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00:59
Is the notation $\cdot k:\mathbb Z \to \mathbb Z$ or $\cdot k: (\mathbb Z, +) \to (\mathbb Z, +)$ correct?
For the latter, we consider the two sets as groups
We just want to map $n \mapsto k \cdot n$
I'd even argue if we want to write it the second way we need to include $\cdot$ in the operations?
01:24
I don't like writing $\cdot k$ to denote a map, that's first of all
The notation is fine
other than that, it doesn't matter which notation you use as long as it fits your context
people also sometimes omit the $\cdot$
that one I like decidedly less, but it's not uncommon
for map $f$ defined as $f(x) = x\cdot k$?
yes
@ILikeMathematics No, not really. The multiplication on $\mathbb{Z}$ is uniquely determined by the fact that it extends addition.
01:28
@BenSteffan It'd be better to write, for example, $-\cdot k$
eh
technically you're correct, but this is something I never see in the wild
the notation $\cdot k$ is generally understood
by whom
people working in algebra and/or algebraic topology
@ILikeMathematics no because what actually matters is, is your map a group homomorphism
and the answer is, yes
the multiplication is irrelevant in this context as an operation on $\mathbb{Z}$
including it, say $(\mathbb{Z}, +, \cdot)$, would indicate you actually want your map to be a (non-unital) ring homomorphism
listing operations means that those are the operations that need to be preserved
If you had a map between $(A, f_1, ..., f_n)$ and $(B, g_1, ..., g_n)$ then that would indicate that you have a map from $A$ to $B$ which maps operation $f_k$ to operation $g_k$
say, if $f_k$ is an $m$-ary operation, $T:A\to B$ is your map, then you want $g_k$ to also be $m$-ary, and to $T(f_k(a_1, ..., a_m)) = g_k(T(a_1), ..., T(a_m))$ where $a_i\in A$
but listing all the operations, I don't think that's really that popular, unless there is not a lot of them, and in some popular subject like rings or groups
so you may usually say, this map $T$ is a group homomorphism, or ring homomorphism, and so on
01:43
if anything, we don't usually list the operations for rings and groups either if they're understood
and just write it as a map between sets
usually a group is just $G$ and a ring $R$, not $(G, \cdot, e)$ and $(R, +, \cdot, 0, 1)$
or $(G, \cdot, ^{-1}, e)$, and whatever that would be for rings
for groups, you just need one operation to be able to determine if something is a group
but to write it in the language of universal algebra, to express all the group axioms, you do need inversion and so on
and same for homomorphisms, to determine if something is a group homomorphism, you just need multiplication
Bob
Bob
02:29
Could somebody here, who is good in math, look at my post:
0
Q: How to determine if there is a unqiue solution for this system of equations

BobSuppose that I have to functions whose range is the set of real numbers. Call these functions $f(x,y)$ and $g(x,y)$, I want to find a solution of this system of equations. By a solution, I mean $(x_0,y_0)$ such that $f(x_0,y_0) = 0$ and $g(x_0,y_0) = 0$. Is there a way to determine if a system of...

Thankks
 
1 hour later…
03:49
Can give ideas for a simple proof of Pi(x^2)-Pi((x^2)/2)>x for x>10. The graph clearly indicates this but how do I proceed?
Oh I forgot to say, pi(x) is the number of primes less than or equal to x
 
4 hours later…
08:13
Hi everyone. Can someone help me with the interpretation of this numerical
Assume that the chances of the patient having a heart attack are 40%. It is also assumed that a meditation and yoga course reduce the risk of heart attack by 30% and prescription of certain drug reduces its chances by 25%. At a time a patient can choose any one of the two options with equal probabilities. It is given that after going through one of the two options the patient selected at random suffers a heart attack.
. Find the probability that the patient followed a course of meditation and yoga?
Here if I want to take out probability of the patient having a heart attack provided he followed the yoga and meditation course what would it be? The question says this course reduces the risk by 30% so does it mean theres still a risk of 70%? Does that make any sense?
