namely if we were putting things in $[0, 1]$ when $A$ is infinite and in $[1, 2]$ otherwise, then $A = \emptyset$ and $A = \mathbb{N}$ would map to $1$
but you can just put things in $[0, 1]$ when $A$ is finite and in $[1, 2]$ otherwise, or just replace $[1, 2]$ with $[2, 3]$
If $n$ is now the least such that it belongs to one set but not the other, say to $A$ but not to $B$, then $g(A)\geq \sum_{x\in A, x < n} 2/3^x + 2/3^n$ yet $g(B)\leq \sum_{x\in B, x < n} 2/3^x + \sum_{x\geq n+1} 2/3^x = \sum_{x\in A, x < n} 2/3^x + 1/3^n$
and this shows that $g(A) > g(B)$, so injectivity follows
Note to self, do not mix pineapple juice with whiskey. Tastes really bad
@Shaun expect update on that ArrowGlue project later on. Just been busy with work stuff, and also have to reconsider things using PEG grammars instead of EBNF, for grammar ambiguity reasons.
Does anyone have a reference to the proof that the model categories of CGWH and sSet are Quillen equivalent? I can only find the one between Top and sSet
Let a die be such that $P(\{i\})=P_i, i=1,...,6.$ Suppose the above die is rolled repeatedly and independently. Let $X$ be the no. of rolls needed to get all the faces appear for the 1st time. So, $\text{Min}X=6.$ Find the probability mass function (pmf).
I tried solving the problem as follows:
L...
If $C$ is the circle $|z|=2,$ taken with positive orientation then show that $\int_C\frac{2z}{2+z^2}dz=4\pi i.$
I tried solving the problem as follows:
Let $I=\int\frac{2z}{2+z^2}dz.$ Then, $$I=\int_C \frac{1}{z-i\sqrt 2}dz+\int_C\frac{1}{z+i\sqrt 2}dz=4\pi i.$$ This is because, for any complex n...
yeah, I tried explaining this a couple days ago (except I then had a minor doubt, but looking back, there was nothing to doubt)
but the Quillen equivalence between sSet_Kan and CGWH_Quillen also follows verbatim for the same reasons as the Quillen equivalence between sSet_Kan and Top_Quillen
if you want a reference, I recommend Chapter 8.6 in Heuts-Moerdijk's Simplicial and Dendroidal Homotopy Theory
(in general, an excellent book to learn about simplicial sets and model categories, you can simply ignore the half of the book concerned with dendroidal sets)
(note that the book uses, as many other sources do, $\mathbf{Top}$ to denote CGWH rather than the ordinary category of topological spaces)
the mere statement (without proof) can also be found in Chapter 7.10 Hirschhorn's Model Categories and Their Localizations (he, too, uses $\mathbf{Top}$ for a convenient category)
I'm having trouble finding the integration extremes of this integral: $\iint_D x dxdy$ where $D$ is the domain between the bisector of the first quadrant, the unit circle and the parabola $y = \frac{x^2}{2}$
In Folland's text in proving $\operatorname{card}(\mathcal P(\mathbb N))=\operatorname{card}(\mathbb R)$, he constructs a surjection from $\mathcal P(\mathbb Z)$ to $\mathbb R$ by $g\left(A\right)=\log \left(\sum _{n\in A}2^{-n}\right)$ if $A$ is bounded below and $g(A)=0$ otherwise. I don't understand how this is surjective. In particular, why would $\sum _{n\in A}2^{-n}$ cover every positive real number?
The other one is instead $\int_{\frac{1}{\sqrt{2}}}^{\sqrt{-2+2\sqrt{2}}} \int_{\frac{x^2}{2}}^{\sqrt{1-x^2}} x dydx$
So: $\int^{\frac{1}{\sqrt{2}}}_0 \int^x_{\frac{x^2}{2}} x dydx + \int_{\frac{1}{\sqrt{2}}}^{\sqrt{-2+2\sqrt{2}}} \int_{\frac{x^2}{2}}^{\sqrt{1-x^2}} x dydx$
@nickbros123 good question. I have no idea how he defines it :) but note the expression with the log thing and the sum is only defined for sets that are bounded from below.
so I'd assume $$\sum_{n\in A}2^{-n}=\sum_{n\in A_1}2^{-n}+\sum_{n\in A_2}2^{-n},$$ where $A_1$ is a finite set of negative integers and $A_2$ a set of positive integers.
Oh yeah, this is just binary representations, kinda i guess? Every real no between 0,1 has a binary representation (may not be unique) that is a sum of this form $\sum \frac{1}{2^k}$. So every real has a non empty inverse map under this map g
Note that MSE stands for Meta Stack Exchange, and Math SE is for the math site. You'll confuse some people using MSE to refer to Math (or, at least, you will confuse me).
