@Matt You could simulate the first by never going to the lectures. As did I :-).

Hmm, group meeting, bye!

@Srivatsan in some contexts, it is 20111213, if it were just 1000 years earlier ;-)

@Srivatsan what do you think of this?

@Matt That it looks wrong...

@Srivatsan Tell me.

@Matt What you are showing is pointwise convergence of $f_n$ to zero.

Pointwise convergence: For any $x$, the sequence of real numbers $f_n(x)$ converges to some real number, call it $f(x)$.

@BenjaminLim Nope, I don't really care for it.

@Srivatsan Yes. Let me fix it. Or were you writing an answer?

@Matt I.e., for any $x$, for any $\varepsilon$, there exists $N$ such that the world is nice after that point. The point here is that the $N$ depends on $x$.

@Srivatsan Yes.

@Matt Uniform convergence: Now, the $N$ cannot depend on $x$. I.e., for any $\varepsilon > 0$, there exists $N$ such that: for all $x$, blah...

Well, you should go ahead and fix the answer.

Uniform convergence is best understood as convergence under the uniform (or sup) norm of a sequence of functions to a function.

@Srivatsan Well, since the guy didn't say anything when I was hoping he'd do so, I'm the first juror.

So, @Matt, what do you think: is the sequence uniformly convergent, or no?

Morning, JM.

Er, it's morning for me. Evening, JM. =)

@JM Thanks. 2 more votes needed.

@Srivatsan Now that you ask this I'm getting unsure. I thought it should because |cos x| < 1 for sin x non-zero so for large n the thing gets small.

@Srivatsan Yep. Good morning to you. :)

@Matt Yep, you should be, for a good reason.

Let's look at what your post says. Ignore the first bit, it's not critical. In the second case, you argue that $\cos x$ being less than $1$, $(\cos x)^n \to 0$. True, but this is a pointwise statement.

We are interested in how large $n$ should be so that $f_n(x) < \varepsilon$, yes?

Yes.

But at the same time, we also want the same $n$ to work for all points. Therein lies the catch.

Let's take a simpler example: $f(x) = x^n$ does not converge to $0$ uniformly. Can you explain why?

On [0, 1)?

Oh yes, on $[0,1]$ say.

It's 1 at 1.

I'm thinking about what happens at 0 in the other example.

Well, $f(x)$ could be $$\begin{cases} 1, &x=1, \\ 0, &\text{otherwise}. \end{cases}$$

I mean, it's quite clear that $f_n(x) = x^n$ does converge to this $f(x)$ pointwise. On the other hand, it doesn't converge uniformly. Why?

I think it can't converge uniformly if the limit function isn't continuous.

@Matt Um, you're right -- in a sense. The issue is not that we cannot speak of uniform convergence when the limit function isn't continuous.

The more precise way of saying what you said is this: Theorem: if $f_n(x)$ is continuous over $[0,1]$ and $f_n(x)$ converges to $f(x)$ uniformly, then $f(x)$ is continuous.

Yes. In the example of the question the limit function is 0, though.

That is a correct explanation, but it's too clever for my purpose here.

@Matt True. But the theorem's converse is not necessarily true, so you cannot apply it in the reverse to conclude that the $f_n(x)$ in that post has to converge uniformly.

@Srivatsan I know, that's why I'm saying.

In fact, this is like a counter-example to show that the limit function could be continuous, and yet the convergence is not uniform.

Anyway, give me a minute to think.

@Srivatsan Yes exactly.

Ok. I hope I am not being too verbose -- I tend to be... =)

@Matt Do you want a minute, or no? Ping me when you're ready to go.

Yes : ) Or two.

Yes, you have just insulted my intelligence 8-).

@JonasTeuwen Um, that would be quite hard, wouldn't it? :P

[sorry, couldn't resist]

@Matt I'm sorry, I don't follow your comment. First, what are you trying to prove or explain?

@Srivatsan You asked me why $f_n$ didn't converge uniformly to $f$.

@Matt I thought we already gave an answer to that. =)

@Srivatsan You said you were looking for a different answer.

@Srivatsan Here.

Oh yes, but I have in mind another explanation which I want to tell you.

I'm trying to guess your explanation, that's what my comment is about.