 
1 hour later…
09:30
@RounakSarkar use PNT and show that the function is strictly increasing
09:52
@sanya do you still need help?
10:20
Yes @mo-_-
Just this question I asked^ not the whole solution
@sanya John replied to you in the other chat
10:50
@Jakobian I think of this as an anomaly of $\mathbf{Set}$ tbh
as you as you work in more general categories, a monoid in which every element is invertible is not necessarily a group
11:08
hello evryone
any fellow math professors in this chat?
I have observed that across all math textbooks, no one really provides a clear, step-by-step process on how to graph the three reciprocal functions. They often avoid discussing vertical and horizontal stretches and compressions. I have noticed this with my own students and wish to change it.
I am about to send my textbook chapter for grammar revision before releasing it to be seen by many other students around the world. Would anyone here be willing to assist me in proofreading my textbook chapter? This is a very important task, and even a small change could improve it significantly. In exchange, you would gain experience, my personal gratitude, and the right to share this document however you wish.
The grammar doesn't have to be checked only the mathematical content
(alsoe note that chatgpt won't help with this task sicne he snuggle himself to go on a route different from the ones of the textbooks)
thanks for anybody interested and reaching otu
11:46
anyone?
 
1 hour later…
13:04
@FedericoRuck What are "the three reciprocal functions"? The reciprocal trig functions, cosec, sec, cotan?
13:20
0
Q: Proofreading a textbook chapter on graphing trigonometric functions

Federico RuckI have observed that across all math textbooks, no one really provides a clear, step-by-step process on how to graph the three reciprocal functions. They often avoid discussing vertical and horizontal stretches and compressions. I have noticed this with my own students and wish to change it. I am...

13:42
Thanks Ben and Jakobian
 
1 hour later…
14:46
`yes
but this version is not the correct one
it lack all the immage sand examples
The real version is in a pdf one
is anybody interested in this opportunity?
hi
hello pizza
15:10
$$\sum_{n=1}^{\infty} \frac{1}{\sqrt{n^2 + 1}} = \sum_{n=2}^{\infty} \frac{1}{\sqrt{(n-1)^2 + 1}} = \sum_{n=2}^{\infty} \frac{1}{\sqrt{n^2 - 2n + 2}}$$ and $-2n + 2 \leq 0$ for all $n \geq 1$, so really $$\sum_{n=1}^{\infty} \frac{1}{\sqrt{n^2 + 1}} = \sum_{n=2}^{\infty} \frac{1}{\sqrt{n^2 - 2n + 2}} \geq \sum_{n=1}^{\infty} \frac{1}{\sqrt{n^2}}$$
Is this correct?
Obviously it needs to be equal, it can't be >
incorrect
Which step?
the conclusion
@Jakobian You agree with $$\sum_{n=1}^{\infty} \frac{1}{\sqrt{n^2 + 1}} = \sum_{n=2}^{\infty} \frac{1}{\sqrt{n^2 - 2n + 2}} \geq \sum_{n=1}^{\infty} \frac{1}{\sqrt{n^2}}?$$
$\sum \frac{1}{\sqrt{n^2+1}} \leq \sum \frac{1}{\sqrt{n^2}}$
@ILikeMathematics well I agree technically since both sides are infinity
$$\sum_{n=1}^{\infty} \frac{1}{\sqrt{n^2 + 1}} = \sum_{n=2}^{\infty} \frac{1}{\sqrt{n^2 - 2n + 2}} \geq \sum_{n=2}^{\infty} \frac{1}{\sqrt{n^2}} - 1$$
Agree now?
I mean, we do have $$\sum_{n=2}^{\infty} \frac{1}{\sqrt{n^2 - 2n + 2}} \geq \sum_{n=2}^{\infty} \frac{1}{\sqrt{n^2}},$$ obviously, no?
