wait

@Gortaur this seems broken to me

@JackSchmidt what do you mean

oh no, never mind, it is all good

)

@JackSchmidt I've found his errata, for the update of 7th october 2011 there is no such entry. should I write him, how do you think?

I've already pointed out one typo in the letter to the professor whose book I was reading, but he never replied. I don't know, maybe it's impolite then to write such e-mails (though e-mail itself was very polite and I was very carefully with the sentence where supposed that there is an typo).

@robjohn @JM I would be of course also happy if you can say your opinion since I don't really have an experience with it

@Gortaur are d\bar{B}^2 and \bar{B}^2 boundary and whole?

yes

I don't know if you have an access to springer

UCLA does not have a license

oh, tb changed his gravatar again )

@robjohn I see

I don't know if it's legal to make photos of this fragments

well, what's the link, maybe I can see.

just a second, please

springerlink.com/content/k80h59/#section=833296&page=1

Chapter: new spaces from old

(my mind originally added another - sign, and changed \bar to d\bar, and changed a 2 to a 1; then i corrected the second - sign, which meant everything was very wrong, until I read what was written, instead of what I saw after grading 350 exams)

@JackSchmidt and what is the result?

@Gortaur thanks, that'll help a lot as the nearest copy of the book is over 10m from me

@Gortaur: two typos in the cone statement, as you say. all else is correct

@robjohn page 67

I was able to download the PDF, lemme look at p 67

oh, that's good

okay, so CX has dimension one higher than X it seems.

correct?

yes

as it is a product

So them why is CS^n homeomorphic to B^n, what is B^n?

B^n is an open ball in R^n

S^n is an n-sphere is R^{n+1}

but CS^n has dimension n+1, right?

right

and B^n dimension n?

yes

as my geometry tells me

so I don't see how they are homeomorphic

I think there should be \bar{B}^{n+1}

at least cone CS^1 seems to be homeomorphic to \bar{B}^2

So it's almost clear for me now. I wonder if I should write him about it

here is the errata

math.washington.edu/~lee/Books/ITM/errata.pdf

there is no such entry

@robjohn are you here?

Gortaur: since it does seem to be a genuine typo, I would recommend sending the author a note.

Does the author's webpage provide a specific e-mail for book feedback? Use that.

@JM thanks, I will check it out.

Sorry. I had to go away for a bit. It seems suspect to me, but I am no expert.

hi

My cat is depressed.

When are cats not depressed?

Often.

@AsafKaragila why is your cat depressed?

I think she wants to go outside.

We don't let our cats outside because of coyotes.

You said something like that before.

I can't really have a system in which the cat comes and go as she pleases, so I prefer her to stay indoors and that's it.

We're considering a second cat, it might cheer her up.

Or it could cause contention. It depends.

Yeah.

We get cats as a pair from a litter, or in the current case a mother and kitten.

This makes things easier.

That way, there is a good chance they will survive together.

It is hard getting cats to bond. Much easier with dogs.

Yeah. It is still possible though.

Yes, it is. It depends on the cats.

@Gortaur: I downloaded the whole book.

I can definitely use it. I am lacking in certain of the topics.

@robjohn good, for what?

ah, I see. it is cool. I am from the phone

@robjohn: Didn't I tell you it was nice a few hours ago? :)

@Gortaur The homotopy and homology sections

What book?

Introduction to Topological Manifolds

Oh.

by Lee

@robjohn tell me, have you slept for last 24 hours?

But check out smooth manifolds and Riemannian Geometry, too, I think they're even better.

But maybe there's less in there that's new to you...

@t.b.: I have a better background in those areas, but I will definitely read those sections, too.

@tb hi, I see you have a new gravatar

@Gortaur, You know that I have an unstable personality )

lol

@Gortaur I napped for a while this morning. That was before I popped in for a few minutes and then left to wake up my wife and take the dog for a walk.

I see you got infected with these space-saving smilies eh?

Just when responding to Gortaur :-P

(to mimick you)

the lines moved when I hit the reply icon

You're a true chameleon... :-)

I do :-) I have to leave you, guys. be careful with doubling cats

@Gortaur and cheshire Marianos

Have a nice evening, @Gortaur, see you!

The Cheshire cat is my favorite Wonderland inhabitant.

For his impeccable argument that Alice was indeed mad.

my too

@AsafKaragila Mariano has that as his gravatar

he was also nice in american mc gee

@robjohn I know.

