Mathematics - 2024-06-23

Xander Henderson

12:51 AM

@SoumikMukherjee sure, but that space isn't even $T_0$.

Obliv

1:29 AM

if a set A has a supremum/infimum not in A, then by definition does A have the greatest upper/lower bound properties

I suppose if it didn't then outside of A the sets aren't ordered?

Obliv

1:43 AM

"Here is a “proof” that every relation C that is both symmetric and transitive is also reflexive: “Since C is symmetric, aCb implies bCa. Since C is transitive, aCb and bCa together imply aCa, as desired.” Find the flaw in this argument". Is the flaw that it's not necessary that C is a total relation but reflexivity is by definition total

user 85795

2:19 AM

Transitive property: if a=b and b=c, then a=c.

leslie townes

2:37 AM

obliv: i don't think it's helpful to encode the flaw in language specific to partial orders. that's specific to your framing. for finding/looking at the flaw it is helpful to think about whether it is helpful or potentially harmful to refer to a relation without specific reference to the set that it is regarded as a relation on (here, what set is C a relation on? why doesn't the "proof" mention that?)

obliv: in plain english, to show that C is a transitive relation on some set S, you need to show that for all s in S, s C s. you'd usually begin a proof of something like that by fixing an arbitrary element of s and using definitions of things to go from there.

so if you do that, and in an attempt to apply the argument fragment, you find yourself wanting to use s as the "a" in the starting point of that fragment, and then you're stuck, because the argument fragment needs a "b" in S for which a C b holds to get the whole argument going.

what the argument fragment shows is that, if a and b are elements of S for which a C b happens to hold, then a C a will also hold. this isn't a statement about all elements of S, it's a statement about specific pairs of elements whose hypothesis includes already knowing that they're related to one another by C.

you wouldn't have that, in general. good examples would be S = any nonempty set and C = the empty relation, which is not reflexive. other example would be S = real numbers and C = the relation of "equality of integers" as a relation on S, i.e. x C y holds if and only if x and y are integers and x = y. and similar examples.

in that last example, you could also regard C as a relation in the set of integers (i.e., C is a subset of R x R that also happens to be a subset of Z x Z). that same set C of ordered pairs happens to be an equivalence relation on Z. but it isn't an equivalence relation on R.

as a relation on R, C is transitive and symmetric, but it isn't reflexive, because if you start with some a in R that isn't an integer, there aren't any elements b in R for which a C b, and the argument fragment can't get off the ground.

Obliv

3:04 AM

Thank you, that's basically what I was trying to get at. I guess I equated C being defined on the entire set as C having every element $a \in A$ in C (so by necessity then it'd be reflexive since no matter what $a$ is related to, by symmetry and transitivity it must be related to itself.)

Another remark/exercise is to show that the intersection of equivalence relations on the same set is also an equiv. relation. This is basically saying at least there would be an equiv. relation which is the set of reflexive pairs that are technically also symmetric and transitive

leslie townes

when textbooks and other sources present false proofs it's always helpful to take the exposition into account. like, why would someone unravel the definition of symmetry or transitivity without any quantifiers? why aren't we talking about the set the relation is regarded as a relation on [which can change whether the same set of ordered pairs is 'reflexive' or not]

Obliv

like (a,a) for all a in A

leslie townes

even good proofs can become confusing and look false if you scrub all of the quantifiers out of the statements you are trying to prove and use

Obliv

yeah I assume that exercise just wanted the reader to understand the necessity of specifying quantifiers/setting of what ur talking about

leslie townes

as i've seen it in textbook exercises, it's usually two messages, (1) proofs need quantifiers, (2) transitivity and symmetry are properties of a relation that don't really depend on what set your relation is regarded as a relation on, but reflexivity absolutely does