We will be dividing by less on the LHS
@ILikeMathematics that's true, yes
Sry I meant -1, not +1 in the first message
From that follows $$\sum_{n=1}^{\infty} \frac{1}{\sqrt{n^2 + 1}} + 1= \sum_{n=2}^{\infty} \frac{1}{\sqrt{n^2 - 2n + 2}} + 1\geq \sum_{n=1}^{\infty} \frac{1}{\sqrt{n^2}}$$
By adding $1$ to both sides
And thus $$\sum_{n=1}^{\infty} \frac{1}{\sqrt{n^2 + 1}}$$ diverges to infinity
Do you agree with everything now, @Jakobian
$\sqrt{n^2+1}\le n+1$ so $\frac 1{n+1}\le \frac{1}{\sqrt{n^2+1}} $ isn't faster?
15:21
True, it is
@FedericoRuck almost all textbook have a section where the authors explain how to sketch $1/f(x)$
15:43
yes but it is incomplete
16:30
@FedericoRuck Honestly, I'm not a big fan of your exposition. When I teach this material, I like to go back to working with sets in the plane. A function is a set of ordered pairs, e.g. $$f = \{ (x,y) : y = f(x) \} $$ (note the abuse of notation). Elementary transformations can be described by working directly with the ordered pairs.
why use an abuse of notation so early in the learning
For example, it is, I think, pretty clear that the set $g = \{(x+1,y) : (x,y)\in f\}$ is the same set as $f$, but translated to the right by $1$ unit. But if $y = f(x)$, a quick change of variables (take $x'=x+1$, then $x=x'-1$) gives that $g(x) = f(x-1)$.
@think_meaning_buildß This is how I present the idea to other mathematicians. I am a lot more careful with students.
If you prefer, suppose that $f \subseteq \mathbb{R}^2$ is a function.
You can define a new function $g$ as $g = \{ (x+1,y) : (x,y) \in f\}$.
Then observe that $g(x) = f(x-1)$.
do you tell your students that a function is a correspondence?
16:34
It is, I think, more clear that $(x+1,y)$ is "just" the point $(x,y)$, translated to the right by one unit.
@think_meaning_buildß What do you mean by "correspondence"?
oops i meant function
Do I tell my students that a function is a function? I don't understand....
Oh, you edited your comment.
I still don't understand what you are asking.
how do you define a function to your students
A function $f$ is a relation which satisfies the property that if $(x, y_1), (x, y_2) \in f$, then $y_1 = y_2$.
ok, what about a relation now
16:38
That is, for any real number $x$, there is at most one ordered pair in $f$ which has $x$ as a first coordinate.
A relation is just a set of ordered pairs.
After building some familiarity working with functions and relations as sets, I introduce the idea that a function can be viewed as a kind of "arrow" which maps things in the domain to things in the codomain (which is what I think you mean by a "correspondence"), but that is not typically how I introduce the topic.
right, and a function is a relation in which different ordered pairs have different first coordinates
7 mins ago, by Xander Henderson
A function $f$ is a relation which satisfies the property that if $(x, y_1), (x, y_2) \in f$, then $y_1 = y_2$.
do you talk about the vertical line test
To talk about a reciprocal function, given a function $f$, define $$1/f = \{ (x,1/y) : (x,y)\in f\}.$$ Observe that $1/1 = 1$, that if $|t| > 1$, then $1/|t| < 1$ (and the bigger $|t|$ is, the closer to zero $|t|$ is), and if $|t|<1$, then $1/|t| > 1$ (and the smaller $|t|$ is, the bigger $1/|t|$ is).
And if $(x,0)\in f$, then $x$ is not in the domain of $1/f$.
@think_meaning_buildß Sure. The definition I gave above is the vertical line test, once you draw the picture.
THEY should draw the picture :-)
16:45
@think_meaning_buildß Okay...