Hm, it sounds something like it, but we will not know unless both of us make ourselves precise.

Let me try again.

@Srivatsan :).

@Srivatsan I can't think of how to make it more precise but I'll try to rephrase it: Fix $\varepsilon > 0$ and pick $x=1$ and any $y \in [0,1)$. Then $|f(x) - f(y)| = |1 - y^n|$. The limit of this is $1$ so no matter what $N$ you pick, this is never smaller than $\varepsilon$ so $f_n$ doesn't converge uniformly.

@Matt Um, close enough.

T_T

But let me give my take, and let's come back to see.

Yes, please. : )

Ok. Imagine that I fix some $n$, and consider $x = 1 - \frac1n$. What does $f(x)$ look like?

It's 0?

$f(x_n) = (1-\frac1n)^n$, right?

No, but $f_n(x)$ is.

@Matt Do you know how this sequence $(1-\frac1n)^n$ behaves as $n \to \infty$?

[I am switching to the $x_n$ notation.]

@Srivatsan It tends to zero.

:) HINT: For a fixed $\lambda$, what does the sequence $(1 + \frac{\lambda}{n})^n$ converge to?

@Matt Alternatively, let's look at the multiplicative inverse of the sequence: $$ \left( \frac{n}{n-1} \right)^n = \left( 1 + \frac{1}{n-1} \right)^n .$$ What does that converge to?

If anything is not clear, I will explain more clearly.

No, it's quite clear. I just find computing this thing elusive. Give me some time.

Sure, take your time.

Oh. Actually, $1 - \frac1n$ tends to $1$ as $n$ goes to $\infty$. So $(1 - \frac1n)^n$ also has to tend to $1$.

Same for $1 - \frac{\lambda}{n}$ and $1 + \frac{\lambda}{n}$.

@Matt =) Let's review our situation here.

@Matt Have you seen the limit $\lim_{n \to \infty} (1 + \frac1n)^n$?

@Matt: I see that you redacted your answer. There are two different answers to that question. I like them both (if I do say so myself :-)

@Matt No.

@robjohn Planning to read and up vote later.

@Srivatsan No. Then I have not. I was going to write that it has to tend to $1$, too.

Oh my sweet Davis. Why should I consider logspace reductions? =)

@Matt See here.

We are kind-of getting off-track from the original question, but it's good. I hope you do not mind.

@Srivatsan The earth always fails to swallow me when it should. So $|1 - y^n|$ from above tends to $e^-y$.

@Srivatsan Not at all. This is so good : ) I'll be eternally grateful to you : )

One second. I am replying to the comment thread in the other question.

I should know this really...

@Matt $e$ is involved, yes, but that expression is not quite correct.

Let's go back. $(1 - \frac1n)^n$ converges to $e^{-1}$. In fact, $(1 + \frac{\lambda}{n})^n$ converges to $e^\lambda$.

But we don't have $|1 - y^n|^n$ so how is $e$ involved?

Jury+executioner needed.

@JM Jury needed: math.stackexchange.com/questions/91080/proof-with-sets.

(This site has a well-coordinated hit squad...)

@JM whistling A-Team tune

Ha, Matt's vote.

What?

@Matt You might also want to look at this answer and the answer linked at the beginning.

@Matt This is the first time I am seeing your close vote.

@rob, can you be the last juror?

@JM ready for the executioner :-)

Who will believe that the best coffee in town is right in my building? And the shop opens in *five* minutes. So get ready, son.

Thanks rob.

Will you take us with you when you go for coffee?

@Matt Sure, come along.

Nice : )

@Srivatsan Anyway, something personal: how do you like your coffee?

@Matt I'm a tea person. I don't do coffee.

@JM he does ;-)

@robjohn Oh. All sorts of tea or just the usual?

@robjohn Heh. :D

@JM I like latte everywhere, but otherwise I have one set drink per shop. It's always an "Italian cappucino" in the cafétaria below. I don't like black coffee. I guess this is the answer you're expecting.

@Matt I like all kinds, but I think my favorite right now is Jasmine green tea.

@Matt I even go for a chai latte once in a while, but usually I don't have milk with my tea.