Obliv

3:12 AM

I think for the intersection question, the only cases are just all reflexive pairs or theyre the same relation. Kinda similar to equiv. classes being disjoint or identical except here instead of disjoint they only share the reflexive pairs

leslie townes

this isn't so much related to your question but one reason i think 'intro to proof' books always go through at least the motions of introducing symbols for quantifiers and symbolic encoding of math statements is a belief that, if you give people symbols for quantifiers, maybe they will be tempted to actually notice them and pay attention to them and maybe use them

i'm not convinced that that is true, but it seems to be one of the ideas

Obliv

@leslietownes so by this you mean reflexivity property requires the relation to be defined on the entire set but the transitivity, symmetry don't

leslie townes

in reality i'm not sure it is any easier to keep track of symbolic quantifiers than it is to keep track of english ones (we were talking about this the other night - the idea that it's easy to keep track of symbolic quantifiers and manipulate symbolic sentences is, i think, a mistaken impression that you only get by only considering simple symbolic sentences)

@Obliv roughly, yes, exactly

at 50,000 feet, the definition of reflexivity is the only one that require anything to be in the relation, and to do so in a way that depends on the set you regard the relation on

the others are more abstract formal properties

Obliv

nvm thats a bad example

ill cook another one up later

leslie townes

another way to think about it is that if R is a relation on a set S, it can also be regarded as a relation on any set that contains S, and from the way transitivity and symmetry are defined, if R is symmetric on S, it will also be symmetric as a relation on any set that contains S. symmetry is determined only by what R is as a set of ordered pairs, and not what cartesian product you regard it as a subset of

reflexivity isn't like this

this is just a more detailed way of pointing out that when you refer to something kinda informal like whether R is "defined on the entire set," that's not a question you can answer just from what R is as a set of ordered pairs

R := {(0,0)} is a perfectly good relation on any set that contains 0. R is reflexive if you regard it as a relation on the set {0}, but not reflexive if you regard it as a relation on any "larger" set (i.e. any set that properly contains {0})

so just saying a relation is "Reflexive," full stop, is either a misuse of language, or something that only makes sense if everyone in the room agrees on the choice of ambient set that you presumably had in mind when you said a relation was "reflexive" without specifying what set it was reflexive on

Obliv

3:27 AM

@leslietownes because the elements not in {0} aren't in R?

leslie townes

please be more specific, there are some type mismatches there. because if x is an element of the set that contains {0}, and x isn't 0, then the element (x,x) is not in R

"the elements" was playing a kind of dual role in your version, both as elements of the underlying set and as ordered pairs

Obliv

I think I get what you're saying. reflexivity has to be specified and kinda redefined if the set S gains new elements but the properties of symmetry and transitivity don't since the old elements that represented those traits can still do so

I thought that's what I said when talking about partial/total relations though. Like if R is sym, trans, on the entire set S and every element of S is in R, then R is an equiv. relation. But if unspecified, we can consider it a partial relation on some subset of S so reflexivity on the entirety of S isn't guaranteed

leslie townes

yeah. reflexivity requires 'more and more' as the set grows, there has to be an exact match between S and the set of t for which (t,t) is in the relation

maybe it is what you said, but i don't think it's helpful to hide what's going wrong with missing quantifiers in a proof by verbal formulas that themselves encode quantifiers :)

Obliv

$S = \{(x,y)| y=x+1 \text{and} 0<x<2\}$

Describe the equivalence relation T on the real line that is the intersection of all equivalence relations on the real line that contain S. Describe the equivalence classes of T . I think it's gotta be all the reflexive pairs of the relation S

no

it's the symmetric pairs

nvm im too tired to think lol

yeah I read that wrong. S isn't an equiv. relation on its own which is where I got confused. The equiv relations that contain S though would have the other half to make it symmetric, and include the reflexive elements etc

i.e., (y,x) s.t. x-y=1

or something like this

leslie townes

3:47 AM

well, for the same reason we had with 0 and sets containing 0 above, if R is a known equivalence relation on any set S, and S' is some known set that properly contains S, then R will be an example of a symmetric and transitive relation on S' that is not an equivalence relation on S'

you can formalize it or just take for granted that any example is going to look like that one for some choice of S and S' and R

