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02:10
actually does $(-\infty,a]$ have the least upper bound property for $a \in \mathbb{Z}$
I have to imagine yes, but it doesn't seem to have a greatest lower bound property
unless these properties are specified only for finite nonempty sets
@Obliv it does
What is the greatest lower bound?
er like for a given subset (-infinity,a)
its not even bounded below to have a largest element in the set of lower bounds
02:31
ok munkres doesn't explicitly say $A$ has to be bounded below , just that it is ordered..
which sounds wrong
well yeah obviously if $A$ has a lower bound this proof becomes trivial. Given $A$ is a chain, has a LUB property, the set of upper bounds of a subset $A_0$ of $A$ in $A$ has an associated set of lower bounds, namely $A_0$, which has a greatest element since $A$ is a chain.
similar argument for the converse
I can imagine this gets nuanced but for now if we guarantee $A$ has an upper and lower bound and that it's ordered, there is nothing to prove anyway shrug
 
4 hours later…
06:24
@Obliv it has both
@Obliv its greatest lower bound in R
@Obliv (-inf, a] doesn't have to be bounded from below but its subset
@Obliv There is nothing wrong here
yeah, obliv if you haven't resolved that confusion already, you might run whatever you think your intuition is about (-infty, a] against the fact that R itself has the relevant property (or if the latter seems controversial focus on that and not the subinterval)
to say that a set X has the greatest lower bound property or least upper bound property is to say something about certain subsets of X, and that something does not include X itself having a lower bound or an upper bound
it's maybe unfamiliar as a matter of language that X having the "greatest lower bound property" (think of this phrase as an atom, not something that is divisible into pieces) does not imply that X has a "greatest lower bound" or a "lower bound"
but that itself is not any weirder than a vice president not being also a president
06:50
@Thorgott Ah ok, I get the point.
i am thinking how to modify the argument
@leslietownes vice president ex-president
real numbers are neither real nor numbers, but you won't find that on wikipedia
07:59
Suppose we have $Y=[0,1]$ with base point $1$. Then we can take $Y\vee Y$ to be $[0,2]$.
@leslietownes 🤯
the natural numbers are not natural to count with
{1, 2, 3,...}
08:53
Now let $\nu(x)=1-x$ on $[0,1]$. First the map $\mu$ maps $Y$ to $Y\vee Y$ by $\mu(x)=2x$. Then we have a map from $Y\vee Y$ to $Y$ such that this map fixes $[0,1]$ and flips $[1,2]$ on $[0,1]$. This composition of maps are homotopic to constant map. So $\nu$ is a co inversion.
So in general we can have a space $X$, then consider its suspension $\Sigma X$. first consider a map $\nu$ on $\Sigma X$ that turns it upside down. Then a map $\mu$ that makes $\Sigma X\vee \Sigma X$ out of $\Sigma X$. Then we consider the map $(id,\nu)$ on $\Sigma X\vee \Sigma X$ that is id on first $\Sigma X$ and $\nu$ on 2nd $\Sigma X$. Similarly we consider the other way. If $(id,\nu)\circ \mu$ and $(\nu ,id)\circ\mu$ are homotopic to a constant map then $\nu$ is a co inversion.
 
1 hour later…
10:06
@SoumikMukherjee Yes, this is always the case.
I see two lines of attack.
If you simply wanna show that neither is homeomorphic to a deformation retract of the other, then I would do this by proving that a) the spaces aren't homeomorphic (you've already done this) and b) no proper subspace of either space is a deformation retract. in fact, the inclusion of no proper subspace in either space is a homotopy equivalence or even surjective on $H_0$ and $H_1$ as you can observe explicitly (e.g. if you remove a non-wedge point from $S^1\vee S^1$, the resulting subspace deformation retracts to one of the copies of $S^1$, whose inclusion is not
The other way is proving the stronger statement that neither is homeomorphic to *any subset* of the other. To do so, observe that the spaces are finite, connected graphs. You can write down an explicit list of all closed, connected subspaces of such a space.
e.g. for $S^1\vee S^1$, you get the entire space, you get the union of one circle with an arc of the other that contains the basepoint, you get the union of two arcs meeting at the basepoint and lastly you get just a singular arc in one copy of $S^1$. then, you can check (e.g. using a local disconnection analysis like you've attempted)
 