@Srivatsan Yep. Interesting. :)

@Srivatsan I like black coffee but unless at home I order Cappuccino because usually the coffee is so bad that I can only force it down with milk : )

@JM - Do you drink coffee? How do you like it?

@Matt I like mine black too. Sometimes I add sugar.

For tea, I mix it up, but I mostly drink oolong.

@JM mmmmm, with ice : )

@robjohn I quite like houji cha.

@Matt You order cappuccino at home? =)

Now, unfortunately, I have to take the dog for a walk. I hope you're all still here when we come back. And maybe we can talk a bit more about uniform convergence, @Srivatsan, if it's not a bother....

See you later, Matt.

@Srivatsan No, at home I make black coffee : )

brb

@Matt Bye, Matt. I hope to be here, let's see. If I am here, then sure thing.

...and I have to step out, too. See you guys in a flash.

Exam done

Bye, JM.

LIKE A BOASS

@Matt I really like Genmaicha, which is made from houjicha

@JM is flashing?

@robjohn There is Genmai cha made from sen cha and "special" genmai cha, made from houji cha (roasted sen cha).

@N3buchadnezzar boass?

(I hope tb has gone to sleep and won't read the transcript when he comes back. It's just too embarrassing that I don't even understand uniform convergence.)

@N3buchadnezzar I presume you did well; great to hear. =)

Mmmmm... coffee

@Jonas: I don't have to pay to write my thesis... did you pay tuition during your M.Sc.?

@AsafKaragila Yes, but I got money for TA (which was not compulsory).

@JonasTeuwen Sucks to be you. I don't pay squat. I take money.

Both scholarship, tuition and wage of war for being a TA.

Back.

@AsafKaragila Nice. I don't have to pay anymore. They now pay me ~2k.

Nice.

I'm not sure about the paragraph that I have added to the intuition about aleph numbers answer. It is probably going to be confusing, but I think it is essential...

The only question I flunked =(

@N3buchadnezzar Wait what? Both limits equal the same thing.

ops

Show $|M - L| < \epsilon$.

$|M - L| = |(f(x) - L) - (f(x) - M)| \leq |f(x) - L| + |f(x) - M| \leq \epsilon$.

Agh, Comic Sans in the CERN webcast...

@ZhenLin Comic Sans?

@ZhenLin You talking about webcast.web.cern.ch/webcast ?

Yes, but the previous presentation.

@ZhenLin Superconductivity and Cryogenics at CERN; from bubble chambers to accelerators?

No, the presentation from the ATLAS group. Now the CMS group is presenting.

@robjohn Or the CERN webcam?

@robjohn Can I ask you something about this?

I don't like how this guy always gives full solutions...

Hi everyone.

@Matt I have to go out for a bit; I should be back in a bit

Bye rob

Hi tb.

Bye robjohn.

@tb I guess my avatar matches the colour of my face nicely.

@tb Why not?

@Matt Okay, what's confusing?

@JonasTeuwen Because getting stuck at such an exercise means you missed something fundamental.

Oh.

hence, I'd rather give a short hint, hoping the OP sees what was missing in the understanding.

This is by far the laziest asker around.

@tb Yes, he is.

@robjohn Nothing is confusing but I was thinking about the exact argument why $\lim_{n \to \infty} \left ( 1 + \frac1n \right )^{nx - \lfloor nx \rfloor} = 1$ because $nx - \lfloor nx \rfloor$ isn't constant. Now I think that it's $1 \leq \left ( 1 + \frac1n \right )^{nx - \lfloor nx \rfloor} \leq 1 + \frac1n$.

@Matt that's right.

Yes, it is sandwiched between $\left(1+\frac{1}{n}\right)$ and $1$

@robjohn I quite like AD.'s answer to the uniform convergence question.

I don't like that he leaves $P_k^m=\sum_{r_0+r_1+...+r_{m-1}=k}\frac{k!}{r_0 !...r_{m-1}!}$ hanging.

and it only works for integer powers.

Are we talking about the same thing?

@Matt Nope :-)

AD also gave an answer to the other question you were asking about.

That we had been talking about before your comment

If you look closely, AD's answer is exactly my answer with $t=\cos(x_n)$

He is showing the work I did to find that $\sin(x_n)=\frac{1}{\sqrt{n+1}}$ is the maximum.

Yes, he says in his answer.

