@JacobBlack I was Jacob first, so I maintain it's your fault ;).

Hi

Hi

if you don't use cardinality

is there a difference betweeen finite and infinite sets

maybe a stupid question maybe i try to ask it in a different way

is their beside the cardinality a difference between finite and infinite sets

@DominicMichaelis Typically one defines a set X to be finite if for some n there is a bijection with {1,2,...,n} for some natural number n.

yeah so there is no difference (as is unterstand)

I

Well, assuming a weak form of the axiom of choice

I'm not sure which we need off the top of my head. We have that a set X is infinite if and only if it can be mapped bijectively to a proper subset of itself.

well i mainly ask cause i answered to this one [finite sets]math.stackexchange.com/questions/307926/…

i guess if my answer would be wrong they would downvote it

@robjohn: apparently

@DominicMichaelis Asaf's answer pretty much nails it I think.

but it is over cardinality isn't it ?

i mean the number of elements in $X$

I too thought so. I am not sure if induction requires cardinality. (He is not making explicit use of cardinality, he is just using the numbers as a placeholder for explanation.)

well he says he makes it for all elements in his set, as the number of elements is the cardinality

Yeah.

and it is a finite induction

Yes.

it isn't well defined if he doesn't use the cardinality (even though he doesn't say it explicitly)

Hmm. I am still trying to think of a different definition of finite/infinite sets that is not based on bijection.

maybe we can try something topological like $A$ is a finite set if $(A,T)$ is hausdorff iff T is the discrete topology

maybe this one is nonsense i didn't listen to topology till now

I don't think that would help. I am starting to feel cardinality $\equiv$ bijection. So, any proof of bijection is implicitly using cardinality at some point.

yeah i thought about using f:X-> Y with x kompakt and y hausdorf bijection and continous implies f^-1 is continous

but this one doesn't really helpo

cause we don't need a continuous bijection

yeah that one sounds good

even thought "I don't want to use any rules of cardinality" isn't well defined

yeah

@Et how are you?

but is my definition of finite sets right ?

@DominicMichaelis what is your definition?

the one i thought about that $A$ is finite if (A,T) is hausdorff, if T is the discrete topology

It may be self-referential. (Defining a metric for hausdorff, or defining discrete.)

This might be somewhat helpful.

I am reading it myself.

@Ethereal How are you?

@OrangeHarvester I am OK, no school from now on: exams start on the 25th.

@Ethereal okay. cool. best luck. :-)

@OrangeHarvester Thanks!

Greetings people!

@orange i got an idea i edit my post working

@DominicMichaelis okay. I will see it.

@Chris'ssisterandpals Hello!

@OrangeHarvester: heyyyy :-)

@Ethereal Yo

@skullpatrol yo back

@Ethereal That would be a Yo-yo...

@orange could you check it now ?

Blimey.

Now what^

@JonasTeuwen Welcome to the jungle.

Coooool song^

@DominicMichaelis I feel your proof is just trying to erect scaffolds on the outside. The core idea will still require cardinality at some level.

Good.

mh could be

I hardly ever read the preface of a book...

@JonasTeuwen That is because you do not suck at math, you swallow the whole thing in its entire monstrosity.

@Jonas: hey!

@Ilya Ha! Ive caught you!

@robjohn hm, there is no point for me to not agree :)

@Ilya I see you in the avatar bar... :-)

@OrangeHarvester 8-).

@Ilya Hi.

@Ilya I think robjohn was waiting for that moment... like a lion.

@JonasTeuwen That's why it took me 50 minutes to reply to his message :-D

like a sleeping lion

" I had my goal to understand nearly every good undergraduate textbook and I think, I finally reached it." How many years do you think he was studying?

@JacobBlack No, he is asking for a book which covers all of those topics in incredible detail with all kinds of problems/theorems/lemmas/appendices etc.

I want to cry.

@JacobBlack More and more I am getting the feeling that the guy is asking for a roadmap to study the topics he has listed by asking for books on each of those topics probably (in other words of course.)

He has apparently done all the undergraduate courses.

@JacobBlack It should also be CW, so I've asked to OP if they have any objection.