2 hours later…
12:40
@SineoftheTime 1148
@Tsundoku then i can beat u :)
12:52
@Thorgott Thanks! I was actually thinking of using graphs, but I didn't think this way.
13:03
what subjects is everyone interested in
At least one of the hummingbird eggs in the nest outside my kitchen window has hatched. Yay!
That's mama bird feeding de babby.
how many eggs are there left to hatch
I made coffee in a coffee maker today. It seems pretty strong
I've put > 1/2 of the cup with milk because there was such small amount of coffee but the coffee has still very strong presence in the drink
Moka pot
@RyderRude math, zoology, paleontology, anthropology
13:19
you should be ready to make theorems soon
or a quick travel to the potty
> Rényi, who was addicted to coffee, is the source of the quote: "A mathematician is a device for turning coffee into theorems", which is often ascribed to ErdÅ‘s.
I agree with the quote, no, I live by it!
> He is also famous for having said, "If I feel unhappy, I do mathematics to become happy. If I am happy, I do mathematics to keep happy."
Alfréd Rényi (20 March 1921 – 1 February 1970) was a Hungarian mathematician known for his work in probability theory, though he also made contributions in combinatorics, graph theory, and number theory. == Life == Rényi was born in Budapest to Artúr Rényi and Borbála Alexander; his father was a mechanical engineer, while his mother was the daughter of philosopher and literary critic Bernhard Alexander; his uncle was Franz Alexander, a Hungarian-American psychoanalyst and physician. He was prevented from enrolling in university in 1939 due to the anti-Jewish laws then in force, but enrolled at...
I could be him if I wasn't myself and was actually a successful mathematician at an academy. We think very similarly
13:27
he only lived to 48 :(
That's alright, I'll probably live that long
@Jakobian oh
so u love two historical subjects + zoology and math
@Jakobian do u do the other three as hobbies and math as primary
@user85795 No idea. There were two in the next a couple of weeks ago, but I don't want to get close to the nest at this point, and risk scaring off the mother bird.
@RyderRude I'd say they all can have something to do with zoology, but yes, paleontology and anthropology have a historical underlining
@RyderRude noted
13:39
@Jakobian I generally don't like moka pot coffee all that much. It isn't that it makes strong coffee, but that it makes coffee which tastes kind of burned (to me).
well, they can have at least, anthropology doesn't have to be about cultures in the past
@user85795 A comathematician is a device for turning cotheorems into ffee.
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@XanderHenderson I was reading an article that you should stop when the coffee it produces seems to be yellow
this is what creates the burnt taste
@Jakobian do u prefer the study of modern world
@Jakobian Indeed, most of anthropology is not about the past, but about the present. Only archaeology (a branch of anthropology in the American tradition, but more closely tied to classics in the European tradition) deals with the past.
13:41
@XanderHenderson oh
@XanderHenderson but anthropology also has human evolution
@RyderRude no, I prefer historical anthropology
its interesting to see people's culture in the past
@RyderRude Oh, sure, there is bioanth, but that subfield also deals a lot with modern humans, and overlaps a lot with archaeology.
(and paleontology).
i have divided history into some categories: pre-earth history (early universe astrophysics), pre-human earth history and human history
13:44
Everything before written records is pre-history. :)
yes. but i like to distinguish early universe history and pre-human earth history
Pre-human Earth pre-history.
there is an interesting thing : out of these three categories, human history is the most dense but only comprises some 100k years maybe
when u read pre-human earth history, there is a significant event every millions of years, like extinctiom events
@RyderRude yes, I just maintain my interest with anthropology etc., I don't study it seriously
@Jakobian makes sense. one cant be an expert in two subjects so far apart
early universe history has big bang and how elements and stars formed
and then earth history has the history of life and extinction events
all these subjects are really juicy
13:50
as in, juicy topics for debates with theists?
juicy for the curious brains :)
debates go nowhere imo
@RyderRude I'm not convinced of big bang
it is kind-of well established i think...but the early parts are not well understood
I think the opinion/agreement that its true is well established. I don't know about evidence