Heh. I didn't look at the top line :-)

So to prove or disprove uniform convergence the way to do it is to find the maximum of $f_n$, possibly as a function of $n$, and then see whether given the function value at that maximum, it's possible that it converges uniformly (or not) to the pointwise limit.

You have to show that given an $\epsilon>0$, you can find and $N$ so that for $n>N$, $\sup |f_n-f|<\epsilon$

Uniform convergence means that all of the points converge "at the same time".

Yes but if you have $f_n(x_n) \to \infty$ at the maximum then the answer is: pick any epsilon and then for some N you get what you want.

Hm. Just got an offer to work on iPhone stuff.

This is saying the same thing. sup or max.

if you know that $|f_n(x_n)-f(x_n)|=\sup|f_n-f|$...

Thanks @robjohn! For a mean square you're not that mean.

Just average ;-)

: D

burp

Hm. Time to reboot, and the large drive is gonna fsck, so it'll be a bit before I return. So long!

fscking hard drives...

OP asked to close this question as a duplicate.

Done.

Alright. I'm not at my home computer, but I have an embarassingly basic question about getting the TeXworks & Viewer (part of MikTeX) working. It won't parse equations inside $ $ signs, but it will make boldface from $\bf{Motoko}$ for example. I have to put some kind of \usepackage{} in one of the beginning lines, correct? (EDIT: woops, I guess naively that latex would actually parse in chat for many with the bookmark thingie posted...)

Try \usepackage{amsmath}

sounds familiar, that must be it. (I've always started from templates beforehand.) Do you know if that would come with the MikTeX or would I have to DL that separately? I'm assuming it's already there.

I don't know anything about MikTex, sorry.

no prob

Drawing from my exam today, I am hoping for extra credits due to the beautifully drawn sombrero.

I had an interesting insight some weeks ago. Both the cutting sticks question and my mario kart question are born from the same family of problems (from a LinAl perspective): solving for the matrix in a linear system Ax=b via a collection of givens x,b and a restricted subset of matrices.

@N3bu: To be honest, that looks more like a witch hat because of the height/width ratio. :)

The text is: Collect all the fishies

Oh lord, someone prepare tomorrow's class for me. I don't want to do that.

Also, does anyone get what PseudoNeo tried to say here?

@AsafKaragila Write a few problems on the board, involving hand grenades, fish, and your subject. Then leave clas

Alright, help me understand the idea behind a picture of the "dual curve." The d.c. is a curve in the space of lines of R^2 consisting of all those tangent to the original curve y. But when people actually put it on paper as in the red curve above, it's viewed as a set of points in the original space (visually). What points correspond to what lines?

@N3buchadnezzar I doubt it'll help.

@AsafKaragila Why? I guess it`s less work

@N3buchadnezzar Yeah... that's not what I'm supposed to teach.

@Asaf: Pseudo is referring to the fact that there are not fields of every finite cardinality, I believe.

@anon Ohhh. That. Well, there are rings of every finite cardinality. The "field" was just an example.

I hate it when people catch onto unimportant specifics. Even more when the person they bother with those specifics is me.

There are ordered sets of every finite cardinality, right? I'm not wrong about that, I hope.

not any of my areas, that's in your court :)

In which classes do one learn to tackle really hard integrals?

Finite stuff? Nah :-)

Complex analysis?

Not complex, just complicated ;-)

Bad example, was the first string that sprung to mind..

Yes, I think complex analysis would help you there, if only in spirit. And lots of background knowledge of special functions like gamma(z) and experience with u-substitition / variable changing. @Asaf: Oh, finite. Couldn't you just order {1,2,...,n} for any n? Or do I not understand orders?

Yes, you could. It was a rhetorical question ;-)

errr...

?

my brain put the squares outside of the trig functions for some reason

Hands googles to anon

for (1), expand \ln as a power series and be able to find \int sin(x)^n dx for arbitrary n, and for (2) I'm not sure if that even converges TBH, but writing cos/sin as complex exponentials would surely be key

Quite sure the second one converges to (2pi)^(0.5)

Indeed

Ugh.

I hate it when people write not 100% clear exercises and then I have to explain them to the students.