@JacobBlack no, I think he has studied rudin and now has got his hands on global calculus. It is also possible that he is quiet unfamiliar about the mathematics education (probably not in a mathematical community and self studying), so does not really know the organization of mathematics, and is just studying topics he finds interesting. He has probably seen the book and realized that everything is given in very short and he needs to understand each one of the topics in more detail.

@JacobBlack Well, each of the books fills some part of the puzzle. Imagine you knowing only rudin's book and then picking up the book which covers the topics he has listed. How would you feel? I personally would feel that it is some kind of condensed refresher course. So, I will try to find books which explain the things in that book in more detail. So, each of the books that has been given does that. If you answer with a book covering lie algebras then he would like even that as an answer I think

Please share his comment >8(

@JacobBlack Asaf's comment was a joke I am sure.

@skullpatrol duuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuude.

I'm not going to chase him.

You are chasing him by asking us to share his comment.

@JacobBlack Let it simmer a little.

This comment kinds of puts things in perspective. I feel he does not know about the topics in mathematics and their relationship. (He is trying to find topology in analysis book.)

@JacobBlack why?

@JacobBlack You are not trying to look at the post from the point of view of an uninformed self-taught student.

Zorich is a russian author right?

I think I had his text on analysis.

@JacobBlack yeah. I have read only one I think.

He is far ahead of me. I have not even seen Helgason. I think some more advanced people might be able to help him.

So, I will sign off now, and go for my evening run.

Good bye.

@Ilya You here?

@MichaelGreinecker: I am, how are you?

@Ilya Good, good. I'm moving into a new flat this weekend.

@Ilya And you?

@Michael: I'm not :)

@Ilya What's the issue?

@MichaelGreinecker congrats! moving in a new apartment is quite cool, though it takes some effort

@Michael: are you familiar with MDPs? I guess, you should since you are even doing stochastic games

@Ilya I know a bit about them.

they are often defined over Borel state and action spaces, with universally measurable policies: history-dependent randomized selectors

yes.

@Michael: a particularly comprehensive study has been done in the book by Bertsekas and Shreve

I am looking into the sufficiency of the stationary Borel policies

have you heard about any results on that?

@Ilya You want to know when optimal policies can be taken to be Borel measurable?

rather $\epsilon$-optimal

for the additive cost

@Ilya Can you make some continuity assumptions?

well, yes - but under some continuity assumptions there are results on non-randomized policies - which is not necessary to me

I expected assumption in case randomized policies are allowed to be milder

@Ilya In some sense, a situation where randomization is necessary is somewhat pathological. If you randomize between two options and put postive probability on both, you should be indifferent between them. So I guess nonexistence of deterministic optimal policies can only be a problem in rather weird cases.

@Ilya Do bertsekas and Shreve have examples where deterministic policies do not exist but randomized ones do?

@MichaelGreinecker I can give you a simple one now: suppose the enviroment state is $a$ with probability $\frac13$ and $b$ w.p. $\frac23$. Whenever you guess the enviroment's state, you get $1$ and you get $-1$ otherwise

@Ilya The optimal policy is simply choosing $b$.

let me see

I was trying to adopt it from the Nash equilibrium of the game, where equilibrium is attained on randomized policies

@MichaelGreinecker perhaps, you are right - it means that you do not expect assumptions to be milder in case randomized policies are allowed?

@Ilya In a Nash equilibrium, a players best response is the set of probabilities over the support of best responses.

@MichaelGreinecker so?

@Ilya I think under reasonable restrictions, you will never have to deal with randomized policies.

@Ilya Putting probability on a strategy in the support of some best response is also a best response.

@MichaelGreinecker hm, but in the zero-sum game where you choose A or B, and I have to guess your choice, the optimal policy is always randomized?

@Ilya The unique NE involves mixing. But a NE is more than just a best response. It is a fixed point to a best response mapping, and for this, we need the convexifying effect of randomization.

icic, so if we fix NE on the one side and turn the problem to be the optimal control one, mixing might not be needed

@Ilya Exactly.

didn't find an example in BS so faar

found

you have the book?

electronically...

