it is reasonable to ask the OP to clarify this, as you have done in comments. i would vote to close in its current form as lacking context. even if "everybody knows" what those symbols mean (and guess what, they don't), the OP saying what they are and what they know about them is part of setting the stage for the problem.
in an ideal world the OP would even explain what a 'p-solvable group' is although this is getting to the level where if you are capable of answering the question you probably know already. the point is to relieve potential answerers of the burden of assuming that you know things that maybe you don't.
the last thing anyone wants to do is respond to a question like this and then get a response "wait, why is that true?" and it's buried in the definitions of one of the things OP assumed without comment.
well if it had something even on mathematical reviews or one of the journals that published book reviews, that might be the first neutral view of the book
if their editorial process wasn't up to snuff in 1974
if it was close enough to current research at the time it's possible that nobody really looked through it critically
one of the profs on my thesis committee had a book where i'm convinced that nobody ever read it (including him)
I'd probably say that on the surface it looks like a good reference, but one needs to be careful, the book isn't peer review and its riddled with errors. Its a bad reference but its good enough if you check whats written there
The exercises are horrible, they literally ask you to solve open problems, some of them need fixing (again, riddles with errors), on which you may or may not spent few hours
Or more depending on if you are lucky
The issue isn't even that the book is terse
It looks like the authors just half-assed the book
I wouldn't recommend anyone without sufficient experience in general topology to read this book - you'll easily end up confused and believe in something that might be false
Leslie: What exactly is a research monograph, to be concrete
I understand the concept on a surface level, but I don't really know
I do actually like that it has exercises, there are things you can learn from them if you are someone like me, and you don't really know much about all the variations of normality-like properties
i mean it means whatever it means to people, as i was using the term i meant, some books are published not to provide students with something to learn from or work through, but to provide a kind of organized tour through relevant recent research
and maybe most of the value of the book is just giving people something to cite other than a string of papers
I don't think it would be as beneficial if I lived in 1974, but nowadays where I can (readily) check facts and information, the book is helpful and it actually does help me learn from other sources (it might not be the best place to learn, but it does give me incentive to research those things)
don't get me wrong, the book is still bad, I wouldn't recommend it to learn from for any undergraduate in general topology
its just something that kind of works for me
and not for any gatekeeping kind of reason, but (sometimes, not always) you just need to have a lot of knowledge to be able to tell what is true and what is bullshit in this book
not always just pure knowledge, but also some kind of skill in reading books
I imagine I'm the second or third person that attempted to read this book
Let $F(x)-F(a)=\int_a^x f(t)\,dt$ where $f\in L^1([a,b],m)$. Is $F$ absolutely continuous then on $[a,b]$? What I can use is the following result;
> Result If $f\in L^1(m)$, then the function $F(x)=\int_{-\infty}^xf(t)\,dt$ is in $NBV$ and is absolutely continuous, and $f=F'$ a.e.
I'm thinking, if I extend $f$ to equal $0$ outside $[a,b]$, then we have $F(x)-F(a)=\int_{-\infty}^x f(t)\,dt$. However, this is not quite on the form $F(x)=\int_{-\infty}^xf(t)\,dt$. There's this silly constant $F(a)$ that bothers me. What can I do about it?
@psie $F$ here is a function $F:[a,b]\to\mathbb C$ and $-\infty<a<b<\infty$.
@psie ok, something like this. Constant functions are clearly absolutely continuous (the definition is trivially satisfied). Also sums of absolutely continuous functions are absolutely continuous (this is a tiny bit more work). So $F(x)=F(a)+\int_{-\infty}^xf(t)\,dt$ is absolutely continuous. Bingo.
psie oen thing to keep in mind with all of this is definitions being in flux, e.g. whatever 'the definition' of absolute continuity is you can find some book where it's something else
as another example of that, something like follands theorem 3.5 is actually taken to be the definition of absolute continuity in some treatments
it generalizes to finitely additive measures in a way that folland's textbook definition doesn't
Personally, I would argue that the best phrasing would be something like "A constant function $f$ is absolutely continuous, as $\sum_k f(b_k) - f(a_k) = 0 < \delta$ for any collection of intervals $\{ (a_k, b_k) )$ and any $\delta > 0$."
