Mathematics

Fargle

12:00

But sin's, cos's, and catenaries still come up as perfectly reasonable answers to perfectly reasonable questions in applied mathematics, which is my point.

Now, does one work with exact solutions in, say, numerical analysis? No. But that's almost tautological.

The whole point of numerical analysis is the approximation of solutions which we only know must exist by the theory of the reals.

Secret

meanwhile, one thing I am still not sure about is why we want power set axioms, alternately, why we want reals to be an uncountable set instead of a countable proper class of sorts (which will be the case if we lack power set)

do the real still be as good if it is never "complete"?

Semiclassical

Right. I think it can nevertheless be an interesting to wonder what happens when we don’t permit continuity to be used

Secret

Or maybe another way to phrase it is, why are we interested in uncomputable reals, is it because we need them to ensure continuity?

Semiclassical

And in some cases that kind of discrete mindset can even be crucial

Kasper

@Fargle Do you really need the theory of real numbers for that? Cant we deduce without the theory of the real numbers that we can find a number a/b so that (a/b)^2 - 2 = epsilon where we can make epsilon as small as want it?

Fargle

12:07

@Kasper Maybe we could do that for the square root of 2. But I think you'd become hard-pressed to prove approximate versions of existence theorems for solutions to initial-value ODEs, or things of that more complicated nature.

Semiclassical

But to uncritically reject the reals isn't much better than uncritically accepting them

Fargle

Why hamstring yourself into that variety of approximate justification when you could just as easily work in the reals in thought, and then approximate by rationals in practice?

Chris Wilson

Hey guys, I see that one of the guidelines of the chat is don't ask to ask, so here's my question: If a question that I've asked in math.stackexchange is a duplicate of another question on another site on the network, should I delete the question? I tried voting to close my own thread but when specifying what the question was a duplicate of I got the message that the question has to be within the same site

Secret

Well I guess that's true

Astyx

Oh damn, are we talking about Wildberger again ?

Fargle

12:15

As another example, it's possible to find functions which are continuous mappings from the rationals to themselves, for which the intermediate value theorem fails.

However, to reject IVT on these grounds would be pretty silly. Try to draw a line from the bottom to the top of a sheet of paper without picking up your pencil, in such a way that you don't go past the middle.

@Astyx Yeah.

anon

@ChrisWilson It'd be a good idea to. I looked for a dupe of your question on MathSE but didn't find one (although I didn't look long, and I still feel like one might exist). If you do find a dupe on MSE you can use that instead.

Kasper

Oh I find it interesting to hear how you guys think about this. And I used to think like that in the past. I'm not a good mathematician to be honest. I did my bachelor, but got my interested in programming and education after that. But I believe that Wildberger has a wide arrange of videos and books where he talks about how to do mathematics like, calculus without needing the real numbers.

The question of course, why would you rebuild everything in mathematics without using real numbers?

In programming, I begin to enjoy the fact, that I could just **see** how everything works. I can stop…

Fargle

@Kasper "...they are just not something we can perceive as human beings." I mean, sure, but mathematics isn't in the business of perception. It is in the business of imagination, and I can imagine a continuous line just fine. As I said before, infinite concepts don't require infinite representation, so we can fit them in our heads without issue.

Oh, just saw the bottom. You have a good day too, @Kasper.

Astyx

Oh damn, are we not talking about Wildberger any more then ?

Fargle

I mean, I'd love to keep talking about Wildberger.

I've seen lots of his videos myself. I just flatly reject the idea that finite systems can't represent or reason about infinity.

anon

12:23

A computer works with too much information to replicate it all in our own minds - we use abstraction and honed intuition to comprehend what is happening in them, even if what is happening in them is more information that we can fit in our heads. (See also: subitizing, magical number seven +/- two.) As Fargle says, same thing with infinities. Similarly, we can't directly imagine things in more than three dimensions but we do math in higher dimensions just as well, using abstraction and intuition.

Astyx

What bugs me is that WIldberger claims it's nonsensical to talk about reals and that they don't exist. I think, as Semi already pointed out, that it's indeed interresting to see what happens if you reject the use of infinity etc, but since you can formally define those, it's not nonsense to use them

Balarka Sen

I see we are having a intellectellectuallect conversation about a wild burger

Fargle

Right. The issue isn't whether we can literally hold an infinite amount of information in our heads. It's whether the actual concept itself is consistent with the pre-existing notions of arithmetic, and it is.

It didn't take Euclid $\aleph_0$ neurons to deduce that there are infinitely many primes, but he still could.

Semiclassical

As an intellectual exercise, it’s interesting. As a philosophical commitment, it seems silly

Secret

@Astyx We sometimes discuss about finitism, "countablism" and ultrafinitism in the logic room

Astyx

12:26

Hi @bsen

Fargle

@Semiclassical For sure. And to some extent I think math should absolutely concern itself with using as little of its machinery as possible to determine the truth value of a proposition, or to solve a problem.

Secret

Actually, is PA and robinson the only known arithmetics we knew?

Balarka Sen

On a tangential note, it's kind of metaphysically surprising to me that humans - finite beings, given that we have a finite lifespan, capable of only storing and juggling only finitely many observations and conclusions, there are only finitely many neurons firing inside us - is nonetheless capable of comprehending that there is a notion of "eternity" or the infinitude of time, I guess, and also is capable of understanding that it probably lies far outside of our threshold of understanding

Fargle

But understanding is not synonymous with storage.

@BalarkaSen There is something very remarkable about the ability of a conscious mind to represent things larger than it could ever store in smaller spaces. It's reminiscent of the Berry paradox.

Balarka Sen

Yup!