So, any help on my elementary proj geo question? :<

I hate when one of the first exercises in the book was first proved as a main result of a 50-year-old paper :)

oh, line coordinates. thanks wikipedia

Last time I was here the library was hosting some kind of kids-doing-piano thing with family/friends as audience, in what I would describe as a secret backroom. I noticed this and snuck in before the one security guy came in to man the front, and was able to obtain a free meal that day. It's amazing how awesome free food is when you're living off of ramen and PBJ.

Does a basis for the uniform operator topology look like this? The set of all balls around all points T in B(X,Y) of all radii epsilon: $$\mathcal{B} = \bigcup_{T \in B(X,Y)} B_T(\varepsilon)$$ where $$B_T(\varepsilon) = \{ T^\prime \mid \sup_{\| f \| \leq 1} \| (T - T^\prime)f \| < \varepsilon} $$ for all $\varepsilon$ in $\mathbb{R}$?

That mod inside the sup expression should've been $\| \cdot \|_Y$

What you said in words is correct, but it's not what you wrote. $$\mathcal{B} = \{B_{\varepsilon}(T)\,:\,T \in B(X,Y),\,\varepsilon \gt 0\}$$ would be what I'd write.

I wanted to spell out the bit that tells me what the metric is. Because next I'm going to have to worry about a basis for the strong operator topology : )

Thanks, btw.

That was a typo. This $\bigcup \{ B_T (\varepsilon) \}$ is what I actually meant to write.

That's better :) However, you still don't bring all the $\varepsilon \gt 0$ into play.

Yeah just noticed when I wrote it down on paper.

Before worrying about the strong operator topology explicitly: given a vector space $X$ and a family $\mathscr{P}$ of semi-norms on $X$ can you write down a basis for the topology?

Here: $\bigcup_{\varepsilon > 0, T \in B(X,Y)} \{ B_T(\varepsilon )\}$

Now I'm happy.

By the way: are the balls really written like that in your class? It strikes me as odd to emphasize the radius more than the center. What I usually see is either $B_{r}(x)$ or $B(x,r)$.

No, I just made that up. I don't think it matters here.

How can a family of semi-norms induce a topology? I think I need to read some more before I answer your question.

The weakest topology that makes all the semi-norms continuous (the initial topology induced by the semi-norms).

Yes but I don't see how to write down a basis. Yet.

The strong operator topology, for instance, is the topology induced by the family $$\mathscr{P} = \{p_{x}\,:\,x \in X\}$$ where $p_{x}(T) = \|Tx\|_{Y}$.

Do anyone know any female mathematicians ?

?

@N3buchadnezzar Emmy Noether.

@N3buchadnezzar Right now, or historically?

Or rather, your favourite female mathematician would be a better question i guess

@Matt Every seminorm $p$ gives you a notion of balls with respect to $p$. A basis is given by finite intersection of such balls.

@N3buchadnezzar I am unfamiliar with the works of any other than Noether, and she's not my favourite.

@N3buchadnezzar Ulrike Tillmann, for example

I just feel like there are extremely few female mathematicians. And sadly I can hardly think of a single one

Time to study more

@N3buchadnezzar Check here

Curie is interesting, although she belongs to the branch of physicists

They don't have Marie-France Vigneras? That seems odd.

Isn't she a bit young to have a biography there?

too old

grins

She's not young.

Probably about as old as Dusa McDuff.

I'd have guessed around 60, yes. Maybe a bit more. Anyway, the list is far from complete.

Of course.

I'm starting to like all that "Further reading material" links.

I started by looking for number theorists.

I'm going now. Goodnight.

Good night, Asaf!

Bye Asaf!

G'night.

I think there is a typo in the lecture notes. They define the strong operator topology on $B(X,Y)$ to be the weakest topology for which the evaluation map $L \mapsto Lx$ is continuous for every $x$. It should be $L \mapsto \| Lx \|_Y$ for the norm on $Y$. But then there is only one norm and no family of semi-norms.

Puppy stroll time. Brb.

Both descriptions are correct. Note that the semi-norm $L \mapsto \|Lx\|_Y$ depends on $x \in X$.

And where is the family of semi-norms? $\| \cdot \|_Y$ is just one (non-semi) norm.

\usepackage{mathtools} is perhaps preffered?