the initial fraction of a second has some degree of unknownness
@Jakobian it is based on the expansion of the universe
13:53
its certainly not as well established as "the earth is round"
there r some related theories, like big bounce
big bang and big bounce cannot be distinguished, i think
@Jakobian yeah
yeah, so I'll remain doubtful about big bang
I don't like how people treat it as if it happened
i dont know the details of this theory, but i think it's based on running the expansion of space backward in time
the expansion is what's observed experimentally
until someone can convince me that we actually have good evidence of big bang
if we observed expansion, then i think it's reasonable to run it backward. it's a standard application of laws of physics @Jakobian
but im not sure. maybe an astrophysicist can give better opinion @Jakobian
13:57
Universe is accelerating, as claimed, because of dark energy
we don't know what dark energy is, so how am I supposed to trust this
@RyderRude That's not true. It just takes a lot of work. It's one of the reasons that serious mathematical historians are so rare.
yeah.. that stuff isnt completely understood
@XanderHenderson yeah..
yeah, I won't trust a theory based on sticks
because it could as easily fall apart
there is a question i think is related : how can we trust that the past really existed
while I'm not saying big bang didn't happen, I'm not convinced that it did happen
14:00
@RyderRude I don't think that solipsism is a useful philosophical point of view...
I am struggling to parse the last two sentences here and am hoping for some help
@EE18 why
@XanderHenderson it's not solipsism though. one can claim that everyone in the universe, along with the universe, was created right now
and we have no way of disproving it
That Cauchy sequences are "translation invariant" in $E$ follows simply from vector space properties, right? i.e. nothing special about their being Cauchy?
im just saying that anything we claim about the past is based on induction
14:02
$(x_n+a)-(x_m+a) = x_m-x_m$, yes
But then the last sentence I don't get at all. Why does this show Cauchy sequences can't be defined using neighborhoods?
well, it doesn't
in fact its wrong
The problem of induction is a philosophical problem that questions the rationality of predictions about unobserved things based on previous observations. These inferences from the observed to the unobserved are known as "inductive inferences". David Hume, who first formulated the problem in 1739, argued that there is no non-circular way to justify inductive inferences, while acknowledging that everyone does and must make such inferences. The traditional inductivist view is that all claimed empirical laws, either in everyday life or through the scientific method, can be justified through some form...
in a topological vector space, a Cauchy net $(x_i)$ for every neighbourhood of $0$, call it $V$, there exists $k$ with $x_i-x_j\in V$ for $i, j\geq k$
@RyderRude That basically devolves into solipsism.
But, if you prefer, last-Tuesday-ism is not a useful philosophical point of view.
14:04
this defines a Cauchy net for any topological vector space, they have a natural uniform structure over which you can define Cauchy-ness
@XanderHenderson it's just an interesting philosophical view to know of, not an interesting philosophical view to have
How do you catch a convergent fish?
With a Cauchy net.
@EE18 what the author might have tried to imply is that being Cauchy is not a topological property, but a property of a metric, or, more generally uniformity, or more generally a Cauchy space
@RyderRude That seems like a matter of opinion. I don't find last-Tuesday-ism at all interesting. :D
i thought it was mindblowing :P
but i cant imagine anyone seriously living their life while having this view
14:07
@Jakobian I suspect you are right. But I still don't see how that follows from the penultimate sentence?
@RyderRude Like I said, it is a matter of opinion. From my point of view, last-Tuesday-ism is right up there with other stoner thoughts, such as "Hey, bruh! Why do they call them fingers if they don't fing?!"
This was in the intro to the section, maybe I need to read more?
@EE18 me neither
In any event, I need to run.
I don't like the author that wrote that
Instead, give an example of two metric spaces on the same set, inducing the same topology (i.e. the same open sets), but with different Cauchy sequences
For example, $\mathbb{R}$ with standard Euclidean metric, and $d(x, y) = |\arctan(x)-\arctan(y)|$
in $\mathbb{R}$ with standard Euclidean metric, every Cauchy sequence is convergent