Secret

12:30

@BalarkaSen I think it might have something to do with the very notion of the concept infinity can be expressed in some finite length of sentence, in that we only need finite amount of information to talk about an infinite concept (at least for those that are nice enough)

Fargle

And to those who say we can't "really" understand infinity, I would either ask them to define "understand", or say flat out that "hey, we have a word for it, we absolutely can think about it for that exact reason".

Astyx

@BalarkaSen But these neurons can be in a infinite number of configuration ...

Fargle

(To clarify: I don't mean that our ability to think about it is predicated on the existence of the word, just that the existence of the word demonstrates our ability to think about it. I don't want to get all Sapir-Whorf.)

anon

I don't think we fully understand important things about infinities. The continuum hypothesis is independent of ZFC, which means there are models where CH is true and models where it isn't, just like how the parallel postulate is true in the Euclidean plane but not in the hyperbolic plane. But we don't fully understand important things about finite situations either.

Secret

@Astyx but neurons have nonzero volume, meaning you can only pack so many of them in a region, meaning there is a maximum number of configuration

Balarka Sen

12:32

@Astyx How? There are only finite number of neurons firing discretely in a finite time (our lifespan), no?

But my point is really not that much about biology really

The standard explanation seems to me to be just extrapolating a dialectical abstract of "non finite", given we understand "infinity". But I don't buy that, really

Fargle

@anon That's a cogent point. But as you say, we'd have to reject finite reasoning as well if we demanded complete understanding. Or for that matter, all of mathematics (thanks Goedel).

Secret

@anon what are the important open questions about finitary that is outside of number theory?

Fargle

@Secret P = NP comes to mind.

Balarka Sen

That's a good example

Astyx

Fargle

12:35

And depending on what you mean by finite, there are still lots of unsolved problems in finite-dimensional topology.

Astyx

There is an infinite number of configurations for the smaller circle to be in the larger one (assuming space is not discrete)

anon

the moduli space of oriented circles (including individual points) on a sphere is $S^1\times_{\Bbb Z_2}S^2$. (Fun fact.)

Balarka Sen

Mm I see why

Cool

Secret

in practice, physical uncertainty (because every atom is ultimately quantum) will mean a nonzero area of positions makes no difference physically, and combined with putting many circles in that big circle, the configuration becomes finite effectively

Balarka Sen

The $\Bbb Z/2$-quotient comes from the fact that every geometric circle in $S^2$ has two "centers"

I think

anon

12:38

yeah

something done in Lie sphere geometry

pretty sure the moduli space of oriented k-spheres in an n-sphere is $S^1\times_{\Bbb Z_2}\widetilde{F}(k+1,k+2,n+1)$ where $\tilde{F}$ is the oriented flag manifold

Balarka Sen

This moduli space is the space of pairs (point, circle) such that point $\in$ circle, right?

Feels like blow-up kinda

anon

no, just the space of circles in a sphere

Balarka Sen

Hmm.

anon

not necessarily great circles

I like to think of $(e^{i\theta},n)\in S^1\times S^2$ as the circle $n\cdot x=\cos\theta$ on the sphere with orientation from the sign of $\sin\theta$

such a circle has spherical radius $\theta$ around center $n$ (or $\pi-\theta$ around $-n$)

Balarka Sen

Oh, by individual points you mean circles of zero radius in $S^2$. The $S^1$ factor comes from the radius, which is determined by an arc going between the two centers of the circle, which are identified in the quotient

anon

12:42

right

Fargle

Y'all done lost me.

Secret

I knew nothing about moduli spaces

(or any algebraic geometry in general)

and I am still at least countably far away before I can start reading an algebraic geometry book

Balarka Sen

God I need to study why am I chatting about infinities now

Fargle

@BalarkaSen Because it's fun!

Secret

yeah, I also need to write up my report

It is fun, but it is also like being intoxicated and you want to stop but cannot

Balarka Sen

12:46

@Fargle A likely excuse

Secret

Perhaps the reason I don't need to take drugs because infinity is my drug

Balarka Sen

@Secret hahahahah speak for yourself suckers

Kasper

Euclid actually avoided the word "infinite" in his theorem.
https://math.stackexchange.com/questions/920406/why-does-euclid-write-prime-numbers-are-more-than-any-assigned-multitude-of-pri

He said something like:
There are more prime numbers than any amount of prime numbers.

Secret

well if given some n, you always have some m such that n<m, then the set has to have at least countably many elements in it (assuming axiom of choice)

Fargle

@Kasper That's entirely orthogonal to my point.

Secret

12:49

if not assuming axiom of choice, then you can have many different possible flavors of uncountable

which will be like meeting Cthulhu

Kasper

I thought you said that Euclid deduced that there were infinitely many primes. I don't think he claimed he deduced that.

Fargle

The language he used is inconsequential. My point was just that Euclid was able to attain a result that in modern language would be called "the infinitude of the primes" with only a finite amount of reasoning.

Secret

depending on the axioms, there might be a set of primes of some infinite cardinality, or you just have a proper class of primes

(actually, can we have proper class of finite size?)

Kasper

The question is, should we use this modern language. Or should we go back to the more humble approach of saying just that you can always find a new prime, no matter how many primes you already found.

Fargle

It's not more humble. It's logically 100% equivalent.

Secret

12:54

If a collection is unbounded, then by definition it cannot be bijected with any natural number, and hence infinite by definition

Fargle

"No matter how many primes you've found, there will always be another" $\iff$ "There is not a finite number of primes" $\iff$ "There are infinitely many primes"

Secret

so they are logically equivalent

anon

"Sit down. Be humble." - Kendrick Lamar

Fargle

All "infinite" means is "not finite". Every set is either finite or not finite. Euclid established that the set of primes is not finite even if he would not have said it like that, and as a result, they're infinite.