@Matt $\{|L|_x = \|Lx\|_Y\,:\,x \in X\}$

@tb Oh. I misunderstood what this means. : ( Thank you!

Do we really need a research-mathematics tag? Doesn't seem to be of much use.

That should have been mathematics-research, of course.

Hmm.

I've always doubted the rationality of the expectation maximization heuristic in decision theory / whatever mathematical discipline (you should choose an option that maximizes the first moment of the outcome). One idea I had is to construct decision heuristics based on which choice's probability distribution for outcomes maximizes a given functional, and investigate what kind of functionals correspond to what kind of heuristic features in various set-ups, compare them and so on.

Anybody know if this has been done? I know I've seen an MO question on replacements for the expectation maximization idea but I can't find it anymore...

anon: I think that's been looked at extensively in economics (with the root of the statement being 'wealth has nonlinear utility'), but I may be misunderstanding what you're going after. (Note that it's still essentially maximizing expectation, but with a nonlinear weighting function applied to the results). You might start hunting under the phrase 'risk aversion'?

(For instance, this paper looks like it might be related: econ.hit-u.ac.jp/~kmkj/uncertainty/DowWerling%20Ec%2093.pdf )

ah, that sounds like the relevant kind of stuff. thanks.

Hey Guys!

Just wondering is it possible to find which questions are about to be closed .. say having more than 2 close votes?

I think with 10k tools, yes.

Oh :/

I thought may be with some kind of query over the data ..

I have had a lot of accepted answers lately. I'm glad.

not sure about querying se, might be possible but I wouldn't get your hopes up.

Hm! Thanks Anon :)

@AsafKaragila Hooray!

I have only stopped now grading homeworks functional analysis :|.

Who told them that $L^1$ convergence implies pointwise convergence?!

I did. I wanted them to be wrong and make your life a living hell of grading bad papers! Muahahaha!!!

@JonasTeuwen Komlós did :)

It is much easier to correct them when they do it right. This sucks.

...almost, of course.

Hm. I answered 50% of the axiom-of-choice :-)

@tb Uh... what?

Oh.

Asaf: it's always interesting the associations one forms with some names reading answers. You're definitely one of my two 'oh, hey - set-theory!' people; whenever I see a foundational set-theory question I expect you to be involved somewhere. :-)

@StevenStadnicki Well... there's a reason why I am the top user in the set theory related tags; and furthermore one of the few near-20k-reputation without a Generalist badge ;-)

nod

(FWIW, the other is Andres Caicedo, and I think I have you two mentally coupled because of the vague alliteration of your names.)

Well, if you read them - I hope you understand my answers. :-)

handwave I'm only a somewhat enthusiastic amateur in the subject (for instance, I have Kanamori's Higher Infinite but understand maybe half of it; my brain starts to shut down somewhere around the usage of elementary embeddings, extendible cardinals etc.) but I really enjoy the topic and your (mre obscure) answers are usually right at the edge my level of comprehension.

I think I'm much weaker in grasping at the 'small' end in strange choiceless worlds than in comprehending most of the large-cardinal principles.

If you're able to understand half of Kanamori you're doing much better than many set theory grad students, I'll assure you that.

@StevenStadnicki It's a doozie, yes. Even when my advisor is trying to help me up with an argument he gets confused with choice-y stuff, even I do - and I've been ramming choice over and over for the past year.

@JonasTeuwen I would say a.e. is implied, bur I guess it's not necessarily true

Well...

"Don't worry about people stealing an idea. If it's original, you will have to ram it down their throats." lol.

@Ilya Even if you have $L^1$ convergence, you don't need to have pointwise convergence.

Actually, since I have you in convenient chat format... :-) I'd love to ask about something that's been nagging at me for a little while now. It feels like the prevailing wisdom in set theory is one of 'permissiveness' in the sense that 'anything not forbidden is consistent' (yes, this is very very vague)...

@JonasTeuwen icic, but I would say that it implies a.e. convergence though

Given that - AD has always seemed to me to be more permissive than AC but of comparable complexity, and that it avoids most of the really strange properties that you can get without choice (for instance, it still offers DC and even stronger principles IIRC), why is ZFC so favored over e.g. ZF+AD? (And I've seen the results saying AD is equiconsistent with some large-cardinal hypotheses...)

