1$. Is not it please?

@Avra Do you assign to $1$ each loop?

@robjohn. I got it! It seems $i++$ takes $2(n-1)$ and $i<n$ takes $n$ since we have to run the loop one additional time.
I got confused about bound popular issue
Thank you

9:10 PM
@Shashaank What makes you think this is a complex vector space?

9:28 PM
@Shashaank Can the two dimensional real vector space with basis $\{(1,1),(i,-i)\}$ be realized as a complex vector space?
think of $(1,1)(1+i)$

9:40 PM
@TedShifrin Ok I see what you mean, H_n is hermitian but iH_n isn’t hermitian. So it won’t be a vector space in the first place
@robjohn I suppose you meant “think of (1+i, 1+i)”. Well that is just equal to (1+i)(1,1)+0(i,-i). Sorry what did you want show there, I couldn’t get. Can you give a further hint or something
@robjohn Also I am confused how do you have as a basis vector (i,-i) when you are considering a real vector space. The thing inside the “(“ and “)” should be a real number but “i” is an imaginary number which is not in R. So how is it a basis vector in the first place….

10:15 PM
@Shashaank The two dimensional real vector space with basis $\{(1,1),(i,-i)\}$ has the element $(1,1)$. This cannot be realized as a complex vector space because if we multiply this element by $1+i$, we get $(1+i,1+i)$ which is not in the space.

@robjohn $11 = 2^3 + 2^1 + 2^0$. $\frac{2^63 - 1}{8} = 2^60 - 1$. $\frac{2^63 - 1}{2} = 2^62 - 1$. $\frac{2^63 - 1}{1} = 2^63 - 1$. $2^60 - 1$ is the largest integer common to all of these quotients, but we can analytically determine that it is not the answer because this value multiplied by 11 is greater than the dividend, hence the issue. The brute force approach would be to decrement this coefficient and test to see if it is less than or equal to the dividend.
This is primarily why I asked what the formal name of the process or algorithm of finding the largest common integer (which I will define as best I can) is, as that is effectively what I'm doing here.

@Shashaank The complex numbers can be considered as a real vector space with the basis $\{1,i\}$. The basis vectors do not need to be in the base field.

I'll begin by assigning a mnemonic for brevity: $\text{greatest common integer} \sim \operatorname{gci}(A, B)$. Let $\operatorname{gci}(A, B) := \operatorname{gcd}(A, B) \text{ for } A[i],B[j] \in \mathbb{N} \text{ where gcd is greatest common divisor.}$

10:35 PM
Is $log(n +n^3)$ of class $log(n)$ please?
i.e. $O(\log{n})$

Can't you answer that yourself? What algebra should you do here with $\log(n+n^3)$?

It's harder and harder to tell these days.

The problem seemed trivial so I was wondering if there was something wrong with the statement
but after I added a solution in the comments it got deleted :/

I find such statements very hard to parse, so I wouldn't even try.

10:40 PM
Oh nvm it got undeleted I think I jumped the gun :/

Someone posted an easy question about geodesics on the $n$-sphere and deleted the question after I'd made some obvious comments that he/she had not previously understood. I think that was more a matter of embarrassment, but we never know now.

@TedShifrin. You are right! Just look for fast verification. It's $log(n)+ log(n) +log(\frac{1}{n} +n)$?

yes I think it was the same here :/

So I see it's $O(\log{n})$

10:41 PM
I'm not quite following, @Avra. $\log(n^3+n) = \log(n^3)+\log(1+1/n^2)$?

@TedShifrin. geodesics is covered in calculus 3?

Why should that have been calculus 3?

I just saw surfaces and distance between points, so I concluded cl 3

Oh, I see, you wrote $n^3+n = n^2(n+1/n)$. I think my way is better.
You might be thinking of a different question. The one I am talking about didn't last too long.

@TedShifrin. The question was to prove that it's $O(\log{n})$

10:44 PM
I understand.
$\log(n^3)+\log(1+1/n^2) = O(\log(n^3)) = O(3\log n)$.

@TedShifrin. I am embarrassed now too

LOL

You just cut through and took the biggest term
So embaressing
I will delete it if I can
can not delete it

Sometimes easy questions (with embarrassingly easy solutions) belong here to help other people. The problem is that most people don't search at all.

@TedShifrin. Maybe laziness ?

10:48 PM
Yes, or just trying to get a quick (cheating) answer :P

@TedShifrin. You are right, at least there has to be attempt

According to justice, we ought to give as much effort as the asker has made effort.
Well, if the asker has put no effort, then I am obliged to put zero effort. Anything more is generosity at that point.

@AMDG. 100% Nice philosophy

There are so many people who want to earn tons of rep by answering every question they can, independent of effort made.

Justice is a universal and fundamental principle that manifests itself in many ways. :)

10:53 PM
I'm not much of a believer in justice these days. Despite having been vaccinated months ago and having taken great care for a year and a half, I believe I'm now the proud owner of a breakthrough case of COVID. Just went for a test. I am so pissed off at the selfish, stupid assholes.

Hope you feel better soon and it's just the flu

@TedShifrin And then there are those who put an unjust amount of effort into their answers or questions.

I don't feel that horrible (thanks to the vaccine), but I don't think it's any flu.

Like SE is some kind of formal peer review site where every nook and cranny and i and t must be dotted and crossed.

There are a few questions for which I spent literally days figuring out answers. Obviously, those questions interested me.
But they get very few upvotes because they're nontrivial questions.

10:55 PM
@TedShifrin. What topics? Math?

Yes, math. One was a line integral over an implicitly-defined curve. The other was a differential systems (differential forms + geometry) question.
I'm sure there are others I've spent nontrivial time on, but those stand out in my mind (because in both cases I got a little help from a colleague or another).

@TedShifrin. I hated implicit term from Calculus 3

The inverse and implicit function theorems are among the pillars of mathematics.

I am scared of everthing that has implicit in it now :/

@TedShifrin Hm, well, case in point: if you received a vaccine, then it should protect against what it vaccinates against. If it does not protect against what it vaccinates against, then it is not a vaccine, but a farce.
There's your justice in principle applied here :)

10:57 PM
Well, protection isn't 100% guarantee when mutants are running rampant.

Exactly, which is why I rely on my own immunity than vaccines.

Mutation is a venerable part of biology.

@TedShifrin. I have been with 95 year uber driver who got the vaccine. He told me that that he felt nothing b/f and after vaccine !

It's quite curious how seemingly volatile microorganisms are.

Well, you're among the selfish, AMDG. I've had cancer and have serious heart disease, so I hope you're enjoying helping to kill me off.

10:59 PM
They're so plastic and malleable.