Or something... I don't remember the definition off the top of my head.
Clearly to prove that constant functions are absolutely continuous, we need to prove that constant functions are absolutely continuous. As its clear that constant functions are absolutely continuous, we have shown that constant functions are (trivially) absolutely continuous
@XanderHenderson I am pretty sure they need to be disjoint, otherwise you end up with something different
the intervals
maybe it was that you end up with constant functions... hmm
@Jakobian I mean, adjust the phrasing to be correct. But in this case, it doesn't matter, as any collection of disjoint intervals is a collection of intervals.
That is, you lose nothing by adding the phrase "pairwise disjoint" between "collection of..." and "...intervals".
both statements claim a fact, but the latter phrasing communicates where that fact comes from ("clear" meaning that you can derive it from the definitions as an exercise)
@Thorgott I disagree. It doesn't immediately convey that it is a direct consequence of the definitions. It conveys that the author thinks that it is obvious or clear.
if I'm reading a paper and come across a statement that I haven't seen before, if it's said to be "clear", it signals to me "hey, think about this for a while and you will see it", but if there's no such designator, it could mean that, but it could also reference a "well-known" fact in the literature that I haven't seen yet
If everyone were as disciplined as you about using "clear" only to mean that it follows directly from the definitions, then I probably wouldn't have a problem with it. But many authors are not.
perhaps not universally consistent, but "clear" communicates to me that I should be able to convince myself of the claim purely based on what the text has already given me
and that's a useful thing to have been communicated
cause when I'm reading a paper somewhat out of my depth, it's not always immediately clear (pun intended) if a statement that's just said is supposed to be "clear" or an incarnation of some well-known theorem that's being quoted implicitly
and I think such a distinction is pratically helpful
I don't disagree that you can be even clearer about what "clearly" means and that it would even be good if people did that, but I still think saying something is "clear" conveys more than just making a statement of fact without any additional information
I've heard the argument before, and I am not at all convinced by it. I don't believe that mathematicians are at all consistent about how they use words like "trivial", "clear", "obvious", etc. They are, as often as not, used to elide important details.
they aren't, but my perspective is that it's an instruction to the reader (sometimes it's a faulty instruction, but that's a separate issue)
when I'm reading math, nothing is more disorienting than a statement of fact that is not immediately followed up with an explanation/argument, a direct reference to another result or an indication that it is "clear"/"trivial"/etc.
@Thorgott Okay, but that's your perspective, which is not universal. Again, if these phrases were used in a universally consistent manner, they would not bother me. But they aren't, which makes words like "clearly" a trap.
@Thorgott I don't disagree, which is why I advocate for more explicit phrases than "clearly", e.g. "it follows from the definition that..." or "theorem 5.3 implies that..." or "it is an exercise to show that...".
yeah, I'm on board with wanting stuff to be more explicit, but I think "clearly", "trivial" and "fill the details in as an exercise" are practically synonymous
yes, I'm not arguing in favor of terseness, I'm arguing in favor of "clearly" being better than nothing at all (which is also an argument in favor of being less terse)
@copper.hat I got mine from the grocery for \$3.50/lb. Not a lot of work---just toss it in the oven and roast it, covered, for an hour or two, then uncover it, crank the heat to crisp it up, and done.
Having a pre-packaged duck means I don't have to remove the feathers or innards or anything. Saves a lot of time.
Honestly, your question is too general to be answerable. But if I need an answer for something, then I am going to try to keep engaging until I understand what I need to understand. But I am also a specialist, and if I am asking a question, it will likely be of another specialist, and we will likely speak the same language. Thus the probability of it being "over my head" is unlikely.
Last exam is finally done. I should probably spend some time grading this afternoon, but I think that I am going to go home and drink, instead. I'm so tired...
@ModularMindset I've been waiting for my hair to go grey since I was in my 20s, but (1) I have been shaving my head for more than 20 years now, so no real hair there and (2) my father didn't start going grey until his mid 50s, so I've maybe got another 10 years to go.