@MichaelGreinecker: p. 201, Example 2 (Chapter 8)

@Ilya I think I can't get it right now. In the classic paper by Blackwell (projecteuclid.org/…) there always exist stationary epsilon-optimal strategies under a given initial distribution over states

@MichaelGreinecker but that with discounts?

@Ilya Yes. Your problem is undicounted?

@MichaelGreinecker nope

@ilya Good. I know little about undiscounted problems.

contractivity makes some things easier, you know

Absolutely. And everything is nice and recursive. Economists like recursive methods.

@MichaelGreinecker recursive? what do you exactly mean?

@Ilya Things like the Bellman equation.

@Michael: well, in fact I am doing an optimization of $\sup\limits_\pi\mathsf P^\pi(A)$ where $A$ is some event

the problem is clearly bounded, and sign-definite, but not discounted

@Ilya It is also static, right?

helppp

$\lim _{ x\rightarrow 0^{ - } } \frac { 1 }{ x } (e^{ 2x }-e^{ x })$

@pourjour $e^x\sim 1+x$

@MichaelGreinecker well, no really. In fact, I start with events of the form $A = \uparrow A_n$ where $A_n$ are finite time-horizon

and then I obtain more complicated events from such $A$

the first problem is dynamic: I have dynamic programming over $A_n$ and a fixpoint equation over $A$ which does not necessary have a unique solutions

@Ilya But it is static as an optimization problem? You do not aggregate over the values over different periods.

@MichaelGreinecker that I don't get. It is as static as an optimal control of the additive cost

@Ilya I think I misundersttod someting. What exactly is $\sup\limits_\pi\mathsf P^\pi(A)$ in your context?

ok, Proposition 8.7 requires continuity assumptions for the Borel-measurable randomized policies that are $\varepsilon$-optimal...

@MichaelGreinecker here $A \subset (X\times U)^{\Bbb N_0}$ is an event, $\pi = (\pi_0,\dots,\pi_n,\dots)$ is a (sequential) policy and $\mathsf P^\pi$ is the induced path measure

@Ilya Ic

@Michael oh, I see now your confusion - did you think $\pi$ is just 1-step decidion, say?

@Ilya Without continuity assumptions, the value-function is usually not Borel measurable, which brings all the complications.

@Ilya Yup.

@MichaelGreinecker yeah

ok, I just need it for the part of the results, so I better state that "under assupmtions Borel stationary policies are $\varepsilon$-optimal, you have this"

and as an example refer people to stronger continuity assumptions

@PeterTamaroff That question (or something very similar to it) came up on both mathoverflow and mse. :)

@Ilya I guess. Whether such assumptions are reasonable will depend on the problem at hand.

I'll provide a computational case study as well, I think

@Michael: in fact, people in my field were struggling with the problem that these problems are not additive: they are multiplicative, or sum-multiplicative, so they were mostly re-deriving the theory.

it appeared, however, that if you just a little augment the state space, you get the additive cost formulation - and zillions things for granted

@Ilya It certainly looks very sophisticated. Is the stuff easy for finite state problems?

it is algorithmic there

model-checkers are doing everything there now

I mean, explicit solutions are available (almost for the whole class of events in the path sigma-algebra) and there is a dedicated software

so people are mostly reasoning about the complexity, compositions of systems etc.

@Ilya Icic

in fact, all this bisimulation stuff was mostly/partly developed for this reasons

to provide "lumped" model which still can be composed as the original ones

@Ilya So the big project is to take a general measurable model and find a good finite copy and then verify that it is still a good copy for various transformations and other stuff you do to the model.

@JacobBlack Can you understand the question by Thomas now?

@MichaelGreinecker sorry, there is a MSc student came to me. And 2 more to come :)

@Michael: thanks a lot for the help - I'll try replying later to your last comment

@Ilya Ok. I go grab something to eat. Will be back later.

It's only complicated when you don't understand it.

anyone know anything about prime sieving

You just missed Eratosthenes.

@JacobBlack I thought you were going to change your username?

@JacobBlack You should try to choose one you're going to keep for awhile, no?