SeekingAMathGeekGirlfriend

5:15 AM

Yo, what's up yall

nickbros123

10:09 AM

@Jakobian I think you were correct here, I was not using any property of R^m in proving that limit point compact ==> closed + bounded, The reverse direction ofcourse uses R^m to say that closed+bounded==> compact==> limit point compact.

psie

> If $Y$ generates $\mathcal G$, that is, if $\mathcal G$ consists of all events of the form $\{Y\in B\}$ where $B$ is a Borel subset of the range of $Y$, then every $\mathcal G$-measurable random variable is a measurable function of $Y$, and conversely (exercise!)

How do you go about this?

psie

10:35 AM

One direction seems pretty straightforward; if $Z=g(Y)$ where $g$ is measurable from say $(\mathbb R,\mathcal{B}(\mathbb R))\to(\mathbb R,\mathcal{B}(\mathbb R))$, then $Z$ is $\sigma(Y)$-measurable as $(g\circ Y)^{-1}(A)=Y^{-1}(g^{-1}(A))\in \sigma(Y)$ for every $A\in\mathcal{B}(\mathbb R)$.

psie

11:15 AM

Never mind my question, figured it out.

Jakobian

12:35 PM

@psie glad to help

Xander Henderson

10

A: Examples and Counterexamples of Relations which Satisfy Certain Properties

1. Reflexivity and Irreflexivity A relation on a nonempty set cannot be both reflexive and irreflexive. This follows almost immediately from the definitions: a reflexive relation on a nonempty set $X$ must contain every pair of the form $(x,x) \in X\times X$, while an irreflexive relation canno...

Jakobian

12:57 PM

:65838673 (removed)

@XanderHenderson shrugs in not caring about giving a good example something something cofinite/cocountable topology

user 85795

1:21 PM

@XanderHenderson if only all the world's problems could be solved with bigger hammers.

Xander Henderson

1:35 PM

@Jakobian I mean, that was kind of my point yesterday: Hausdorfiness is a sufficient condition for ensuring that limits are unique, but I don't know (nor care) if it is necessary. I suspect it is not. But, like I said, I don't really care.

@user85795 Oh, they can.

If a giant asteroid wipes out all life on Earth, first all of your other problems become irrelevant, and then all of your problems cease to exist.

A big enough hammer solves all problems, though perhaps not in the manner you would like.

user 85795

🤔 how are we defining a "problem" here?

> an inquiry starting from given conditions to investigate or demonstrate a fact, result, or law.

Jakobian

2:10 PM

@XanderHenderson depends. Limits of what

If a space is not Hausdorff, and x, y are points such that all their neighbourhoods intersect, then filter generated by all intersections of such neighbourhoods converges to both x and y

So its equivalent for convergence of nets/filters to Hausdorff

But for sequences? No

Xander Henderson

@Jakobian We were talking about sequences when the topic came up.

But, again, I really don't care enough to think more about this, so I'm going to stop now.

Jakobian

For cofinite topology, sequences with infinitely many distinct terms converge to every point

On the other hand, for cocountable topology every convergent sequence is eventually finite

Let those topologies be on R

Then they're both not Hausdorff, both are T_1, but in one we have uniqueness and in the other non-uniqueness of convergence of sequencess

user 85795

2:43 PM

@copper.hat The Oilers are trying to become the first team in 82 years to win the Stanley Cup after falling behind 3-0 in the Final. The 1941-42 Toronto Maple Leafs are the only team to successfully complete the comeback, defeating the Detroit Red Wings.

psie

When someone writes $E[w(X,y)|Y=y]=g(y)$ and then $g(Y)$, what does $g(Y)$ mean? Is it $$g(Y)=E[w(X,Y)|Y=y]?$$

It is done here, although $g(Y)=E[w(X,Y)|Y=y]$ is my assertion.