In any event, the above is really all you need in order to sketch any reciprocal function. The first examples are $x \mapsto 1/x$ and $x \mapsto 1/x^2$. Then, months later, reciprocal trig functions are no biggie.
and how's that working out for you?
17:13
months later
17:23
@think_meaning_buildß you shouldn't expect an answer
pie
pie
Is there a concept in math called 'surgery theory'?
Is this a field like 'number theory'? And honestly, this seems like a terrible name for a math topic!
@pie yes. It even has a wikipedia page
wikipedia calls it 'collection of techniques' which means its not an actual field of mathematics
17:55
I think fully normal implies strongly collectionwise normal
@pie yes
I suppose you could call it a sub-field
18:18
@pie you will quickly understand where the name comes from once you look into it
I think it's a great name, honestly :)
as to whether it qualifies as a field: what qualifies as a subfield of mathematics depends on how you feel on any given day, but as a topic it's big enough to warrant having a 1000-page book just written on its foundations link.springer.com/book/10.1007/978-3-031-56334-8
but certainly it's not a field on the level of "number theory", or "algebraic topology", or "analysis"
it's subordinate to geometric topology
18:44
Just proved CSB with some training wheels
Now I won't have a moral issue with using that theorem
pie
pie
@Thorgott Any subfield will eventually become a field. Honestly, I want to specialize in this field just to be called a "manifold surgeon.":)
Are all of you mathematicians or specialized in math? I feel out of place here as an undergraduate student.
Let $F\in BV$ be real-valued, and denote by $T_F(x)$ the total variation function of $F$. Then we can write $F=\frac12 (T_F+F)-\frac12(T_F-F)$, where $\frac12 (T_F\pm F)$ are increasing. Now, Folland claims $$\frac12 (T_F\pm F)=\sup\left\{\sum_1^n [F(x_j)-F(x_{j-1})]^\pm :x_0<\cdots<x_n=x\right\}\mp\frac12 F(-\infty),$$where $x^+=\max(x,0)=\frac12 (|x|+x)$ and $x^-=\max(-x,0)=\frac12(|x|-x)$. Could somebody help me verify this equality?
Attempt: Let's just assume $\frac12 (T_F+F)$. Then $$[F(x_j)-F(x_{j-1})]^+=\frac12 |F(x_j)-F(x_{j-1})|+\frac12 (F(x_j)-F(x_{j-1})).$$Now, $\sum_1^n (F(x_j)-F(x_{j-1}))=F(x)-F(x_0)$. But this is where I'm stuck. How do I simplify $$\sup\left\{\frac12\sum_1^n |F(x_j)-F(x_{j-1})|+\frac12 (F(x)-F(x_0)):x_0<\cdots<x_n=x\right\}?$$
@pie no
Man, my small handful of PhD theorems seem trivial. My dissertation feels like it'd get me a Master's at best. I don't know whether it's imposter syndrome or what. My supervisor thinks I'll prove a conjecture soon but I doubt it heavily. I have until this time next year of supervision, maximum.
@pie I'm an undergraduate too
19:01
That's not enough for me to prove anything worthy of a doctorate.
@Shaun It definitely sounds like an imposter syndrome. Just continue on what you're doing, you'll prove something you could be proud of, or you won't. Not a tragedy either way
> My supervisor thinks I'll prove a conjecture soon but I doubt it heavily.
Thanks, @Jakobian. The last time I proved something new was three or so weeks ago, after months of getting nowhere. At this rate, I just won't have enough.
pie
pie
BTW may I ask how one can get a master's degree? I know I need to have a CV to apply, but aside from my grades, what else should I do? Do I need to solve a conjecture or something to be accepted into higher education?