But if you take a sequence $x_n\in\mathbb{R}$ divergent to $\infty$, then $x_n$ is Cauchy in the second metric
(this has something to do with the fact that completion of $(\mathbb{R}, d)$ is the extended real line)
14:17
@XanderHenderson the reason i think the philosophical arguments is useful to know (especially for philosophers) is that it reveals the nature of reasoning and intuition. this is one of the things we take purely on intuition and usefulness, without any proof or evidence required
the problem of induction is a similar subject
the thought experiments are only a tool. the point isnt to entertain them as truth @XanderHenderson
14:34
I will think on these, thanks @Jakobian
15:09
I'm hoping for help in the proof of Theorem 6.6, with the step beginning "indeed, taking the limit $m \to \infty$ in (6.1) yields...".
It appears that AE (the authors) want us to argue as follows. Fix $n \geq N$ and then consider the sequence $u_n(x) - u_m(x)$. By basic limit laws we have $\lim_m [u_n(x) - u_m(x)] = u_n(x) - u(x)$. Thus by Proposition 2.10 (which we can use per Remark 3.1(c) we have $\lim_m ||u_n(x) - u_m(x)|| = ||u_n(x) - u(x)||$. Further, we have $||u_n(x) - u(x)|| < \epsilon$ for that same fixed $\epsilon$ from (6.1) because ?????
That last bit I can't convince myself of
you're taking a limit of a sequence in which every term is $\le\varepsilon$, then the limit is $\le\varepsilon$
The key part is that it seems that the $N$ and $|epsilon$ in (6.2) is the same as (6.1). I feel like I can maybe push the theorem through if I choose some smaller $\epsilon$ ($\epsilon/2$ with $m \geq M$) for the taking of $m \to \infty$ part and then take $\max(M,N)$, but that seems not what they are doing
Ah OK. I see. And i don't need strict inequality here (which wouldn't hold in general anyway) because we're only showing boundedness with this step?
@EE18 the question marks are really distracting
yes, that is correct
and, before you ask about the next part, you can always achieve a strict inequality for a given $\varepsilon$ by starting out with a smaller $\varepsilon'<\varepsilon$ instead
@Thorgott ;) read my mind
To confirm, you're saying if we wanted to be totally formal then we would have taken $N:=N(\epsilon/2)$ way back at the start?
15:19
the exchange of limits is because of continuity, the inequality is because limits preserve non-strict inequalities
no, I'm just saying that's something you can do if you want strict inequalities
it does not make a difference (neither for boundedness nor convergence) whether you define everything in terms of strict or non-strict inequalities
Ah OK I see. In this book we have used strict inequalities for convergence and non-strict for boundedness as a matter of convention but it need not be so, got it
but I guess, you don't like me mentioning continuity and instead want to justify using some of the already proven propositions
You know me too well Jakobian! :)
which may or may not be equivalent to showing that the norm $x\mapsto \|x\|$ is a continuous function
15:24
I'm gonna go out on a limb and assume that it has already become clear in the arguments made in the preceding 176 pages of that book that the difference between strict and non-strict inequality in that definition is entirely immaterial
(in fact, Lipschitz continuous)
Prop 2.10 which I use tells me $x_n$ convergent implies $\norm{x_n}$ convergent
@Thorgott It has yes
@EE18 there is a difference between this, and between saying that if $x_n$ is convergent to $x$ then $\|x_n\|$ is convergent to $\|x\|$
I assume you mean what I wrote
so on that basis, I would not agree that this proof "lacks formality" for using one over the other. it is a perfectly formal argument and my comment merely explains (as you already knew) how to convert it into an argument with strict inequalities if you so desire.
Perhaps I am too flippant with "formal" and instead should say "exactly what the first definition of the thing the book gave is" :)
@Jakobian Yes that's what I mean my bad
15:26
so, yes, proposition 2.10 is just a statement that the norm is continuous
this is equivalent to continuity for metric spaces
@Jakobian yup I recall this from my previous experience, sequence characterization of continuity
@EE18 yeah, but at that point we have to understand that that would not be a desirable thing to achieve in general
perhaps interesting is that its not equivalent to continuity for topological spaces in general
which boils down to sequence not being a strong enough concept outside of metric spaces
in this case we replace the word "sequence" by "net"
continuity still implies that $f(x_n)\to f(x)$ for a sequence $x_n\to x$, but the converse is not true, it becomes true once we reach for nets
 
