It's actually $2^{\aleph_0}$ For the lower bound: For every pair $(2i, 2i+1)$ we can either swap them or not, the subgroup formed by such permutations has cardinality $2^{\aleph_0}$
@Jakobian No, I claim that this subgroup is uncountable ($2^{\aleph_0}$), which gives a lower bound for the cardinality of the whole group of bijections.
Sorry, at the first reading it seemed like you are trying to take subgroup generated by all the translations of $2i$ and $2i+1$. The phrasing was a little unclear to me
> Let $(a_n)_{n \geq 1}$ and $(b_n)_{n \geq 1}$ be two bounded real sequences. Then we have $$\limsup \left(a_n+b_n\right) \leq \limsup \left(a_n\right)+\limsup \left(b_n\right).$$
Would it be correct and an improvement to say at least one of the sequences need to be bounded?
Rudin says in his book that this identity holds provided the right hand side is not $\infty-\infty$, which I think would be accomplished if at least one of the sequences would be bounded, or?
Actually I think Rudin's formulation is better than saying at least one of the sequences is bounded. He singles out the (single) case where the sum could be undefined, which is more general than at least one of the sequences needing to be bounded.
I imagine this is for the same type of reason as that for which it's convenient to define dimension of $\emptyset$ to be $-\infty$, or degree of polynomial $0$ to be $-\infty$
@Thorgott I agree that $\sup\varnothing=-\infty$, but is it clear that there are no chains of prime ideals in the trivial ring? I would have thought that the empty set is a chain of prime ideals of the trivial ring (though I'm not sure what its length should be). The empty set is certainly a "chain" in the order-theoretic sense, i.e. it is a totally ordered subset of the collection of prime ideals of $\{0\}$ (which also happens to be empty).
@Joe I would say that's right, and its length should be $0$
sorry, I see now that they define length of a chain of prime ideals as number of strict inclusions
in this case it'd be more appropriate to say the length is $-1$
(to be honest, this just seems to be a weird definition of length where they try to adjust it to what they want to say... which is something I dislike)
@Joe both union and intersection of chain of prime ideals is a prime ideal OR the whole ring. I'd say this definition is somewhat biased in that it chooses the descending direction
so it's not necessarily a good generalization of Krull dimension
also note that here you also define $-\infty$ as convention
@Joe From what I read online, $\dim \emptyset = -\infty$ is defined solely so that $\dim(X\times Y) = \dim X + \dim Y$ formula holds. Lets compare it to polynomials: you want $\deg(pq) = \deg(p)+\deg(q)$ to hold. Does it hold if you're not in an integral domain? Lets compare it to topological spaces: you want $\dim (X\times Y) = \dim X + \dim Y$ to hold, does it even hold in general (no it doesn't)?
The more natural definition of $\dim \emptyset = -1$ gets replaced by what's convenient (!). Thankfully, the topological notions of dimensions are free of this because 1) $\dim (X\times Y) = \dim X + \dim Y$ doesn't always hold and 2) $\dim \emptyset = -1$ allows us to prove Urysohn inequalities (and gives better definitions of inductive dimensions)
I, personally, dislike defining things per convenience. But you might adopt $\dim \emptyset = -\infty$ if you like
@Joe $\sup \emptyset = -\infty$ in the extended real line. But you might say $\sup \emptyset = 0$ in $[0, \infty]$ for instance. It's all context dependent so it doesn't solve this
I think you (@Thorgott) meant to say that since Krull dimension is supremum of all heights of prime ideals, and there is no prime ideals, it must be $-\infty$. This works if we take this definition, but the problem is still there - where are we taking the supremum?
@Jakobian: Okay, I agree that whether $\sup\varnothing=-\infty$ depends on what ordered set is implicitly under consideration. It's just that in a number of contexts, even outside of analysis, that ordered set is the extended reals. But I like your point about the height of prime ideals. This gives us a reason to exclude the empty chain after all, I think.
@Joe well, it's just that the two definitions shouldn't really be equivalent for the trivial ring
@Thorgott I am taking my $\sup$ in $[-1, \infty]$ and so the definition is that the Krull dimension of the trivial ring is $-1$ in each case shrug debate over
I now remember the reason why somebody said the trivial ring should have Krull dimension $-\infty$: it means that the formula $\dim(A[x])=\dim(A)+1$ is true for any commutative ring $A$.
@Joe I suspect that considering tensor products might lead you to something similar to Urysohn inequalities from topology i.e. $\dim(A\otimes B)\leq \dim(A) + \dim(B) + 1$ so it might be why you might want to actually define dimension of the trivial ring to be $-1$
If there existed some induction proof similar to that of Urysohn lemma, then allowing dimension of trivial ring to be of dimension $-1$ (as per definition 1 of wikipedia article) would make this proof easier so it'd be "a natural choice" in this case, but so far I see no reason why would such proof exist.
As I read just now, $\dim(X\times Y) = \max(\dim X, \dim Y)$ so the previous math.se post must have been wrong
The problem is that the ring of functions on the empty set is the zero ring, and it's unclear what the Krull dimension of the zero ring should be since it has no prime ideals (it's the only commutative ring with this property), so the supremum you're taking to define the Krull dimension is a supr...