@Jakobian I feel like this is an issue with your supervisor tbh, not with you. They're putting unnecessary pressure on you by saying that. You'll explore whatever you are working on further, and maybe you will prove the conjecture you want. Maybe not. Either way, it might be pretty useful to do
Contact who you think might supervise your dissertation, express an honest interest, and perhaps they'll vouch for you in the application process, @pie
19:08
pie it really depends on the program. literally every program is different
or maybe not, there may be some countries or regions where the process is so systematized that a bunch of programs within that country or region use basically the same process
but there is no generally applicable answer to this question at the level you have asked it
@Shaun clear your mind, and attempt it again from another side
or maybe the same side, who knows - maybe you missed something
pie
pie
@Shaun That's it? I thought I would need to write a few books, solve a few conjectures, and publish some research to get accepted. Does this apply to people from outside my country or is it a country specific thing?
@pie write books just to get a master degree? Nah
@pie You definitely don’t have to do that much. Books are reserved for people who've been lecturing for a while, as far as I can tell.
To write a book you ideally want to have really good knowledge of a field, and I meant it as, very throughout knowledge, not just confidence that you do
19:14
@pie I don't know how it is elsewhere, but here you just sign up for a Master's program and that's it
unless its say, educational
you can theoretically write another calculus textbook, well, there is a huge amount of them already
the prerequisite being a Bachelor's degree, of course
I have a conjecture that a particular function of groups takes on a value almost always for $\operatorname{SL}_2(q)$. It's based off of GAP calculations for $2\le q\le 59$ and what we know of other groups the function applies to. I have proven lower bounds that agree with my conjecture exactly, two thirds of the time; and I have a technique for proving an upper bound that hasn't worked yet.
pie
pie
@Jakobian I can't explain how relieved I feel right now. I thought I had to do all of that in the next three and a half years, and it was stressing me out.
I have proven two small theorems that find the values of functions vary the definition of said function.
I also conjectured something that was proven here, with me not knowing its originality to me.
pie
pie
19:20
Another question would be: Do you guys work while studying for this degree, or are you part of a scholarship program or something? And if you have to work, how can you find a job in the field of math with only a bachelor's degree?
what is a "job in the field of math"?
a job as a mathematician will require a Master's degree
My first attempt at a PhD was funded by a scholarship, @pie. But then I fell ill with schizophrenia and the pandemic happened, so I left with an MPhil. Now my current attempt is funded by a government doctoral loan that has very favourable repayment plans, like not have to repay if I'm earning below a certain amount. I tutor for the department and train AI to do mathematics.
This is in the UK.
pie
pie
@Thorgott I meant how do you earn living, if you are a math major.
Funding for a Master's is rare.
many of those who don't stay in academia go into software or finance
19:24
I had an integrated Master's: an MMATH. For that, I had an undergraduate student loan.
pie
pie
@Thorgott So how do you earn living while studying for master's if "a job as a mathematician will require a Master's degree".
The repayment is the same as the doctoral loan.
@pie you generally don't i guess?
The MPhil was not a failed PhD by the way; instead, I downgraded early to strengthen my dissertation, relatively speaking.
Im on a scholarship and can barely make ends meet. Forget about saving any cash
19:27
My current supervisor thinks the MPhil dissertation was nearly a PhD.
personally, I don't earn a living. generally, regular work. there's also specific arrangements called 'work student jobs', tailored for these purposes.
outside of academia, the notion that one works in one's degree area is amusing
pie
pie
@nickbros123 Hmm... that would be hard for someone like me since I plan to be an international student (studying outside my country because the education in math here is poor)."
Student loans are not bad i guess
Public Education in india is pretty cheap
@nickbros123 not really:"
19:30
I pay rs 20k effective, a year, that takes care of tuition, hostel, food
I guess the fees at iisers are not that much but afaik iits charges a lot
Im usd that's 200
One thing my supervisor said was that, in one's first year, all the preliminary material is learnt; in the second, small theorems appear; and in the final year (of a UK PhD), the "big" theorems show up in time for the writing up.
@SoumikMukherjee there's scholarship that pays 80k a year
200 dollar a year??
19:31
@nickbros123 which one?