@psie this condition not only ensures convergence of the sum (if you cared about that), but tracks what we usually think of as decimal expansion. e.g. while you can certainly consider 'bi-infinite' sequences of digits 0-9, the sequences that represent real numbers have finitely many [nonzero] digits to the left of the decimal point
@psie i see no requirement that a_k not be B-1 for infinitely many indices, only a name for when this does happen. the likely intent is just to establish language for talking about the situation where two different base b expansions represent the same real number. one will be "proper" and the other one won't
e.g. 1.0000... and 0.999.... arise from different sequences of digits, and happen to represent the same number, the first one is proper and the other one isn't
@Thorgott Oh yes the explanation was very clear. I just wanted to add a reference to this to the report I am writing and the sources I could gather only show it for Top. But now that you mention that they use Top to mean CGWH, maybe that's what they do in the sources I have. Thank you!
@Mad pick an element on the left, say $y_1$; this will be greater than $a - g(y)$ by definition, so there is some $r \in \mathbb{Q}$ with $y_1 > r > a - g(y)$, so it's in the right, qed. :)
@leslietownes ah ok, makes sense. So in the "proper" expansion, the representation is unique, whereas in the "improper" one, we'd have duplicate representations of some numbers. Grazie!
psie: well, you will always have the issue of duplicate representations (for some numbers anyway). the notion/language gives you a way of talking about it, and getting uniqueness if you want it. (e.g. you could just as easily make expansions unique by choosing the non-proper expansion when you have to choose)
this nonuniqueness issue, and the arbitrariness of the choice in making a representation unique, are at least one of the background unconscious vibe reasons why most analysis books and other math books do not deal with decimals all that much, outside of one-off exercises
in particular, you don't commonly find books defining real numbers in terms of digit expansions, although you could certainly take that approach. you end up having to spend a lot of time proving that what you're doing isn't meaningfully affected by either the lack of uniqueness or the choices you make to enforce uniqueness
and once you're "identifying" elements of a large space of sequences with one another, you might as well just use all cauchy sequences, so you can e.g. define the arithmetic more easily :)
I am quite confident that the set $B:=\{x=(x_1,x_2 \cdots) \in \ell_2: |x_j|<\frac{1}{j} : j \in \mathbb{N} \}$ is not open. (It definitely is not closed im sure) How do I show this
with 0s everywhere else? yes, something like that would work (you could also put a value like that in any 'later' coordinate)
if you fix some number less than epsilon (whether it has the form 1/j or some other form) and look at the sequences you get by having 0s everywhere except in one coordinate, where you put that fixed number, all of those sequences will be within epsilon of (0,0,...), but at most finitely many of them will be in B, in particular, there will be a lot of them that aren't in B
note that while your example is certainly enough to show that B isn't open (because it shows that B has a single point, (0,0,0,...), that is not an interior point), you could use a very similar idea to show that B has empty interior
i.e. you can do a similar thing starting with any point of B, not just (0,0,...)
obliv: that example doesn't make it fully clear. what is an example of a situation where you'd want to use that language? could you not just use series when you use series?
obliv: i might refer to that union as being "indexed by" the positive integers, or as a union that employs the positive integers as "an index set"
but your use of the term "iteration" maybe suggests that there might be more going on here than just indexing, however, hence my follow up questions
yeah, i'm not sure what work "iterative" is doing in that question. just in the specific context of union, i think of union as something that might be a binary operation in some contexts (where you union a pair of sets), or might accept an arbitrary set (perhaps regarded as a family of sets, but in any case, not necessarily a set with two elements) as an input in other contexts
i don't see the definition of union is being inherently 'iterative' in any informal sense
although i guess when you're indexing a union over positive integers, there's a natural enough notion, pun intended, of "well, first you could union sets 1 and 2, then add in 3, then add in 4, then add in 5, ..."
if M is a set, \bigcup M is, by definition or something close to it, {y : exists x in M, y in x}. if M is a set of the form {A, B} then this definition of \bigcup M specializes to the usual definition of A union B. if M is a set of the form {A_1, A_2, A_3, ...} it's what you might write \bigcup M as bigcup_{n=1}^infty A_n
but the ordering of positive integers or the idea of taking increasingly big members of a set of subsets of the positive integers is just our own mental layer on top of that
it's not like \sum_{n=1}^{infty} a_n (for a sequence of real numbers a_n) where the ordering and the sequence of partial sums "really matters" to the definition
this is actually a good counterpoint example because if you want to define sums of arbitrarily indexed sequences of real numbers, you very much do have to add structure that gives you orderings on the index set, or some exhaustion of the index set by subsets, to have that general notion specialize to the usual one of series convergence
@leslietownes Yeah, we're just doing very handwavy beginner things in this intro to real analysis course. Though to be honest, I doubt we'll cover convergence, orderings, set theory, to any appreciable depth.