nickbros123

3:49 PM

I am trying to prove lim point compact $\implies$ compact in metric spaces, so I consider an open cover {$G_{\alpha}$}, of set $K$ which is limit point compact, so for each point $x$ in $K$ there's an $\epsilon_x$ ball around x is in $G_{\alpha}$,

so $K \subset \bigcup_{\epsilon_x}B_{\epsilon_x}(x)\subset\bigcup_{\alpha}{G_{\alpha}} $ , Now I wanna know if this construction is useful: I consider a random $x_0$, and it's $\epsilon_{x_0}$ ball, and I choose $x_2 \not\in B_{{\epsilon}_{x_1}}(x_1)$; similarly, I choose $x_3 \not\in B_{{\epsilon}_{x_1}}(x_1)\bigcup B_{{\epsilon}_{x_2}}(x_2) $ and inductively, $x_n\not\in \bigcup_{i=1}^{n-1}B_{{\epsilon}_{x_i}}(x_i)$.

This set has a limit point $x^*$ in $K$ (from Hypothesis). I am not sure how I will argue further, but some ideas include a subsequence of $x_n$, and choosing a small $\epsilon$ ball around $x^*$ so that a infinite terms fall into this ball.

dont take the word "random" above too seriously

EE18

4:28 PM

Dummit and Foote's presentation of presentations has me beyond stumped

Think I found my answer on main. Good thing DF is popular...

Joe

5:34 PM

I'm trying to solve Exericse 7.7 of Aluffi's Algebra: Chapter 0, but I'm having trouble understanding what (i) even means:

I know that there's a convention that for complexes of R-modules, we omit "tails of zeroes" from both ends. But there isn't a $0$ on the right hand side of the sequence ending in $\operatorname{Hom}_{R-\operatorname{Mod}}(M,L)$, and so I don't know what it is meant to be on the right of it. Does anyone know what the convention in this situation?

So I'm not sure what it means for the sequence to be exact at $\operatorname{Hom}_{R-\operatorname{Mod}}(M,L)$, as I can't even figure out what the intended sequence is.

Thorgott

there is no convention to omit zeroes

the sequence is complete as it is written there

if there was an additional $0$ term to the right, exactness would additionally mean that the last map is surjective

Joe

@Thorgott: Well, I know some authors define allow a finite sequence to be a complex, but I believe there are others that do not. For instance, Eisenbud defines a complex to be infinite in both directions in his book

Thorgott

which is generally not the case

sure, but complexes =/= exact sequences

adding zeroes to the left/right genuinely changes what it means to be exact, so there is no convention to omit them in this context. it would simply not be well-defined

Joe

5:49 PM

In the books that I've been reading, an exact sequence is defined as a particular kind of chain complex. Do you mean to say that this definition is wrong/confusing?

Sahaj

hi @nickbros123! long time; no see

nickbros123

6:15 PM

@Sahaj hi Sahaj, hows it going

Thorgott

6:31 PM

@Joe if you define a chain complex to be infinite, then that definition is incorrect technically speaking and confusing practically speaking

ephe

"If $M$ is a filtered $A$-module the $M_n$ are a basis of neighborhoods of 0 for a topology on $M$ compatible with its group structure." Does it mean here that the $M_n$ form a basis for the topological space and they are neighborhoods of 0 or does it mean that they form a neighborhood basis i.e. a fundamental system of neighborhoods? Does anyone know which of these statements is true? I've read the latter one in Bourbaki but I am confused.

Or alternatively does it mean that they actually form a base for the set of neighborhoods of 0 which makes my job a lot easier.

EE18

7:08 PM

Are there any techniques for actually evaluating sums other than praying they telescope?

I know seems like a vague question but all the sums I'm solving for right now fit exactly that bill

Seems artificial

nickbros123

@XanderHenderson reminded me of the proof that compact sets are bounded xD

Thorgott

7:21 PM

@ephe the latter

ephe

@Thorgott Thank you! You don't know how much of a relief that is, as I have no clue about how to prove the former or produce a counterexample.

