@Asaf Ah, that's nice. The ultrafilter theorem tells us there are at least continuum many, since we can just extend filters containing some combination of singletons and co-singletons.
But I see that's already enough to imply the full result!
Also, would you happen to know why $\kappa$-complete means that infima/suprema of sets of cardinality $< \kappa$ exist, rather than sets of cardinality $\le \kappa$?
@ZhenLin It's easier to talk about $\omega$-completeness as being a filter, and $\omega_1$-completeness as being countably closed. Otherwise you would have to always say "closed under finite ..."
Furthermore, where can I learn mathjax such that I can take advantage of the notation on this site? Perhaps there is a tool that builds the expressions for beginners?
I've been thinking of learning TeX for some unrelated purposes.
I have no strong academic background or any requirements from an ivory tower. I'm that guy who finds out years after school finds out he's really into math.
For what it's worth, I do write code daily, so learning TeX is likely within my grasp.
This is the gap I'm trying to cross: software and math.
Trust me, I feel your frustration :). Hah.
I solved project euler #1 using my understanding of math only to find out later that I have a critical flaw in my general form because I don't understand this problem.
No, I appreciate all your effort. I could just as easily apologize for being so terrible with math :p.
All I ever learnt in highschool was sumation of a single variable, so I could understand things like sum for i, but not mixing the variables or constraints.
Okay, so that bottom line of the equation, if I have my 3 nested for loops that step through each-other... what do I do with the term, what does the "..." indicate, and that final term means I'm adding or subtracting the intersection of all sets depending on n being even or odd?
it gets worse. The next term is a quadruple-nested loop, which is summing over terms like |A \cap B \cap C \cap D|. It adds all them, and then this quantity is subtracted.
Hopefully this last sentence didn't shatter your eureka...
Assume there is a complex dynamical system which has only a one degree of freedom whose trajectory varies linearly with time. Given below is the variation of the generalized momentum ($p$) of the system with time.
Neglect the small wiggles in the graph (they were due shaking of hand while drawin...
I meant it as a joke, I wasn't being impatient. I'm reading a book and if I sit next to the open chat window in this noisy chat room it's too distracting : )
@Matt Either way I have to head out to the university and I might have some time later (before coming back home) today. I left you a comment with a mysterious and unexplained reason. I'll be happy to discuss this further in the gap between coming home in the evening and refill night.
I'm taking it in summer. I thought I'd try something new. Rather than ignoring the course and then doing one term in 5 weeks I thought I could cancel it and actually understand some things properly this time.
I am also currently reading "Godel. Escher. Bach." which is a really fascinating book and I would recommend it to any person without even a slight sign of hesitation.
@AsafKaragila I have another question for later: If (X, <) is a partial order, why is the collection of all partial orders (Y,<) that are order isomorphic not a set?
Basically, when you write down something like $\{ x : \varphi(x) \}$, what you have is a priori just a class. To prove it is a set you need to show that there is a set $X$ such that $\{ x : \varphi (x) \} \subseteq X$.
So that only thing is to see whether there is a formula $\varphi(x,y)$ such that: a) $\varphi(A,\{A\})$ b) if $\varphi(x,y)$ and $\varphi(x,z)$ then $y=z$.
This is how I understand the notion of class function.
In this case I think you have $\varphi(x,y) \equiv (\forall z) (z\in y \Leftrightarrow z=x$.
I should have written that for all $x$ you have exactly one $y$ fulfilling $\varphi(x,y)$ instead in b)
But I think it's just technical details. Usually, they're not so important.
A purely semantic way of looking at the concept of "being well-defined" is to think about "presentations" or "names". So we have sets $X$ and $Y$ and classes $\tilde{X}$, $\tilde{Y}$ together with (class) functions $p : \tilde{X} \to X$, $q : \tilde{Y} \to Y$. We think of $\tilde{X}$ and $\tilde{Y}$ as being the class of "names" of elements of $X$ and $Y$, respectively, and the functions $p$ and $q$ take names to the thing that they name.
Now, if we are given a (class) function $F : \tilde{X} \to \tilde{Y}$, this does not necessarily descend to an honest function $f : X \to Y$.
Checking that it does is what people usually mean by checking that we have a "well-defined function".
For example, if $X = \mathbb{Z} / (p)$ then we can take $\tilde{X} = \mathbb{Z}$ and name residue classes mod $p$ by any integer, instead of arbitrarily restricting to integers $\{ 0, 1, \ldots, p - 1 \}$ as we usually do.
Ah, here's a non-trivial example: there is a set $X$ such that every (Hausdorff second-countable) manifold is isomorphic to an element of $X$. Then we can take $\tilde{X}$ as the class of all manifolds (which is a proper class!) and have a map $\tilde{X} \to X$ taking a manifold to an isomorphic manifold in $X$.
But I suspect showing that such a map exists requires global AC...
Or maybe not, if you're meticulous enough about the construction of $X$. Hm. shrug
For example, if the continuum is a countable union of countable sets then there are no $\aleph_1$ many reals, that is $2^{\aleph_0}$ and $\aleph_1$ are incomparable.
@Matt No, $\aleph_1$ real numbers, that is to say a collection of real numbers which is of cardinality $\aleph_1$.
Ah, that's missing a "t" then. Makes more sense : )
But how does that prevent me from having an injection? Why can't there be an injection from the countable ordinals into a countable union of countable sets? Do we know that the countable ordinals are uncountable?
Essentially there is a theorem that if there is a set of $\aleph_1$ distinct reals then there is a set of reals which is not Lebesgue measurable.
@Matt The set of all countable ordinals is $\omega_1$, it is of cardinality $\aleph_1$.
In models where every set of reals is measurable (i.e. Solovay's model, or the continuum is a countable union of countable sets) the continuum is incomparable with $\aleph_1$.