@JacobBlack I can respect that pal...

hi

hi

@Jacob thanks a lot, may i ask why ?

appeal?

ok :)

tooooooooooooooooooooooooooooo deep

mixed with fate?

i made myself a new defintion of finite set :D

please share

We call a set A finite if the topological space (A,T) is hausdorff iff T is the discrete topology.

well it's about this topic here [no bijection]math.stackexchange.com/questions/307926/…

at first i thought that is totally impossible, because i thought finite sets are always defined with a bijection to {1,....,n}

I didn't listen to any topology lecture but as far as i can see my definition should be equivalent

you like this one ?

How do you differentiate between a discrete topology and a trivial topology or any other topology in between?

how you mean?

the topology of compact hausdorff spaces are incomparable (i hope thats the word)

since the set is finite every topological space will be compact

so we only will have one hausdorff

since the discrite topology comes from a metric its always hausdorff

and taking X compact and hausdorff and $f:X \rightarrow X$ continuous and bijective

you can see f^-1 is continuous cause images of compact sets are compact and because of hausdorff closed

with the pre-image definition you see f ^-1 is continuous

@DominicMichaelis the proof of the inverse being continuous depends on bijections etc. By the time we go to compact sets, the proofs are littered with bijection on finite sets I am sure.

mh maybe as i said i did never have any topology lecture :(

Its okay. :-) It was an interesting approach though.

This seems to be related.

i am so excited about topology lectures this will be so great

You will be attending topology this semester? cool.

the next one

my semester has finished just some boring exams

but as i don't care about marks it's ok

@MichaelGreinecker enjoy :) yes, that is the purpose: with finite models (even if they are probabilistic) you can do pretty much everything algorithmically. Thus, we are looking for abstracting general models via finite ones with some guarantees. I am in particular interested in problem-free guarantees, i.e. you do construct an approximate model once and do with it whatever you want

@Ilya Something like an uniform error-bound over all imaginable problems?

Geez... This is one of those rare days. I've capped and now had 10 upvotes without reputation. Yesterday was similar, but not so many votes without reputation.

I know... no sympathy :-)

@MichaelGreinecker ideally, yes - I described this issue in a paper I sent you. More practically, I am working with TV-like bounds on the difference between transition kernels and trying to propagate in time such bounds over some classes of properties

Anyway, it's off to the park. BBL

@robjohn I was typing, sorry

@robjohn don't leave!

Icic

we all regret

@rob biie!!

@Ilya Lilly's friends are waiting for her. I will be back in 30-45 minutes

@robjohn ok :)

@MichaelGreinecker the most hard part is now infinite horizon, it's pretty unstable to small perturbations

@Ilya I can imagine.

@MichaelGreinecker I need to finish a work on a couple of more "practical" papers, like those discussing optimal control over certain events

and then will focus on this co-algebra stuff. It appeared to be quite cool

@Ilya Good to know.

@MichaelGreinecker: what you are currently working on (besides moving)?

i like this way to show the theorems [add]math.stackexchange.com/questions/308082/…

sin and cos

@Ilya Epistemic game theory. I try to do the whole hierarchy-of-beliefs-thing when the player space is a probability space.

Greetings

hello :)

hello

@MichaelGreinecker does it have to deal with common knowledge?

hi

@Ilya Essentially, yes. Everything is formulated in terms of beliefs, not knowledge. But the two approaches are closely related.

@domin @ilya @michael hello :)

icic

@charlie hello sweetie :)

hello @charlie

@DominicMichaelis haha :P

How you guys going ?

procrastinating :D

@DominicMichaelis good good

@DominicMichaelis what part of germany are you from?

Hessen

Aah

everyone is so quiet

everyone working except for you :-)

hi

@Ilya: Back from the park, making breakfast

@DominicMichaelis aaaaaaaaaaasasssassssssss

mh

whats the problem

@DominicMichaelis you said it was quiet, I made some noise

ah right

charlie calling dom an ass... curious

I did not call him ass

be at peace; I jest

@robjohn cool

@Ilya finished my english muffin, sipping my chai now.