Kasper

Another option would be, we can not construct the set of prime numbers. There doesn't exist something like the set of prime numbers. That is what Euclid probably would have said.

Secret

13:00

but we can still have a proper class, right?

I cannot think of any scenario where a proper class never exist

Fargle

@Kasper But I ask again, why hamstring yourself like that? There's literally no reason that I can see to reject that which is not finite.

anon

Reminds me of the MSE question "what does Gauss think about infinity?"

Probably would have been horrified for $15$ minutes. Might have taken him a day or two to really make good use of the new tools. — André Nicolas May 4 '12 at 22:18

Secret

is there a scenario where proper class don't exist?

Fargle

It shouldn't be blindly accepted either, but in my experience, it never is. It just turns out to be perfectly logically consistent with everything we've ever done.

I honestly can't wrap my head around being so avoidant of the concept of infinity. By all means I think people should be allowed to take that position, but it just seems like a loooooooooong walk for a short drink of water.

Kasper

@Fargle I don't think we necessarily have to talk about sets that are infinite. Why can't we just say, hey this number has a property. The number $2$ has property to be natural and prime. We don't have to put it in some infinite set.

anon

13:07

It is if you want to do set theory. And mathematicians use infinite sets to define reals. And applied mathematicians use real numbers in models that fundamentally rely on properties of real numbers.

Fargle

@Kasper Again, why? What does only reasoning about finite sets actually do for us?

It doesn't buy us any more logical justification, because as I've said, infinity is logically consistent.

We don't have to use computers, either, but it's very practical for us to do so. I'd hate to do a Taylor approximation every time I want the sine of 15 degrees.

Secret

@anon but we don;t have such thing as the "set of all cardinals" because it is inconsistent, yet in set theory we also don't specifically isolate a notion of proper class. So how do we describe this container of all cardinals in set theory, or we just accept that it is unbounded?

Fargle

On the same token, we don't have to reason about infinity at all, but if we bar that door for ourselves, we lose calculus, we lose functional analysis, we lose almost everything.

anon

We could have a society without modern technology, asking "why can't we just do stuff without all these fancy newfangled things?" but that would be silly. I'm not going to tip-toe around or walk on eggshells talking about real numbers, and I'm not going to qualify everything with complicated and unnecessary disclaimers about "approximations" of things some people believe don't exist.

Kasper

@Fargle You have seen Wildberger videos right? So you know that he is rebuilding calculus, and other topics, without using infinite set. You do have the same results, you just have another foundation.

anon

13:12

@Secret Cardinality talks about sets having bijections between them, and there is no set of all sets. Sets, after all, are sets of things, and we have no a priori mathematical definition of "all the things." (Essentially, the issue is with the axiom of unrestricted comprehension.)

Kasper

And of course, some results, you don't have, but are the results that can not be computed anyway.

Secret

right

anon

@Kasper Which is something he has to do because of his philosophical commitments to finitism in the face of the utility of calculus. Totally unnecessary to deliberately handicap ourselves.

Fargle

For my money, $\sqrt{2}$ and $\pi$ and $e$ all exist in the same sense in which 2 and 3 exist, and $\{2x\;:\;x \in \Bbb Z\}$ exists just as surely as $\{1,2,3\}$ exists.

nbro

Does anyone here have a formal definition of a connected component of a polygon? By components of a polygon, people sometimes mean its vertices, edges or faces. Connected components of a polygon do not seem to have something to do with connected components of graphs.

anon

13:16

the only way that would make sense to me is if you allow the word "polygon" to refer to potentially disjoint collections of connected polygons

Fargle

@Kasper Can you give an example of a result in our calculus which Wildberger's calculus is unable to attain, and explain why that result is worth rejecting?

nbro

In particular, a monotone polygon is a polygon such that there's a line which intersects the polygon in one connected component.

A simple polygon, apparently, may have more than one connected component

Perturbative

Does a countably infinite set need to have cardinality of $\mathbb{N}$?

nbro

For example, look at the following picture.

Fargle

@Perturbative This is the definition of a countably infinite set, if I remember right.

nbro

13:18

The first two polygons are monotone, because the horizontal lines intersect the polygon in one "connected component" (according to some authors).

Fargle

So yes.

Perturbative

Ahh okay

nbro

Intuitively, I can see what they mean by "connected component" by looking at the pictures above. However, I would like to have a formal definition of "connected components", because I am not actually sure that if what I think is a connected component is what these authors mean for connected component.

Semiclassical

In the case of intervals on the real line I’d imagine it’s “if a, b in S and c in (a,b) then c in S”

Maybe that’s more like convexity though

nbro

@Semiclassical Is this an answer to my question?

Semiclassical

13:29

Yeah. Not a very good answer, tho

Kasper

@Fargle Good question. I'm watching his videos, I will tell you if I find such an example, he gave one interesting example that I forgot, try to find it now.

nbro

So, what is the connected component in your example?

Semiclassical

What I had in mind was that a subset S of the reals would be connected if it satisfied the above

Fargle

@Kasper Thanks! I do still feel like I'm being harsh, and I apologize if that's the case. I don't really even think you're wrong to reason only in finite sets, just that it's at best an alternative.

nbro

I understand that, but a polygon is a two (or three) dim object.

Fargle

13:32

My problem with Wildberger isn't in the new stuff he comes up with, I just think he doesn't have to reject what he rejects, whereas he thinks he does.

Semiclassical

Sure, but any intersection of a line with a polygon in the plane is 1D

And that’s what your figure seems to be about

Fargle

Ah, I see what you're getting at now, @Semi

Semiclassical

But it’s still not a good answer on my part

Fargle

The lines are the things whose connected components you're worried about, not the polygons

Semiclassical

Because the 2D version of my statement is the definition of a plane convex set not a connected set

Fargle

13:35

But it's equivalent for 1D, and the question is asking about the connected components of 1D things.

