@JohnZimmerman I mean, part of your job is to convince other people that the problem is well-motivated and interesting. YOU need to explain to others how the problem fits into the corpus of mathematical knowledge, and why anyone should care about the solution.
Right now. it just reads like a bunch of gibberish to me (likely because I am not up-to-speed on whatever part of mathematics is relevant, but possibly because it is gibberish).
@JohnZimmerman The more salient question is "In what context was this question 'answered'?" Knowing who the answerer is might matter, if that person is some widely acknowledged expert in the particular field in which a problem is asked, but the context and content of the answer is much more important.
Enderton makes me the following promise (so nice of him! Apparently it has to do with needing to invoke regularity which hasn't been done yet, but that's not the crux of my question). My question is, does the following summarize what he will do (the English language can be informal and i want to be sure I understand): he will construct a function card with domain set of all sets (but this can't be right, there is no set of all sets!) which is such that it obeys the properties (a) and (b)
He answered the following question @XanderHenderson. What is the volume of the largest surface of revolution with constant Gaussian curvature that can be placed inside the unit cube?
@EE18 Enderton is saying that the tools to properly present a definition don't exist yet, but that they will be introduced later in Chapter 7. Until then, you are to take the definition given (called a "promise") as a given.
@JohnZimmerman That looks like a completely different question to me, and you still haven't addressed the context and content of the answer. But I am also not really interested in this problem that you are so very excited about, so I am going to stop replying, now.
@EE18 Basically, Enderton is saying "Trust me. There does, in fact, exist a function $\operatorname{card}$ such that $\operatorname{card}A = \operatorname{card}B$ when $A\approx B$, and if $A$ is finite and $A\approx n$, then $\operatorname{card}A=n$."
(I presume that $A \approx B$ means that there is a bijection between the two, and that $n$ is a von Neumann ordinal, or something similar).
OK, that makes sense. Only thing is that I guess no such function can actually exist, right? Else its domain is the set of all sets, a contradiction...
@EE18 Why is the domain automatically the "set of all sets"?
That seems like a bit of a leap of logic...
In any event, the whole point of that passage is that you are not supposed to think too deeply about it right now, and that it will all be made rigorous in Chapter 7.
I have no idea what the content of Chapter 7 is, but you could, I suppose, read ahead...
@XanderHenderson well said. Coming up with a problem that can't be answered without possible whole lot of mind stretching is not really what interests most people
You can define for any set $A$ a thing depending on $A$, even though this definition won't yield a "function on the set of all sets" (cause there is no such set)
@Thorgott Yes, the distinction is important to understand, but the whole point of that passage is that the correct tools to provide understanding are currently out of reach. Don't think about it now, it will be explained later.
They seem to apply to different things. A definition is just a name for something right? This seems to be saying that there is a thing card A for each A?
@EE18 In principle, you can write down any definition you like, but you do not know, a priori, that any actual object of any kind will actually satisfy that definition. I think that Enderton is actually making a kind of existence claim, but he is not proving or justifying that claim (yet).
Enderton is saying that there is a way of associating to each set $A$ a set $\mathrm{card}(A)$ and this will satisfy certain properties, but this will be done by specifically defining $\mathrm{card}(A)$ at a later point
But your message before that makes sense. We will show that there is such an association (probably many possible) and then we will define card A as a particular choice
there is a sort of equivalence relation of sets where $|A| = |B|$ means there is a bijection $f:A\to B$. Cardinal numbers are the "canonical representants" of this equivalence class in the sense that for each set $A$ there exists a unique cardinal number with the same cardinality as $A$
@Thorgott Am on computer now, but fail to see how my order of quantifiers is wrong. Isn't Enderton basically leading us to "for all sets $A$, there exists a set card $A$ with the following properties, and that we can speak of a specific card $A$ will follow from later arguments when we get more specific about card $A$ properties"
@Jakobian I find it difficult to either agree or disagree with Enderton without knowing how the theory has been developed up to that point, or how it will be developed after that point.
And, according to the pdf I just found, Enderton doesn't formally define ordinals or orderings until chapter 7. So that may very well be exactly the definition he gives.
@EE18 he doesn't list properties of the set $\mathrm{card}(A)$, he lists properties of the association rule $A\leadsto\mathrm{card}(A)$
I guess the way of phrasing this is harmlessly interchangeable if we have uniqueness, but in the absence of uniqueness, this way sounds wrong
also, for what it's worth, if you set up in a context in which it is reasonable to talk about the "collection of all sets" (and this is done regularly), then card will indeed be a function from this collection to itself. it's a hard pill to swallow (so I probably shouldn't philosophize about it at this point), but "size issues" are tyically only a problem of relative nature, not of an absolute one.
Consider the vector space $\mathbb{R}^3$ and the following system of vectors: $$\mathbb{B} = {(k+1,k,1-k),(-1,1,1),(0,k,k)}, \ \text{where} \ k \in \mathbb{R}.$$Determine for which values of $k$, $B$ is a basis for $\mathbb{R}^3$. \begin{pmatrix} k+1 & -1 & 0 \\ k & 1 & k \\ 1-k & 1 & k \end{pmatrix}\begin{align*} \text{det}(\mathbb{B}) &= (k+1)\begin{vmatrix} 1 & k \\ 1 & k \end{vmatrix} - (-1)\begin{vmatrix} k & k \\ 1-k & k \end{vmatrix} \\ &= (k+1)(0) - (-1)(k) \\ &= k \end{align*}$\text{basis means}$: spans entire space & linearly indepdnent.
I tried to do this myself, can anyone confirm if this is correct?
@Thorgott Maybe my phrasing is poor, but I think I am talking about a sort of implicit uniqueness right. I am saying that eventually Enderton will specify a unique association rule between $A$ and the sets which could have served as card $A$ ?
Oh OK, that makes sense. on that I definitely agree
you're just saying that it doesn't even make sense to write down card $A$ (since that suggests/requires some unique thing) until we specify more properties
I can also add these notes above. In $n$ dimensions: - less than $n$ vectors are never enough to cover the entire space (therefore never a basis). - more than $n$ vectors are always linearly dependent (therefore never a basis). - if there are exactly $n$ vectors, then not even one all vectors are linearly dependent AND DO NOT cover the entire space OR they are linearly independent and span space.
EE18 yeah, enderton is using this as a kind of informal guide. at this moment in the book, i would think of "Card" as an "an object to be defined later with these properties," not "an object that enderton has just defined with these properties," let alone "an object just defined that is characterized by those properties"
and in mild defense of enderton, i would think of it that way because that is what he expressly tells you to do
if there are exactly $n$ vectors, then not even one all vectors are linearly dependent AND DO NOT cover the entire space OR they are linearly independent and span space.
What is the mayer-vietoris sequence for $H^0$ of $M$ being the disjoint union of 8 circles? I have a feeling this is related to that of the cohomology vector space of a torus which gives a MV sequence of $M:=T^2$ to be $0 \to \Bbb R \to \Bbb R \oplus \Bbb R \to \Bbb R \oplus \Bbb R$.