On the other hand, I do find the occasional grey hair on my face. Particularly on my upper lip, just below my nose.
"Textbook" because they have exercises (non-textbooks don't typically have exercises), and "in mathematics" because the exercises involve proving things, which is what mathematics is all about.
> I must say that I've gained insights from this book that I did not get from any other text. Prof. Johar manages to deliver the perfect balance between rigor and eloquence, producing a text that is both enjoyable and rigorous, inspiring and exhaustive. The subjects are well concatenated, and I never felt lost due to the structure of the book (only due to the fact that it is a hard subject!).
> With this book, I felt that you have all-in-one; you will not need to refer to another book for a long time. It really goes from numbers to measures. This one takes the crown in my analysis book collection
from someone who apparently looked at a lot of analysis books - of course everyone likes something different but yeah
speaking of higher category theory, anybody know an intrinsic proof that filtered colimits and finite limits in the $\infty$-category of spaces commute?
Any complex measure $\mu$ can be written uniquely as $\mu=\mu_d+\mu_c$, where $d$ stands for discrete and $c$ for continuous. Furthermore it is claimed that $\mu$ can actually be written like this $$\mu=\mu_d+\mu_{ac}+\mu_{sc},$$ where $\mu_{ac}$ is absolutely continuous with respect to Lebesgue $m$ and $\mu_{sc}$ is mutually singular with Lebesgue measure. It is being said this follows from Theorem 3.22 above. How?
I understand why we can write $\mu=\mu_d+\mu_c$, but not why $\mu=\mu_d+\mu_{ac}+\mu_{sc}$.
A measure $\mu$ is said to be continuous if $\mu(\{x\})=0$ for all $x\in\mathbb R^n$. It's discrete if there's a countable set $\bigcup_1^\infty \{x_j\}$ and complex numbers $c_j$ such that $\sum |c_j|<\infty$ and $\mu=\sum c_j\delta_{x_j}$, where $\delta_x$ is the point mass at $x$.
A follow-up question; let $\sigma=\sigma_{n-1}$ be the surface measure on the unit sphere $S^{n-1}\subset\mathbb R^n$ that is defined through $\sigma(E)=n\cdot m(E_1)$, where $E$ is a Borel subset of $S^{n-1}$ and $E_1=\{rx':0<r\leq 1,x'\in E\}$ (here $m$ is Lebesgue measure). Is this measure continuous and mutually singular with respect to Lebesgue measure?
Attempt; if we take $\{x\}\subset S^{n-1}$, then I think $E_1$ is just...what? A line? If so, it has Lebesgue measure $0$. For the mutual singularity part, I don't know what set to choose such that $m(B)=\sigma(B^c)=0$.
@XanderHenderson yeah just saying "clearly" is not enough. I prefer phrases like: "As is immediately apparent to even the most casual observer, constant functions are absolutely continuous"
"If it is not obvious to the reader that $x^n + y^n = z^n$ admits no positive integer solutions for $n > 2$ then we urge them to reconsider their choice of career path"
@copper.hat yeah, it can be frustrating. But what is the difference between $A \mapsto \sigma (A \cap S^{n-1})$ and $\sigma(E)=n\cdot m(E_1)$ where $E\subset S^{n-1}$?
I think I will be happy if we only manage to show mutual singularity :)
@copper.hat remember the polar integration formula we discussed some weeks ago? I think it can also follow from that. Namely $$\int_{\mathbb R^n}f(x)\,dx=\int_0^\infty\int_{S^{n-1}}f(rx')r^{n-1}\,d\sigma(x')dr.$$
@psie Note that $m B(0,r) = r^n B(0,1)$ and $B(0,1) \subset \bar{B}(0,1) \subset B(0,r)$ for all $r>1$. Taking limits of nested sets shows that $m B(0,1) = m \bar{B}(0,1)$.
i learn something new every day, co-area.
@psie as a general rule, i would suggest, in general, look first for the simplest solution.