Semiclassical

Yeah

Connected is easier in 1D

Fargle

What I mean is: they're not talking about connected components of polygons

They're talking about the lines

So the definition of "connected component" for polygons is actually irrelevant.

A polygon is always a connected component, except possibly with weird conventions.

nbro

So, how do you interpret the sentence "The first two polygons are monotone, because the horizontal lines intersect the polygon in one "connected component"?

Semiclassical

However, I’ll also note that the Wikipedia article for connected space does have the general definition of “connected component”

Fargle

I interpret that as "the parts of the lines which lie inside the polygon aren't disconnected".

Look at one of the red lines.

There are two disconnected pieces that lie inside of the star.

nbro

13:38

@Fargle Yeah, that was more or less my intuitive understanding...

Semiclassical

Question, though: is the polygon just the boundary there, or the interior as well?

This won’t change the answer much but it does change the statements a little

Fargle

@Semi I assume it'd be the interior too, because any line that's not "tangent" (bad word, but close) to the polygon would always hit it in at least two distinct points.

Abra001

@LeylaAlkan for further understanding, the op counted permutation cycles of order 2 included in second section of the question since lcm(2,4)=4, while he didnt for order 3 since lcm(2,3)=6 =/= 3

Semiclassical

Point.

Fargle

That is, I think the "connected component" definition of a monotone polygon is talking in terms of the interior.

@nbro More formally, a set is connected if it's not made up of two disjoint non-empty open subsets.

nbro

13:42

@Semiclassical I am not sure. If you're interested in a specific use-case of that expression, have a look at page 51 of the book "Computational Geometry: Algorithms and Applications" (3rd edition) by Berg et al.

Fargle

Or more simply in the case of 2D space, if there aren't two disjoint open sets that separate the set into two parts.

For example, the unit disk is connected, and so is the unit circle, but two unit disks that are far enough apart are disconnected.

Semiclassical

I think Fargle is right, on the grounds that the definition doesn’t make sense if you only look at the 1D boundary rather than the 2D region

Abra001

good still some websites are left directory browsable...

Fargle

The connected components of a set are in a sense the "largest connected subsets".

nbro

The definition of a connected component from en.m.wikipedia.org/wiki/Connected_space doesn't seem consistent with your definition "the parts of the lines which lie inside the polygon aren't disconnected".

Fargle

13:47

If a set has one connected component, it is connected.

nbro

In the picture above, the stars would be just one connected component according to the definition from en.m.wikipedia.org/wiki/Connected_space

Abra001

speaking of polygons does anyone have a slight-to-partial knowledge about this question

Fargle

When the definition says "there's a line which intersects the polygon in one connected component", they mean "intersection" as in "the set of all points which are both on the line and inside of the polygon".

That set is what has to have only one connected component.

Not the polygon. Those always have only one connected component.

So, for instance, the part of one of the red lines which lies inside the star is made of two connected components--the part going through one "leg" of the star, and the part going through another.

nbro

@Fargle Please, have a look at page 51 of the book I mentioned above.

So, the definition of a connected component, in this case, would be a segment, formed by this (sweep) line, which lies inside the polygon

Fargle

Right.

nbro

13:53

It's the most formal definition I can think of for this case.

Fargle

That exactly agrees with the topological definition.

nbro

Hm, why is that? Is it because the whole space would be the line and the connected components the segments inside the polygon?

Fargle

@nbro The whole space would be the collection of all the segments inside the polygon, and the connected components would be each single segment.

(Or point, if the line happens to hit the polygon at a vertex.)

nbro

@Fargle Right

Would you be willing to give a more formal answer, if I asked a question on the main website? I think it would be useful to other people

Fargle

I could try.

user193319

13:59

Let $\mathcal{F}$ be a family of measurable functions over some measurable set $E \subseteq \Bbb{R}$. Define $\mathcal{F}^+ = \{f^+ \mid f \in \mathcal{F} \}$, and define $\mathcal{F}^-$ similarly. Is the following true: $\mathcal{F}$ is uniformly integrable if and only if both $\mathcal{F}^+$ and $\mathcal{F}^-$ are uniformly integrable. I have a proof, and I can't spot any errors in it, but I just want to make sure it isn't obviously false.

nbro

This question would be better asked on cs.stackexchange.com, though.

Computational geometry is more a sub-topic of CS.

Even though the problem arises because of mathematical formalisms

So, in any case, it isn't a bad choice to ask the question here

Fargle

I'm not sure. Whichever you decide makes the most sense.

I think it's perfectly fine for MSE, because it is a question about a mathematical definition.

Kasper

@Fargle Okay, I got it.
There are of course many things we already talked about. Like that non-paralel lines don't have to have an "exact" intersection. In principle, the results are the same, you just think about it in a more applied way. You can not draw the graph of x^3+y^3=1, nobody can, no computer can, but you can teach a computer to draw pairs (x,y) so that x^3+y^3-1 < 0.0001. That is just a more applied way, a more "real" way of thinking about calculus.
But beside that he mentions the Banach–Tarski paradox, non-measurable sets, non-measurable functions, space filling curves. He say…

Fargle

@Kasper I don't necessarily think that mathematics is predicated on a practical notion of usefulness, which is why I choose not to reject them.

Even if you don't think Banach-Tarski is applicable to the world though--and I certainly don't, except for maybe in some contrived situation--it seems like throwing the baby out with the bathwater to therefore reject anything whose logical conclusion is Banach-Tarski.