@SoumikMukherjee inspire
@Thorgott yeah lol
cheap is relative to income, etc
there must be some misunderstanding here. you are not saying your cost of living for a year is covered by 200 dollar, are you?
Sometimes I get to the point where I just . . . I just want to know the &$^#ing answer to my research question and not have to bang my head against a wall, trying to find it.
i bought 3 burritos a few nights ago (my 21 yo son returned from Europe), it came to over us$45. can't imagine us$200 covering much
banging one's head on a wall is part of mathematics
4
19:33
@Thorgott Yeah there's a misunderstanding. All in all, including tuition, food, hostel, I have to pay close to 100k rs. I get 80k rs from the government, so I have to put 20k of my own money, extra. That's ~200 usd
@copper.hat True.
@nickbros123 That's probably the best scholarship out there, you can even get a PhD with inspire
India produces some excellent engineers/academics
@SoumikMukherjee my eyes are on the NBHM scholarship exam 👀
That would cover all my costs, I don't have to put in my own money
atb, when is the exam? in jan?
19:37
I get benefits from the English government because I have schizophrenia. It's a lot. It can be pretty debilitating to live with sometimes, so it helps with costs; like with my avolition symptom, sometimes cooking feels like climbing a mountain, so I get takeaways or eat out, so it comes in handy. Normal jobs would be difficult unless they have the flexibility of a research degree, again because of the avolition. This time last year, I was getting out of bed at 16:30 most days.
Jan 20 i guess
I have to start preparing for it 💀
tbh, nbhm is much easier than net or get if you don't panic in the exam hall
A change in medication made the sleeping in no longer necessary.
@psie how was my attempt? Still struggling understanding this :(
My forehead is calloused from all the walls I've broken.
19:41
@Shaun I can't imagine how debilitating schizophrenia must be. Hope ur doing better now
@nickbros123 Thanks. It got scary a few times. But it's been over 9 months since I heard a voice, the longest it's been so far. They used to accuse me of horrible things, some of which taboo and others extreme, like murdering someone. At first, I thought it was the police interrogating me.
It felt real.
I tried to flee the country once, thinking the voices were real. I almost dropped £1000+ on a last minute ticket to New York from London. I must have looked suspicious, because security intervened. The next thing I knew, I was in a mental hospital.
oof :(
No one should go through this :(
It's nearly 20:00 here. I got to do something for a while. See you all some other time :)
19:57
cya
Do you get annoyed when Desmos doesn’t let you do infinite sums: desmos.com/calculator/6pekkk2sfw
20:16
If anyone missed what I wrote above, I elaborated here.
 
1 hour later…
21:19
Let $NBV=\{F\in BV:F\text{ is right continuous and }F(-\infty)=0\}$. Here $BV$ is the space of functions of bounded variation. Then it is claimed that $G(x)=F(x+)-F(-\infty)$ is in $NBV$, where $F(x+)=\lim_{t\searrow x}F(t)$. I see why $G\in BV$ and why it is right continuous, but why is $G(-\infty)=0$?
pie
pie
Why isn't there a site that lists the best-known algorithms for approximating functions? I think it would be a great idea to create something like this, but I’m not sure who to approach about it. Maybe this community could build something like it? I posted about this idea long ago here.
21:34
because nobody bothered to make one :)
there are few websites in mathematics of this kind
as for "could this community build something like it", I think the question you linked answers that to completion; it's as much as you're going to get out of the "community"
of course, if you want a website like this it's primarily your job to build it, and then maybe people will come and participate, or maybe not
and that's why there's few such websites out there :)
pie: even within pure mathematics there is no universal notion of 'best for approximating functions', right? or even 'approximating.' like, a ton of choices go into variations on what you might want or mean by that. and even if you fix the pure math choices, the "best" implementation of a specific algorithm in specific software or hardware usually depend not just on the math but on the software and hardware.
you're basically asking "why don't places exist on the internet for discussing math, and algorithms, and their implementations" well, they do. just not in a simple form of like, here's the problem, and here's the top 10 list of ways to solve that problem
@psie I think I understand why. $F(-\infty)$ is the limit of $F$ as we approach $-\infty$ from the right. $F(x+)$ denotes the limit at $x$ as we approach $x$ from the right. Now, if we "plug in" $x=-\infty$ in $G(x)$, we should get $0$, since $F(-\infty+)=F(-\infty)$. I don't know what's a more appropriate way of showing this.
