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23:00
@Ted, there's at least a 35% chance that one of the "many things" is preparing dinner?
@TedShifrin Anyway thank you!
No, @Paul. My computer is in my office, not in the kitchen. And I'm injured. :P
@Ted, sorry to hear that. I guessed dinner because that seems to be a constant theme when you sign off.
damn, @Paul, I need to stop being predictable.
I was just about to ask you that, Ted. So, instead of writing $u^{p}$ as $u^{2+p-2}$, I should write it as $u^{(p-1)q$, and then apply holder?
23:02
@Ted, To clarify, I'm sorry about your injury. I didn't mean to be sorry that you don't have a computer in your kitchen -- I don't either.
Yes, with magnitudes in there, @Jessy.
LOL, it's ok, @Paul. :)
Why the hell did my professor give me that crazy hint, then??
His hint is fine.
Okay...I'll have to see how it all fits together. Thank you for being so patient with me, Ted. I don't have as strong a background as a lot of people in my class, so sometimes people don't think I'm worth helping.
You did a real mitzvah! ;)
Small question if I may, when they say B=f[A] (set theory) they mean that B is the set of all the images of the function on the set A, right?
23:05
I never mind helping people who make an effort. My office is full at office hours :P
I think you're close, @Jessy, and thanks :)
I wish you were my professor! LOL
Well, plenty of my students would probably rather not have me. :P
The American system and the European systems are quite different.
@KarlKronenfeld the top answer here spoils how to embed $\omega^\omega$ if you didn't want to work it out yourself
So I've heard from some of my classmates!
@Mike: I thought you were fleeing set theory as fast as your tiny legs would take you :P
23:07
@Mike He already got it from another answer, got me wondering about how far you can take N
Anyhow, @jessy, keep working. You're almost there.
@Studentmath An embedding is provided (or hinted at) of $\epsilon_0$
@Ted Good point. Time to start running.
LOL@Mike. Let me know how you and Jacob get on.
@Ted, Besides helping students who are making an effort, it is also right to help students who are unmotivated and thus lacking in effort. Such students can be guided and advised towards greater industriousness.
@Paul: But they don't show up to be guided. That's the biggest change I've noticed in the last 20 years. I've always motivated students to work harder and succeed, but that's not happening any more. A good portion of my class just doesn't care enough ... or is too busy with other things to make passing my class a priority.
Hence my having 1 1/2 feet out the door to retirement.
23:12
Ted, where do you teach, out of curiosity?
It's in my profile, @Jessy. University of Georgia.
Georgia the state or Georgia the country?
the state in the USA :P
@Ted, Sure, you may get a student whose main ambitions are outside academia, perhaps chess or badminton might seem appealing. So they focus on that, and ignore the maths. Such students may be frustrating sometimes, but I wouldn't say they're necessarily irrational. Badminton's a wonderful sport after all -- you almost get the feeling that you're flying when you hit the shuttle well.
um, yeah, right @Paul.
23:14
;) nice! I'm in NJ myself, and as such, I have to take a break from working in this problem to go help my husband shovel snow. Thanks again!!
@Ted, obviously you're more of a tennis man than a badminton man.
I never used the word irrational, btw.
Ah, @Jessy, somehow I thought you weren't in the US ... perhaps the mitzvah :P
@Ted, Yes, I didn't mean to say you thought or said they were irrational.
@Paul: We have people in college who don't really want to be there. And we have people majoring in math who don't have the work ethic to make it through a real math major. But we keep passing them with "gentleman's C's."
@Ted Yes, I've heard such complaints. It's kind of difficult for me to understand subjectively, because I'm super-motivated mathematically. I can't wait to read those pdfs
23:18
Yeah, @Paul, I don't think either of us is the issue.
Good luck with Szemeredi :P
See everyone later....
My, I am really not getting this recursive definition in the set theory..
I hate it when you just stare at the book trying to figure out what they are saying and stay completely clueless..
@Studentmath, I can help
Thanks @PaulEpstein, yet I don't even get it to the level of having a concrete question regarding it
@Student, Well, why don't you say what you don't understand and I'll explain what it means?
23:27
I'll try to translate their definition, if you could clarify what they mean in there (and if not too) I will really aprreciate it, thanks!
Let <A,<> be a well-ordered set, and B a non-empty set. If g is a function from the group of partial sets to B, then there is only one function f:A->B that fulfills the following:
f(a)=g(<f(x):x<a>) for every a in A
I do not get what they mean in that definition. I can get the proof of it, but not the definition itself. And I think I need to use it for my current question, but since I don't get it, can't get how to use it ;P
But the definition must be in the text. How are partial sets defined?
I think what I don't get is what do they mean with the "fulfills the following:" (as in, what does the f(a)=... means). And to your question - it's a function <f(x):x<a>, from a subset of A to B
where a belongs to A
coming to think about it, it's a function <f:seg(a)> isn't it?
yes but the definitions in your book don't seem particularly standard -- "group of partial sets" is strange terminology.
Can you read a different book, or is that the assigned text? There's hardly a shortage of set theory books.
It's probably due to my rough translation. It's the assigned text, have to use the says and all in there for every proof. But I think I am starting to get it.
Merely trying to explain the definition got me further than staring at it for an hour or so, will try to go on with it, thanks!
Ok, that's the problem. Axiomatic set theory is very sensitive to exact wording. Can you reword your question?
23:40
I will try, yes:
Let <A,<> be a well-ordered set, and B be a non-empty set. If g is a function from the group of functions from A to B, so that <f(x)|x<a> (a belongs to A), then there is one and only one function f:A->B so that:
for every a that belongs to A, f(a)=g(<f(x)|x<a>)
And they call it the recursive definition, or defining by recursive.. or so.
Well on small initial segments of A, it's obvious.
So by well-ordering, and contradicting existence, get the smallest segment of A where no such function exists.
You readily obtain a contradiction.
Then uniqueness is also clear.
Recursive means you define stuff in terms of simpler stuff.
23:56
Huh, I see.. let me try that. Already got most of it clarified now, thanks a lot!
Like the induction in the natural numbers, correct?
You define f in terms of smaller and hence simpler elements -- that's recursion.
I see, yes
Okay, that clears it up, also clears up how I can use it. It's really rather useful.. thanks a lot Paul!
When you explain something deep using terms somewhat crude, that's recursion.
When you fly through the sky, the simple terms, they won't lie, that's recursion.
But it still holds true, if you can prove it of course, for the deeper things, correct?
I'm making a song about it to the tune of "That's amore"
23:58
I'm gonna need youtube for that, Paul.
Recursive definitions are extremely simple once you've understood the recursion concept.

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