As well, infinite results that seem useless (like space filling curves) do often find practical application with a slight shift in perspective. youtube.com/watch?v=3s7h2MHQtxc

Lozansky

How can I integrate this: $\int_{\mathbb{R}} \dfrac{iw-w^2}{1+w^6} dw$

user21820

14:12

@anon No we don't need sets at all. See higher-order arithmetic with a separate sort for each order. Also see reverse mathematics, which has so far shown that ACA suffices for practically all real analysis with practical real-world implication. ACA is very low in the hierarchy, being predicative second-order arithmetic.

nbro

Here you have the question.

0

Q: A polygon is monotone if there is a line that intersects it in one single connected component

In the following picture The first two (top) polygons are monotone, whereas the last two (bottom) polygons are not. At page $51$ of the book "Computational Geometry: Algorithms and Applications" (3rd edition) by Berg et al., the authors state Since $P$ is not monotone, there is a horizonta...

user131753

2

Q: Connectedness in Product Topology via Simple Chain

It is well known that if $(X,\tau_X)$ and $(Y,\tau_Y)$ be two connected topological spaces then $(X\times Y,\tau)$ is also connected where $\tau$ is the product topology on $X\times Y$. However, all the proofs that I have seen either uses an argument like this or this or this. But after seein...

general-topology connectedness product-space

Kasper

I personally don't go that far either. I think there is a place for mathematics that is not (yet) useful. I think there is a place for things that you can only imagine, but not write down (like an infinite sequence, or an infinite set).

I just think that an approach without infinite sets, and with only things you can compute, only mathematical objects, that you can actually write down, without needing three points at the end of your statement, I think that approach may be very well suited for education and applied mathematics.

I just started watching his course on Calculus:
Algebraic Calculus One

Invitation to a more logical, solid and careful analysis

►

Fargle

@Kasper I'm not so sure, but I don't teach applied mathematics. I do think more care should be taken with blithe mentions of infinity in primary and secondary math.

user21820

@Fargle If the foundational system cannot be justified philosophically somehow, I do not think it would be a good idea to just accept it at face value.

After all, given any consistent system S that can be reasonably claimed to be foundational, S+¬Con(S) is consistent as well, and proves everything that S does, plus a bit of rubbish that we can never empirically refute...

Fargle

14:26

@user21820 Agreed. I said earlier that blind acceptance isn't the goal either. But I don't think "physical significance" is a necessarily favored justification.

user21820

@Fargle I don't (can't) claim that physical significance is the only way to justify it, hence my phrase "philosophically". The problem is that apart from the real world we do not have any well-defined notion of soundness to begin with!

Fargle

Right, it is sticky. But I accept the consequences of ZFC, and my justification of that fact is that I don't find anything objectionable about its axioms.

user21820

We could restrict to arithmetic soundness if we assume that there is a physical model of PA, but the issue remains that we can never tell whether our foundational system proves rubbish like its own inconsistency.

@Fargle I object to replacement.

And regularity (on philosophical grounds that it was snuck in just to make the set-theoretic universe appear nice).

Fargle

@user21820 That the image of a set is always a set?

user21820

@Fargle That's not what it is...

Fargle

14:30

@user21820 Is that not what replacement is? I just looked it up.

user21820

@Fargle From that I think it's fair for me to presume that you don't actually know the technical details of ZFC?

Fargle

Yeah, that would be fair. If I could be permitted to be more precise, I find nothing objectionable about the heuristic notions encapsulated by ZFC.

user21820

Replacement is a schema, which in more philosophical terms states that every definable function-symbol with domain already an object can be reified as an object.

Unrestricted comprehension, in those same terms, states that every definable predicate-symbol can be reified as an object. That led straight to Russell's paradox.

user131753

@Fargle: There had been several discussions regarding the "philosophical justification" (mostly about the "soundness" of the axioms of Peano Arithmatic) with user21820. If you are willing to go through it then I may provide you the links.

Fargle

In that language, I reject unrestricted comprehension but not replacement.

"with domain already an object" being the fundamental reason.

user21820

14:33

@Fargle The problem is that restricted comprehension is just a hack.

It was never philosophically justified.

Neither was replacement.

Fargle

In what sense do you mean it was never philosophically justified?

I'm not trying to be obtuse, just honestly asking.

user21820

@Fargle If you check the history of ZFC, you will find that nobody ever gave any cogent philosophical justification. In fact, professional logicians today agree with the general points of my view, namely that ZFC cannot be justified without already assuming the existence of ZFC-like ordinals.

Circular, in other words.

Some are even doubtful of ZFC's consistency.

in Logic, Mar 5 at 7:07, by user21820

> I'm not certain whether I agree with the statement "However, when we use an axiom in a proof, we normally know whether it holds for standard integers or not." For example, ZFC proves the consistency of theories which I'm not fully confident are consistent, so I'm not fully confident in ZFC's consistency, let alone its arithmetical soundness. But the ZFC axioms are of course widely used (including by me!).

in Logic, Mar 5 at 7:07, by user21820

> So I'm not sure that we are always justified in our confidence in the arithmetical soundness of the axioms we use. – Noah Schweber 3 hours ago

@Fargle: You are welcome to come to the Logic room for further in-depth discussion on such matters.

Silent

What is negation of this (that is, how to state that $H$ is not normal subgroup of $G$): For all $h\in H$ and all $g\in G$, $ghg^{-1}$ is in $H$. My first thought was: 1. 'There exists a pair $h,g$ such that $h\in H$ and $g\in G$, such that $ghg^{-1}\notin H$ ', but then I recalled that negation changes 'and' to 'or', so I had this second thought: 2.'There exists $h\in H$ or $g\in G$, such that $ghg^{-1}\notin H$'.

Fargle

@user21820 Fair enough. I guess all I mean is that I don't necessarily care whether it's true or corresponds to anything.

Other than its own constructs.

user131753

@Fargle: See also this.

user21820

14:39

15

A: Is V, the Universe of Sets, a fixed object?