Hm. Made a politically dumb decision earlier today. I left a comment on Mathemagician's answer in order to say that I didn't understand what he was getting at. Of course, Adam chimed in and voted him down and in the end the Mathemagician was getting angry at me and shouted conspiracy!

Hah.

When you have enough reputation it's easy to cause these things.

With great power there must also come a great responsibility, as you know. Reputation implies power...

Now you see the violence inherent in the system :)

I feel so insignificant ;-)

Help Help I'm being repressed!

@tb: I did vote for you, though. :P

@t.b.: what question?

this one the worst comments were removed, though.

@tb Yes, I remember reading that question and Arturo's answer. I didn't see your comment until just now.

Nor Mathemagician's answer.

Well it's the n-th exchange between the two I witness...

Suffering the "somone's wrong on the internet" (xkcd) defect, I presume.

Oh, I gotta run, see you guys later!

Finally!

A question about the axiom of choice.

whats your view on the axiom of choice

Fun!

You can't spell "action" with AC.

Sometimes new users (or even worse, transient users) ask a question, and you answer and they thank you in the comments. It seems to me very rude to tell them to upvote/accept the answer as well.

I always wish someone else will do it for me :P

@asaf: which q?

I'll let you deduce this on your own from the front page. In the meantime I'm gonna grab me some food. Be back in a moment.

Interestingly, I had one person say "thanks loads!" and accept, but got no votes at all.

If it is this one, then it has been upvoted. I feel weird telling the OP to upvote when they may have already.

@anon I have gotten several accept but no upvotes.

I've gotten 8 it would seem, it's just the added comment that intrigued me.

are they newbie users? they may not know the ropes.

Dunno. And no, you've only gotten one accept with zero score, you liar. :P

There should be a newbie manual telling them things like "upvote early, upvote often"

@anon: I have gotten a lot of upvotes on old answers in the last week. Probably the ones I was thinking about were covered.

Honest.

Mmmm... food.

eyes robjohn sketchily

I've had to go back and look at the questions just to make sure that I did answer them.

and what I said.

@robjohn There's a meta thread with that name.

@AsafKaragila Ah, so it does exist. :-)

@robjohn Yeah, meta.math.stackexchange.com/questions/662/vote-early-vote-often

It says "Vote early, vote often"

close enough

Modulo "up" it's the same :-)

Ah the old 3D to 2D projection, mod up

lol

does this projection make me look flat?

Every projection makes you lose 10 lbs.

and about 5'10"

Hah.

So far my vicious plan to answer all the questions on [axiom-of-choice] until there are 100 questions in the tag is going well.

If I can wing it to 400 votes by then (which is reasonable) then I get two specialist badges at one shot. It's gonna be fun on the bun.

Maybe I'll get to 7000 by then. :-[

or as my son would say -_-

Hey Jay!

that reminds me, I forgot to come here and shout OVER 9000 when I hit 9k rep.

Hay Jay?

Achoo!

for those with hay fever

Did you say "A Jew"?

Hello, robjohn and Asaf, just seeing what is here.

@Asaf: A user named Jay entered the room I believe.

Yeah, I noticed that.

Mel Brooks for the win.

Just a bunch of us, flat on the floor.

testing out a new projection.

@anon: you're halfway from there to 10k

@Jay: I keep finding myself at that AoC question that you asked a couple months ago. Some days I understand it, and some days I don't.

yup. I was stagnant a couple days then my latest answer almost made me hit cap this morn

@Asaf A long time ago I studied set theory mainly from Schoenfield's paper in Vol. 13 of some series of books, but remember very little. Have you answered a question on how to construct a model in which every filter in a Boolean algebra can be extend to an ultra filter but the axiom of choice fails to hold?

I do not recall that I have. I can write such answer within a few hours though.

If you want to answer that I will ask it.

Oh, I don't mind it will give me a nice thing to do for the rest of the night. However I wouldn't want you to do it just for me...

reaches out for The Axiom of Choice by Jech

@anon: I am doing rather poorly in the rep dept recently. Spending too long on a single answer.

Dr Spivey's back.

once I hit 10K I'm probably going to thin out and stop trying.

@robjohn: Even my students don't call me "Dr. Spivey." :)

Prof. Spivey?

"Mike," believe it or not. We're pretty casual here in the Pacific NW.