21:49
one reason this stuff isn't kept in list form is because the more precise it is at the level of problem specification, the more quickly it is likely to become irrelevant to anybody as there are changes in software and hardware and the problems people care about
And the notation $F(-\infty+)$ is...well, awful :D
at the high level you have stuff like 'numerical recipes in c' and other books and resources as people were commenting on in that post you linked. harder to go lower level than that without running into stuff with a very short shelf life
even a lot of stuff in that example book is not particularly relevant if, for example, you aren't working in C, or more generally if you have different processing or storage considerations than the ones that formed a kind of background understanding of what scientific computing had or needed at the time that book was written
22:08
@Shaun I think avolition, or at least something similar, describes me as well
Joe
Joe
22:21
I'm struggling to understand the statement of Theorem 2.18 in Basic Algebraic Geometry I (Shafarevich), page 125. In particular, it's not particularly clear to me what an "affine open set" $X'$ is, or what $k[X']$ should mean.
@Jakobian That's unfortunate. I'm sorry. It sucks, huh? Have you seen a doctor about it?
Maybe it's depression. It's the same symptom that the depressed feel after all.
Yeah it might be, apathy or just depression
It differs from laziness or a simple lack of motivation. Often, I want to do things and there's stuff I can do alright, like my studies or volunteering here, but with stuff I have to do, there's a huge barrier in the way. Does that sound familiar?
It's everyday things, like tidying up, cooking, getting out of bed, socialising, washing, laundry, etc., that I can't do so well.
Its a bit like, I was supposed to do this thing but I'll just give up on it because I don't have the energy
Exactly.
It could be a symptom of something worse.
You never know. Get yourself checked :)
22:32
Well, I have pretty good clarity of mind, so its probably nothing serious
Clarity of mind is independent of a lot of mental health issues. They're not always deficiencies of intellect.
I think so, but I also think that if something was wrong with me, then I would notice
Like I said, I thought my voices were the police.
. . . at first.
I'm self-aware, I think about how I act and how it might look to others
Its not that I am saying that a person can't deny their mental illness, its just that me in particular would notice
at least if it was something serious
Don't be so sure! But hey, you know you more than anyone else. I hope you feel better soon :)
22:42
@psie $\lim_{x\to -\infty}F(x+) =\inf_{x\in\mathbb R}\inf_{t>x}F(t)$. I wonder how them $\inf$'s simplify...
@psie what do you mean
well, my intuition says $\lim_{x\to -\infty}F(x+) =F(-\infty)=\inf_{x\in\mathbb R}F(x)$, but the definition of the expression $\lim_{x\to -\infty}F(x+)$ is $\inf_{x\in\mathbb R}\inf_{t>x}F(t)$
your intuition sounds correct
what do you mean by definition?
rather, limits become infima because of monotonicity
And clearly, $\inf_{x\in\mathbb{R}}\inf_{t>x} F(t) = \inf\{ F(t) : t > x, x\in\mathbb{R}\} = \inf\{F(x) : x\in\mathbb{R}\}$
right, I should have probably said $F$ should be increasing. So my intuition is correct, that's great to hear :) yeah, it's that last equality that I'm investigating
the one I just wrote?
22:49
$\inf\{ F(t) : t > x, x\in\mathbb{R}\} = \inf\{F(x) : x\in\mathbb{R}\}$
yeah
alright, so do you see how those are infimum of the same set
yeah, actually I do now
that feels good

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