As you noticed, the iterative conception of sets requires a pre-existing universe of sets, and ordinals with which we can label the stages. So if you work within ZFC itself, in other words within an existing model of ZFC, you can perform that iterative construction to obtain $V$. Like Asaf Karagi...

Fargle

I unfortunately have to bow out, gotta write a paper. Good discussion, though.

Take care!

user21820

@Fargle: Sure! If you are interested later, see the above where I cited Boolos who was one of the few logicians who publicly shared the same view on replacement.

@user170039 That's a super-long article! It may be worth pointing out a few key points relevant to our discussion on whether Zermelo's set theory (a subsystem of ZFC) can be philosophically justified.

> I have not yet even been able to prove rigorously that my axioms are “consistent”, though this is certainly very essential; instead I have had to confine myself to pointing out now and then that the “antinomies” discovered so far vanish one and all if the principles here proposed are adopted as a basis. But I hope to have done at least some useful spadework hereby for subsequent investigations in such deeper problems. (1908b: 262) − Zermelo

user131753

@user21820 Sure.

user21820

> (d) The last line of objection was to a general feature of the 1904 proof, which was not changed in the second proof, namely the use of what became known as ‘impredicative definition’. An impredicative definition is one which defines an object a by a property A which itself involves reference, either direct or indirect, to all the things with that property, and this must, of course, include a itself. There is a sense, then, in which the definition of a involves a circle.

> Both Russell and Poincaré became greatly exercised about this form of definition, and saw the circle involved as being ‘vicious’, responsible for all the paradoxes. If one thinks of definitions as like construction principles, then indeed they are illegitimate. But if one thinks of them rather as ways of singling out things which are already taken to exist, then they are not illegitimate.

> [...] In short, Zermelo's view is that definitions pick out (or determine) objects from among the others in the domain being axiomatised; they are not themselves responsible for showing their existence.

> In the end, the existence of a domain B has to be guaranteed by a consistency proof for the collection of axioms.

And that last point is where everything falls apart if we cannot justify Z's consistency non-circularly.

> Zermelo initially had doubts about the Replacement Axiom (see the letter to Fraenkel from 1922 published in Ebbinghaus 2007: 137), but he eventually accepted it, and a form of it was included in his new axiomatisation published in 1930 (Zermelo 1930).

@user170039: I was not aware of this claim that Zermelo himself doubted replacement at first, and I also have not seen his letter to Fraenkel. If you have seen it, feel free to post the relevant contents here.

Zee

14:57

Is the axiom of choice essential outside of analysis ? Is zorns Lemma essential outside of analysis?

user21820

@Zee If you like to be able to do things in real analysis easily, all you need is dependent choice.

As far as we know, Zorn's lemma (for arbitrary posets) has no apparent real-world significance. Similarly for transfinite induction/recursion. Note that we do not need these for the nice structures that we have so far found to be relevant to the real world. For example C is the algebraic closure of R and can be proven so without any ordinals or choice. Same for algebraic closures of finite field.

If you do not accept Zorn's lemma in general, you will lose some 'elegant' theorems, such as "every field has an algebraic closure" and "every vector space has a basis".

But the core mathematical facts will remain: Every well-orderable field has an algebraic closure. Every well-orderable vector space has a basis.

Zee

does Zorns Lemma imply choice ?

user21820

@Zee Over ZF, AC and Zorn's lemma are equivalent.

Zee

That’s too bad , choice didn’t used to bother me but now I see it’s kinda artificial

It’s like being born in society , “I didn’t agree to the rules, I was just brought here without permission “ , it feels kinda like that going into math with choice

user21820

@Zee: I suppose you will find yourself still quite comfortable with DC (dependent choice).

user131753

15:09

@user21820 In the book Ernst Zermelo: An Approach to His Life and Work there is a subsection entitled The Fraenkel Correspondence of 1921 and the Axiom of Replacement. Probably that would be a good place to search for the letter. But I don't know for sure as I haven't read that book.

user21820

@user170039 Thanks for that.

user131753

@user21820 I have just noted that the reference you have given here is the same to which I have given above. Sorry for not noticing it earlier.

user21820

@user170039 I did not notice it was the same either.

Zee

DC seems interesting, how powerful it is ?

user193319

If $A \subseteq \Bbb{R}$ has finite Lebesgue measure, does this mean $A$ can be fit into some compact interval?

Alessandro Codenotti

15:12

$A=\Bbb Z$

user21820

@Zee Good enough over any reasonable foundational system for you to recover almost all measure theory. What you lack is the ability to conjure weird stuff like the Vitali set.

user193319

@AlessandroCodenotti Dang...Thanks!

user21820

22

A: Does constructing non-measurable sets require the axiom of choice?

In the 1960's, Bob Solovay constructed a model of ZF + the axiom of dependent choice (DC) + "all sets of reals are Lebesgue measurable." DC is a weak form of choice, sufficient for developing the "non-pathological" parts of real analysis, for example the countable additivity of Lebesgue measure ...

Zee

And how is DC related to zorns Lemma ?

user21820

@Zee Over ZF, DC is strictly weaker than AC, since AC gives you the Vitali set, and AC is equivalent to Zorn's lemma.

DC is the ability to pick existential witness at each step in a countable sequence of steps, depending on the earlier choices.

Zee

15:15

Interesting, I don’t see why DC isn’t adopted

Non measurable sets are annoying as fudge

user21820

The typical textbook proof that continuity is equivalent to sequential continuity actually uses DC (but usually no teacher/student notices).

@Zee Hahaha...

Zee

What about Hahn Banach ?

Silent

@AkivaWeinberger, will you please look at this?