Joe

7:49 PM

@Thorgott: Okay so I suppose the way to go is to allow complexes to be finite, or infinite in one/both directions. Then "exact" means what we want it to mean, I guess

psie

8:05 PM

I am considering placing a bounty on this question. Proving the identity $E(f(X,Y)\mid Y=y)=E(f(X,y)\mid Y=y)$ has been on my mind for some days now, but I'm inclined to believe that the RHS does not make any sense...

leslie townes

8:27 PM

psie it might make more sense to ask your own question. it is an old question from a long inactive user, at first glance (i am not spending any time on this) it seems at least partly a question about definitions given in some specific textbook and partly not (looks like some commenters interacted with OP about this, but i don't see it as a fully resolved issue).

at least, if you are not specifically interested in that textbook, i would write something else so you can start on a clean slate and don't have to engage with what OP was thinking about (who knows) and what that author intended (who knows)

psie

good idea, I'll write a draft

leslie townes

conditional expectation is a tricky concept and a lot of the sort of 'mathier' probability (e.g. early 20th century measure theoretic formulation) could be regarded at least in part as putting it on a less ad hoc basis where you have it some of the time by some formulas and not in other cases. many mathematicians would regard this as an question about an appropriate formulated version of a 'radon-nikodym theorem'

although how or even whether that would fit into any textbook's treatment of probability could totally vary from author to author and model to model

copper hat has a funny anecdote relating this. it's a shame he isn't here to tell it

user 85795

:(

Thorgott

8:46 PM

@Joe or you just define exactness without reference to complexes, either works

Jakobian

9:12 PM

@EE18 yes

copper.hat

9:44 PM

i think Raydon Nykodym should be taught in kindergarten.

CroCo

Hi everyone,
Please recommend a good dictionary of mathematics (if there is one). With my background in engineering, I sometimes encounter the same concept with different terms. Some writers reduce their level of formality for educational reasons.

I believe the idea of math dictionary is doable but I couldn't find a good one.

I found in Amazon "The Concise Oxford Dictionary of Mathematics" with 512 pages with no preview pages. The page numbers make me suspicious. It is very brief.

Novice

10:17 PM

possibly the princeton companion books?

leslie townes

10:30 PM

croco, if one is designing a reference like that, you can get "formality" (by which i might mean: precise definitions and internal consistency within the book about what means what), or you can have a general reference that attempts to reflect variations in how people in the world outside of the dictionary use the material. but probably not both. the former kind of thing is usually just a textbook limited to a subject more specific than "mathematics." wikipedia is a good example of the latter

there just isn't enough consistency across all disciplines that use math (or even within, like, the "pure mathematics" corner of people who use math). we run into this all the time on this chat or on the main site, where people come in with questions and the first question is, well, how are you defining that thing that you call X, or what are the exact hypotheses of the theorem that you're calling Y

leslie townes

10:46 PM

and anything resembling an explanation of why two different-looking appearances of something are 'the same concept' usually does depend on what the definitions are and what forms of reasoning beyond definitions you have access to

for the vibe of why different-looking appearances of something are 'the same concept,' there's wikipedia :)

Novice

Wu–Yang dictionary

In topology and high energy physics, the Wu–Yang dictionary refers to the mathematical identification that allows back-and-forth translation between the concepts of gauge theory and those of differential geometry. The dictionary appeared in 1975 in an article by Tai Tsun Wu and C. N. Yang comparing electromagnetism and fiber bundle theory. This dictionary has been credited as bringing mathematics and theoretical physics closer together. A crucial example of the success of the dictionary is that it allowed the understanding of monopole quantization in terms of Hopf fibrations. == History...

Jakobian

11:44 PM

@CroCo Encyclopedia of X, is probably what you're looking for, where X is a given field of math

E.g. there exists Encyclopedia of General Topology

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$Mathematics$ Mathematics