obliv: one useful aspect of the union (however defined) is that the union of a sequence of sets doesn't depend on the ordering of those sets, even if we naturally think of the objects involved in the union (and/or the union itself) as the result of some iterative process
@leslietownes Iterative processes. It always makes me wonder: there's no "passing of time" in math. A sequence, it's already there, the whole sequence. It's our thing to think as $x_1$ comes first, then AFTER we have $x_2$, etc
for an example, in this problem set, we have $\bigcap\limits_{n=1}^{\infty}(1-\frac{1}{n},1)$ which should be $\varnothing$ but idk what the justification is supposed to be because I never learned it.
obliv so, baked into the definition of \bigcap, we see that an element of that intersection would be a real number x satisfying 1 - 1/n < x < 1 for all n. so far so good, but specifically why that can't happen might well depend on how you axiomatize real numbers
In an elementary calculus class, it is generally not possible to give a rigorous treatment of the real numbers, so the extreme value theorem kind of needs to be taken on faith (since it relies on the least upper bound property). I also don't have a problem completely skipping epsilon-delta definitions for limits (though I think that if you talk about the epsilon-N definition for sequences, you can give a rigorous treatment).
The key is to be clear about what the omitted results are. It shouldn't be "hand wavy" so much as "[this] is out of scope, you are going to take the statement of the theorem on faith".
okay nvm, I see what you mean. If we take $\lim\limits_{x\to 0} 1 - x = 1$ then yeah, that follows. I just don't ever know when it's okay to "simplify" limits or whatever.
I'm not convinced that any of this is out of the usual even for a rigorous course. Finding your way around tools and definitions is often left to the exercises.
it's a fun little statement proved by Atiyah in '61 that allows one to prove a version of PD for extraordinary cohomology theories
basically it says that $M^{-TM}$ is equivalent to $DM$, the Spanier-Whitehead dual of $M$ whenever $M$ is a closed smooth manifold (here $M^{-TM}$ is the Thom spectrum of the virtual bundle $-TM$ over $M$)
using the only tool available for such things in the 60s: brown rep theorem :)
on an unrelated side note I learned from the green book that apparently people even set up things as basic as products of spectra via brown rep back in the day. the classical setting really doesn't sound fun to work in
xander: you don't want anything from me because i've been coopted by the establishment and go to the conclave, but i think of "symbolic expression" when i hear that absent context
@leslietownes Coopted by the establishment is fine. :D
Would anyone object if I claimed that a formula is a mathematical model, expressed in notation? (@leslietownes I think that this aligns with your idea of a "symbolic expression", with a bit of "well, okay, but an expression of what?").
@BenSteffan Well, I disagree here, in that a formula usually expresses the idea that it is an equation which means something. I wouldn't call $3x + 7 = 2y - 8$ a formula, for example.
I think that I am going to avoid the word "formula" in this section. I don't like it. It doesn't have a very precise meaning in mathematics, and the book's use of the word doesn't even seem to be very consistent.
They give 9 examples of the form $y = f(x)$, then finish with $\alpha + \beta + \gamma = 180^\circ$ (the angles of a triangle add to two right angles).
I think that I'll just teach the whole thing as "models".
Or "relationships". Or something.
Thanks for the input. It helped to clarify my own thinking.
I posted about this recently, but in proving $\operatorname{card}(\mathcal P(\mathbb N))=\operatorname{card}(\mathbb R)$, one can construct a surjection from $\mathcal P(\mathbb Z)$ to $\mathbb R$ by $g\left(A\right)=\log \left(\sum _{n\in A}2^{-n}\right)$ if $A$ is bounded below and $g(A)=0$ otherwise.
I have doubts about whether or not $A\mapsto \sum _{n\in A}2^{-n}$ is well-defined or not. Call this map $f$. I recognize $f$ as the binary expansion of a real number, but it was some time ago I looked at this stuff. Basically I want to verify that for the pairs $(A,r_1)\in f$ and $(A,r_2)\in f$, we get $r_1=r_2$. What are some ways I can do this?
true :D I'm just worrying about stuff like 1.000... and 0.999....
but maybe I shouldn't, since these are equal
If there was a set that mapped to 1.000... and 0.999... and, in some universe, maybe far far away, 1.000... was not equal to 0.999..., then we'd have a problem.
@psie The problem is that $1.000\ldots$ and $0.999\ldots$ are just symbols. They are both defined in a particular way (in terms of decimal expansions, which are in turn defined by series), and those definitions ultimately require that they are equal to each other. So this "other universe" of yours would simply require that the notation $1.000\ldots$ and $0.999\ldots$ be defined in a different way.
@psie you said well-defined. Any series of non-negative real numbers has a sum (possibly infinite), no matter the way you order them, equal to supremum over finite sums