2 hours later…
17:10
0
Q: Why is $(x^2+x+1)$ a factor of the minimal polynomial over $\Bbb R$ just because $(x^2+x+1)$ is a factor of the characteristic polynomial?

Thomas FinleyI was studying about minimal polynomials in a Linear Algebra course. I am using the book Linear Algebra by Stephen H Friedberg, Insel and Spence for this purpose. I was doing a problem but while reading a solution for the same, I came across a concept about which I need a little clarification. Su...

I need some help with this.
@Jakobian Can you please take a look at this? I am so badly stuck with it...
17:21
I don't want to
@Jakobian oh! Sorry, if I have upset or disturb you. Coz, I really didn't mean to. Again, I am sorry
17:40
But the problem remains. Is there anyone willing to take a look at the post?
17:51
I don't know why every comment in the main post, always says "they will be the roots of p(x) blah blah blah"when I just wanna know why will they be the roots of p(x). Seems funny to me.
18:03
thomas, just because you've stated a result which would only apply to real roots if you take F = R in your result, does not mean that the complex roots are irrelevant to any analysis of the minimal and characteristic polynomials, or that you could not state a result that would also apply to them and show you what you want. you're sort of seeing a circularity that isn't there, because you're imagining the only route to analysis of this problem is through your statement with F = R plugged in.
i'm not going to engage with a full solution, MSE seems to have a number of people engaging with it there and presumably one of them will post an answer if you just wait a while.
if it is un answered in a week, let me know and i can talk you through it
 
3 hours later…
21:03
@Dragonrage o/
So, what is it that you are worried about?
heyo
oh, i just wasnt sure if this was a question that would work on your site since im not familiar with your guys site. basically I was trying to figure out how to do extrapolation for a graph with 6 given points. I know how to do linear extrapolation given 2 points, but dont know how i would go about using 6 non linear points
@SoumikMukherjee "You can't fall out of bed if you sleep on the floor" Sun Tzu
not sure if you have a question about that already (i couldnt find one), or if it is on topic (i think it is?)
@Dragonrage You might just search the site for "Lagrange interpolation". But it seems kind of broad...
11
Q: Explanation of Lagrange Interpolating Polynomial

user87274Can anybody explain to me what Lagrange Interpolating Polynomial is with examples? I know the formula but it doesn't seem intuitive to me.

@XanderHenderson oh, that sounds familiar. i knew i had done it at one point, but its been like over 10 years at this point
21:07
Though I am not quite sure if that is what you are trying to do...
well, the question i was given to do was something along the lines of a person was studying for a test and their practice test results were 65, 75, 90, 70, 85, 75 in that order. if we assume their actual test result will follow this pattern, what would we expect their test score to be
Oh, that is a stats question.
I see.
You are looking for a tendline of some sort. E.g. linear regression.
Less appropriate for Math, but maybe a better fit on CV.
i vaguely remember doing something like that in high school, but couldnt recall the correct approach
@XanderHenderson cool, thanks. ill reach out to someone from their site and ask
The naive answer is "linear regression".
21:59
@XanderHenderson Their chat room hasn't seen a comment for 3 days
not with that attitude, it hasn't :)
would that be a stattitude?
3
23:09
My textbook says that a function defined on an interval with at least one jump discontinuity cannot have an antiderivative, and it considers the example $\varphi(x)=1$ if $x \ge 0$ and $\varphi(x)=0$ if $x<0$. I tried to work out the details of the example, can someone check if my work is correct please?

Solution: if by the sake of contradiction there exists $a>0$ and $f:(-a,a)\to\mathbb{R}$ such that $f'(x)=\varphi(x)$ for each $x \in (-a,a)$, then we have $\text{im}(f')=\text{im}(\varphi)=\{0,1\}$ and we have $f'(-a/2)=\varphi(-a/2)=0$ and $f'(a/2)=\varphi(a/2)=1$. Since the derivatives
23:46
Looks good to me. Why were you concerned?
copper, funny correction on this story: "In an earlier version of this article, Dean Morgan and John McNamara were described as being Irish nationalists. That has been corrected. They are "Irish nationals."" cbsnews.com/colorado/news/…
@copper.hat Thanks! Sometimes I just do silly mistakes I don't recognize and feel inadequate :/ so I need comparison a lot of times
everyone makes mistakes
Yeah, I agree, but the thought: "You're not good enough no matter how much you study; maybe you should do something you enjoy less but you're better at" is always in the back of my mind. It's really draining sometimes...
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