Ho hum, Mike's back.

:-)

:-)

      (-:

anon has been hit by inversion.

nocitome

his emoticons are backwards

˙spɹɐʍʞɔɐq pǝdʎʇ pooƃ ʇɐɥʇ ʞool ʇ,usǝop 'ɥǝ

@Jay: the proof is not simple at all.

It requires some heavy machinery, and the only thing good is that I already have an answer with the construction of the Mostowski ordered model.

@Asaf When I was working on set theory I was trying to modify the way forcing works. The idea was that the conditions would be maps from an initial segment of some singular cardinal to $2$. The difference would be that given a condition and a statememnt there would be

Hit the wrong key.

Also, the basic Cohen model in which there is a Dedekind finite set of reals is a proof for BPID -/> AC

there would be a contion that was a finite extension of the first that decided the statement. When you construct a generic extension one would not only use individual conditions but also sets of conditions that had cardinality less than the the singular cardinal. Has anyone done anything like that or is it an absurd idea?

So you're adding a new unbounded subset of a singular cardinal?

That was the idea. Then by using the sets of conditions one would not add sets the smaller cardinals.

Sounds somewhat similar to the idea behind Prikry forcing, only with less hassle of needing a measurable cardinal and a normal measure.

I would believe this is not something new. Although I cannot be certain.

@anon: I assume that was you that upped my comment :-)

and how did you get the upside down characters?

the first up, yes. very sleuthy of you.

google it.

ah, hadn't refreshed to see the second.

Ah just a unicode converter.

@Jay: Either way, singular cardinals are vicious and untamed beasts. As for the proof you asked, the usual Cohen model should be sufficient for this proof.

That's interesting. Prikry moved from Wisconsin to Minnesota while I was a graduate student at Minnesota. I don't think I talked to him about this though. He liked my proof that you could not force the existence of a measurable cardinal from $V = L$. It is simple. Suppose you can force the existence of a measurable cardinal from $V = L$...

Do you use Scott's theorem about a nontrivial embedding or are you more inclined to the 0# method? :-)

Then there is a least ordinal that can be forced to be measurable. This ordinal is definable in the constructable universe. By a result of two people whose names I do not remember any any such ordinal is countable in the presence of a measurable cardinal.

Kanamori perhaps?

I recall seeing that if 0# exists then all definable ordinals are countable (which is quite trivial actually)

And of course a measurable implies 0#

That is not the person I am pretty sure it was a result by two people. I don't know anything about $0#$.

I think that 0# is one of the most beautiful results in set theory.

It sets such a sharp boundary in the consistency of V \neq L

Hah. I just got my own joke! 0#... a sharp boundary... Sharps!

Maybe I should get a set theory book. Is there a good book that assumes naive set theory, does forcing fairly early on, and then does whatever?

Well, you can read through Jech's Set Theory in fair jumps if you have the basic knowledge.

I'll have a look at that. Thanks, Asaf.

Sure thing.

¡ʇı sǝop uouɐ ʍoɥ ǝǝs I

i!i!i!i!

@Jay: Perhaps Kanamori's book too if you prefer to stay in the large cardinals related areas.

He's a better writer, in the sense that the book has more proofs in it (although a lot less theorems and topics)

Kanamori might be better. It's been a long time, about 40 years, since I worked on axiomatic set theory. I am sure there is a lot of stuff I just don't remember. For example I took compex analysis but do not have a clue as to what the Cauchy integral theorem is about.

Don't feel bad, I took complex analysis two and half years ago and I can't tell you that answer either :-)

algebraists!

:-)

Where?

I was just looking at you guys not remembering the Cauchy Integral Theorem.

And this is what you had to say? Algebraists?

set theorists?

I couldn't very well say analysts!

Logicians? Set theorists?

Ah...

analysts are like t.b., algebraists are like arturo

okay. I should be more broad in my non-PC thoughts :-)

I am not sure t.b. counts as an analyst.

yeah, couldn't think of an example off the top of my head

Willie?

I am an analyst, or was...

You're a fruity programmer :P

Apple to the core!

I don't know if they collected my badge when I went to work for Apple.

Hmm, I wonder if something like (renormalization) or (regularization) should be a tag. I could bump up one of my questions then, but it might be too specific to make a tag. Thoughts?

Old analysts never die, they just lose some of their functions...

ba dum tss

Set theorists never die, they are just forcing into the empty set.