38 mins ago, by Silent

What is negation of this (that is, how to state that $H$ is not normal subgroup of $G$): For all $h\in H$ and all $g\in G$, $ghg^{-1}$ is in $H$. My first thought was: 1. 'There exists a pair $h,g$ such that $h\in H$ and $g\in G$, such that $ghg^{-1}\notin H$ ', but then I recalled that negation changes 'and' to 'or', so I had this second thought: 2.'There exists $h\in H$ or $g\in G$, such that $ghg^{-1}\notin H$'.

user21820

@Zee I don't know much about that. Wikipedia lists it as equivalent to AC over ZF.

Alessandro Codenotti

@user21820 weird, Hahn-Banach is strictly weaker

BPI should be enough to prove Hahn-Banach

user21820

15:19

@AlessandroCodenotti Oops Wikipedia got it right. I didn't read the heading carefully.

Alessandro Codenotti

even if the standard proof uses Zorn's lemma

Zee

Hahn Banach is essential to analysis, if we can get that , your half way there

user21820

@Zee Um. Wikipedia says "In ZF, one can show that the Hahn–Banach theorem is enough to derive the existence of a non-Lebesgue measurable set."

Sounds like fudge? =P

Zee

God damn it

Xander Henderson

mmm fudge

I like fudge made with hazelnuts

Zee

15:22

Perhaps there is a weaker form of Hahn Banach theorem still as useful

user21820

But in general, just look at the proof and see whether it can be done using a transfinite recursion over the structure in question. If so, then the core fact remains unchanged given a well-ordering of that structure. In general, I'm quite sure that the applications you actually want to use all these general theorems for can be done without using those in full generality.

For example, most vector spaces people are most interested in are nice ones and easily shown to have a basis.

Zee

Yes yes , am thinking the pathology comes from wanting to do that to all vector spaces , but the vector spaces we care about in analysis are tinnny subset

user21820

@Zee: Anyway I got to go soon. You're welcome to the Logic chat-room for further discussion next time!

♦ $Mathematics$ Logic

This room is meant for discussion about logic, including found...

1

Jannik Pitt

16:30

If I have a first countable topological space, then sequential continuity implies continuity. What doesn't work about the proof if every neighbourhood has an uncountable neighbourhood base? I don't quite understand why you have to have a countable neighbourhood base (or do you have to have a at least countable neighbourhood base)?

Alessandro Codenotti

The problem with points without a countable nbhds basis is that they can't be reached by sequences

Jannik Pitt

16:52

So convergence of sequence can't be defined in those spaces? Is that the reason why you generalise them into nets (I don't know anything about nets)

Alessandro Codenotti

You can still speak about convergence of sequence, but that's not enough to describe the topology of the space

For example there could some subspace $A$ such that there is an $x\in\overline{A}$ but such that there is no sequence of points in $A$ converging to $x$

0celo7

you need nets

Alessandro Codenotti

Because the sequential closure is smaller than the actual closure in general

Akiva Weinberger

@Silent Nah you were right the first time

Negation of $\forall h\in H,\forall g\in G,ghg^{-1}\in H$ is $\exists h\in H,\exists g\in G,ghg^{-1}\ne H$

Silent

@AkivaWeinberger Wow! now it is so clear!

Jannik Pitt

17:06

But what I don't understand why sequential continuity doesn't imply continuity in a space with an uncountable nbhd basis. Why do you need nets for that?

Akiva Weinberger

Technically $\forall h\in H,P$ is a shorthand for $\forall h,(h\in H\implies P)$

and $\exists h\in H,P$ is a shorthand for $\exists h,(h\in H\land P)$

Leyla Alkan

Hi @Abra001 , yes exactly!

Akiva Weinberger

But the negation of $\forall h\in H,P$ is still $\exists h\in H,\lnot P$ because the negation of $h\in H\implies P$ is $h\in H\land\lnot P$

Alessandro Codenotti

@JannikPitt Are you familiar with ordinals? In particular $\omega_1$?

Akiva Weinberger

@Silent

Jannik Pitt

17:13

@AlessandroCodenotti Yes, roughly.

Alessandro Codenotti

Ok, consider the topological space of ordinals$\leq\omega_1$ with the order topology

Silent

Thank u!

Alessandro Codenotti

Uhm, I wanted to go through the fact that every sequence in $\omega_1$ is bounded but maybe there is an easier example

Rick

17:29

This question has been annoying me all day, can someone help? If $(6-12x+12x^2)^n = \displaystyle \sum_{r=0}^{2n} T_r x^r$, prove $ T_r = (-2)^r 3^n \bigg[\binom{2n}{r} + \binom{2n-2}{r}\binom{n}{1} + \binom{2n-4}{r}\binom{n}{2} + \dots \bigg]$

I tried by writing $(6−12x+12x^2)^n= 6^n(1-2x+2x^2)^n = 6^n\displaystyle \sum_{a,b,c}^{a+b+c = n} \frac{n!}{a!b!c!}(-2x)^b(2x^2)^c = 6^n \sum_{a,b,c}^{a+b+c = n} \frac{n!}{a!b!c!}(-1)^b 2^{b+c} x^{b+2c}$ which gives two conditions, $a+b+c = n$ and $b+2c =r$

So the (a,b,c)'s possible here are (n-r,r,0) ; (n-r+1,r-2,1) ; (n-r+2,r-4,2) ...

user193319

Claim: Let $f : \Bbb{R} \to [-\infty, \infty]$ be integrable. If $\int_U f = 0$ for every open set $U$, then $f=0$ a.e on $\Bbb{R}$. Attempt: Since $\Bbb{R}$ is a nonempty open set, $\Bbb{R} = \bigcup_{n=1}^\infty U_n$, where the $U_n$ are pairwise disjoint open intervals. Hence $\int_{\Bbb{R}} f = \sum_{n=1}^\infty \int_{U_n} f = 0$....Unfortunately, to my chagrin, this doesn't imply that $f = 0$ a.e. on $\Bbb{R}$, since $f$ is not necessarily a nonnegative function. I could use a hint...