That is a very good question.

@Ilya Don't you only have a.e. pointwise convergence of a subsequence?

First you need to remember that historically AC came first, and tradition has a very important role in mathematics. By the time AD appeared we already know that AC is consistent with ZF, and all that shizzle.

@JonasTeuwen maybe - I wrote that I guess that it's not necessarily true

nevermind

On the other hand, by the time we formulated AD we could show that its consistency strength is much stronger, now people don't like too much the fact that ZF may not be consistent. We don't talk about it in parties, but deep down we all know that we may wake up one morning to a world without ZF.

"The reader may be disturbed by 'probabilities' that do not sum to one. It should be stressed that the probabilities, together with the utility function, provide a representation of behavior." I was indeed disturbed, but that is a valid point. Perhaps this is like Greeks with irrationals or ??-century mathematicians and complex numbers, and the idea of probability will be radically more general than the common person conceives of them [contd]

This been said, if the world were to accepted AD then it would be impossible to wake up to such world, since ZF+AD would imply that ZF is consistent.

@Ilya 8-).

[contd] like how mathematicians think of numbers as elements of algebraic structure that bear some numbery resemblances with the original ideas of numbers.

Isn't the same essentially true of ZFC, though?

Lastly, AD is far more technical in its nature. The axiom of choice is very intuitive in the sense that we've been using it all our lives; AD is less intuitive and works very differently. We can no longer make sense of the world, and the intuition we want infinitary objects to inherit from finite stuff will be mostly annihilated.

@StevenStadnicki No, Con(ZF) $\leftrightarrow$ Con(ZFC) is a true sentence, while Con(ZF+AD) is much stronger than both.

Ahhhh, all right. (ZF implying ZFC is basically Godel's result showing choice in L?)

Exactly.

That last point really makes a lot of sense to me - the fact that there exists a book showing all of the equivalencies of various results with various principles of choice (not to mention the fact that there are so many readily-defined distinct choice principles) certainly suggests that AC is a very well-connected axiom in some deep sense.

Yeah.

It's even worse. You can easily understand the axiom of choice: nonempty sets, choose one from each.

The determinacy first needs to tell you what there is to be determined.

I do agree, however, that the more you give it thought the more AD makes sense to the "conventional" mathematician.

nod I think a lot of the 'choice is intuitively obvious' situations are ones where you're not using the full power of AC at all, just some fragment like DC.

It implies nice features of the Lebesgue measure; the consistency of categories (via universes axiom); DC still holds which is what most mathematicians use anyway...

Choice really becomes very non-obvious at least to me once you start getting into the 'corners'.

nod

However, if you also think about it, most mathematicians also sit in L. They consider the constructible model as their universe, because it's so small and everything is so nice and cozy... but there's choice and no determinacy there.

Do they sit in L or L(R), though? Certainly virtually all of the 'real' real numbers that get used are in L, but I also think you'd find a lot of mathematicians take the reals as a given in some vague sense...

I am no expert on determinacy, but if I recall correctly AD implies that there are no aleph_1 many real numbers. Which is a really strange situation... it implies one of the craziest things you can think of:

There exists an equivalence relation on the real numbers such that the cardinality of all equivalence class is strictly larger than the cardinality of the real line.

@StevenStadnicki To be completely fair, mathematicians work in ZFA+AC most of the time, the atoms are usually the field (real numbers, p-adics, etc) and then there are sets and stuff.

Very very true. (And huh - I'll have to chew on that former for a while to really digest it, I think. I thought AD was consistent with choice for sets of reals, but I'm more than willing to believe I'm misremembering.)

And ack, I'd better run here. This has been a wonderfully informative discussion - thank you immensely!

Stop me if I'm mistaken here, but If you can choose from all sets of reals you can well order them.

Sure :-)

I should probably hit the hay as well. Goodnight!

Good night, Asaf.

Gutenacht!

Also, you should know - before I leave - that because you liked it so much I've decided to keep the "Further reading material" whenever I can think of related answers or questions.

Cool! : )

If those Chinese would make their homework in Mandarin it wouldn't be less readable.

I need a Glenmorangie now.

You just have to learn Mandarin.

That wouldn't help.