Rick

17:49

So substituting those a,b,c's I get the coefficient of $x^r$ as $6^n \bigg[\frac{n!}{(n-r)!r!0!}(-1)^r 2^{r} + \frac{n!}{(n-r+1)!(r-2)!1!}(-1)^{r-2} 2^{r-1} + \frac{n!}{(n-r+2)!(r-4)!2!}(-1)^{r-4} 2^{r-2} + \cdots \bigg] =$

$\displaystyle 6^n(-1)^r2^{r} \bigg[\frac{n!}{(n-r)!r!0!} + \frac{n!}{(n-r+1)!(r-2)!1!} 2^{-1} + \frac{n!}{(n-r+2)!(r-4)!2!} 2^{-2} + \cdots \bigg]$

$ = \displaystyle (-2)^r 3^n \bigg[\frac{n!}{(n-r)!r!0!}2^n + \frac{n!}{(n-r+1)!(r-2)!1!} 2^{n-1} + \frac{n!}{(n-r+2)!(r-4)!2!} 2^{n-2} + \cdots \bigg]$

I guess I have to prove that $ \displaystyle \bigg[\frac{n!}{(n-r)!r!0!}2^n + \frac{n!}{(n-r+1)!(r-2)!1!} 2^{n-1} + \frac{n!}{(n-r+2)!(r-4)!2!} 2^{n-2} + \cdots \bigg] = \bigg[\binom{2n}{r} + \binom{2n-2}{r}\binom{n}{1} + \binom{2n-4}{r}\binom{n}{2} + \dots \bigg] $ now, which I'm not able to..

Akiva Weinberger

18:14

@MeowMix 12tone answered one of my questions again

but he mispronounced my name :(

0celo7

@BalarkaSen probability and algebra overlap

what a shit

@Semiclassical I'll take stat mech. Rip my plans of having no homework

Semiclassical

18:32

Stat mech is fun

Was for me anyways

0celo7

18:48

I could take ANT

but that might kill me

Xander Henderson

19:17

ANT if offered in the entymology department, right?

Rick

Ok I posted my question on the main site: math.stackexchange.com/questions/2715143/…

0celo7

@XanderHenderson algebraic nanoparticle theory actually

physics

Xander Henderson

You're like some kind of chat bot... just stringing words together

I mean, all of those words are english

and it seems like that phrase is grammatical

but it is utter nonsense...

0celo7

:(

I thought we were friends

Xander Henderson

Me, too. But how can one be friends with a chatbot? :P

"Algebraic nanoparticle theory..." double ewe tea eff?

rschwieb

19:34

Any French speaking algebraists in the house today?

0celo7

well, you use algebra to study nanoparticles

Xander Henderson

I feel like this is a thing you have made up

just to pester real mathematicians, who all do analysis

/me runs and hides

Daminark

@XanderHenderson rekt by nonexistence of IRC

Xander Henderson

I'm old. Habits from the days of IRC die hard. :'(

quallenjäger

19:53

Can I actually find a inverse function for $g$ if $g$ is continuous and only monotoneic?

not strictly monotonic.

Antonio Vargas

Is it just me, or are these kinds of questions almost as bad as "guess the next number" questions?

rschwieb

@AntonioVargas Even worse: at least guessing the next number suggests some form of pattern seeking, but the thing you're pointing to is of the form "why are things the way they are?" as if there were some sort of deep interpretation.

Rumplestillskin

20:10

Hello all! Can someone help me with some index notation: I have the following: (V^j)^2 V^j \delta_{ij} ..... when I equate the indices in the Kronecker delta does the expression become (V^i)^2 V^i or (V^j)^2 V^i?

0celo7

@quallenjäger of course not

@Rumplestillskin that's awful notation that means nothing

Rumplestillskin

@0celo7 It could well be. My original expression was (V^j V^j)^2 and I wanted it's derivative with respect to V^i. This is only in Cartesian coordinates.

quallenjäger

Lol I thought first my chatjax is broken

Akiva Weinberger

@quallenjäger What if $g$ is constant on some interval

or, hell, what if it's literally just the constant function

quallenjäger

Thats what I am wondering. I have the following lines in a book

Suppose that $I=[a,b]$. Put $J=g(I)$. Calculus gives us the following facts. If g is continuous and monotonic increasing, then 1. $J$ is the interval [g(a),g(b)]. 2. $g$ has an inverse function $h:J\rightarrow I$ such that $g(x)=y$ if and only if $x=h(y)$.

The second point seem to me so wrong. I never learned this theorem from calculus.

Akiva Weinberger

20:24

I think they're using "monotonic" to mean "strictly monotonic" to be honest

quallenjäger

I don't think so,. In the proof they have the following lines:

If $g$ is monotonic increasing, then $g(X) \leq g(h(y))$ if and only if $X \leq h(y)$

Akiva Weinberger

That's still true

It's just not a definition

In fact, it's false for nonstrictly monotonic things, isn't it?

quallenjäger

You mean the line in the proof?

Ok I see, it makes sense. Thanks

cpx

20:55

A = min⁡{x │ y = Pmax }. Set 'A' is a set of all x such that y is Pmax. What's the min in front and how to read it?

Leaky Nun

@cpx set 'A' isn't the set of all x, A is the minimum of that set

♦ $Mathematics$ Logic

Transcript for

$Mathematics$ Mathematics

♦ Logic

Transcript for

♦ $Mathematics$ Logic

$Mathematics$ Mathematics