I believe here Qiaochu Yuan actually means tensor product?
Despite appearances these general questions on dimension have nothing to do with algebraic geometry, but are pure general topology.
Here are two results which answer vastly more than what you ask.
A) If the irreducible components of the topological space $X$ are the closed subsets $X_i, i\in I $...
found this which also has corresponding result for topological dimensions but not the same thing
more or less, for separable metric spaces it holds that if $X$ is union of closed subspaces $(F_n)$ then $\dim X = \sup_i \dim(F_i)$
what would union of affine schemes correspond to for rings?
oh this always holds?
I suppose the reason it works is that $X, Y$ are both affine schemes
But if we just considered one affine scheme $X$ and its subspaces then it wouldn't be like this. But then we are moving away from what interests us
@Jakobian the union of two affine schemes is generally not affine
anyway, caring about closed subschemes is a bit unnatural in this setting. any scheme is the union of its irreducible components and its dim is the sup of their dims (as in the linked answer), but you cannot write any irreducible component as union of closed subspaces without one of them being the entire component, so that's that
you do have a sup-formula for a cover by open subschemes
even better, $\dim(X)=\sup_{x\in X}\dim(\mathcal{O}_{X,x})$
I think you can't really argue that it's natural to define $\dim 0 = -1$ like this. So because of how nice everything is, I think there's really nothing against defining it to be $-\infty$ (and it does have its merits to define it as such)
magma returns the dimension of the empty scheme as $-1$ apparently (dunno why), but I've never seen any algebraic text actually make a convention other than $-\infty$ for $\dim(\emptyset)$ and I also do not see why any other convention would make sense
Assuming that in this case topological dimensions (say covering dimension) and Krull dimension should agree, it would be at least somewhat justifiable to say that Krull dimension of $\emptyset$ is $-1$
Because for covering dimension it is justified
At the end of the day it's probably best to just use what others use. And if it's that as you say that people pretty much uniformly agree that Krull dimension of $\emptyset$ should be $-\infty$, then that's what people learning the subject (like @Joe) should take
besides it's not like it really matters, we're talking about the empty set
I see. Then I don't know why would they define it to be $-1$ either
The only possible other reason I can think of is that they might impose some axioms on what a dimension function should be, and to obtain analogy with inductive and covering dimensions they decided to define $\dim(\emptyset) = -1$. But that's probably a huge jump and unlikely their reasoning
For the record, the book I am studying from defines the Krull dimension of the trivial ring to be $-1$, but I can't say that this is a well-known or popular book.
I liked it because it covered a lot of topics not covered in other commutative algebra books, like field extensions and linear algebra over a commutative ring.
It also assumed familiarity with things I actually was familiar with, like basic category theory and set theory. This meant I didn't have to skip much.
(But it also says on page 132 that a linear endomorphism of a general vector space is injective iff it is surjective, which is quite a serious error... Oh well. At least the proofs are detailed enough that I can usually figure out whether they are correct or not.)
Magma uses Hartshorne as reference, but Hartshorne doesn't comment on what the dimension of $\emptyset$ should be (probably considers it not important)
@Joe I think they really just applied definition 1 on wikipedia to the trivial ring and there is no need to overthink this
Empty set is weird enough that you will want to exclude it for some theorems to hold, or ignore its existence completely
Trivial ring, I suppose it's sort of the same thing but for rings
I have found a number of occassions that thinking about the empty case actually leads to more elegant and natural definitions. There is some discussion of this on the nLab at Too Simple to be Simple.
Example: if $A$ is a commutative ring, then why is $A$ not a prime ideal? Well, an ideal $I\subseteq A$ is prime if its complement in $A$ is closed under products. In particular, $A$ is not prime, because $\varnothing$ is not closed under products – the empty product $1$ does not belong to $\varnothing$. (By the way, some people would take "closed under products" to be closed under binary multiplication. Here, I mean $n$-ary multiplciation, since I want to avoid a bias towards the number $2$...)
Well, this definition is the same as saying that $I\subseteq A$ is prime if its complement is multiplicatively closed. And since localization plays such a large role in commutative algebra, I find it very natural
@Jakobian in AG this is actually kind of important at times, you have to exclude the empty set from being irreducible (but a bunch of sources neglect this edge case)
@Jakobian how can the union of a chain of prime ideals be $A$? no prime ideal contains $1$
I often find that when a theorem has an "edge case" where it fails, it has not been stated correctly. For example, "every ring has a maximal ideal" – false for the trivial ring. Better is to say "given any ring, every ideal is included in a maximal ideal".
Here's an argument for why $A$ should be a prime ideal. Arbitrary intersections of chains of prime ideals should be prime. The intersection of the empty chain is $A$.
and, to be honest, after all those talks about definitions, I came to the conclusion that arguing about definitions never leads to any good conclusions
> Theorem 2.26 If $f\in L^1(\mu)$ and $\epsilon > 0$, then there is an integrable simple function $\phi = \sum_{1}^{n}a_j\chi_{E_j}$ such that $\int |f - \phi|\,d\mu < \epsilon$.
Proof. For every measurable $f:X\to\mathbb C$ we can find a sequence of simple functions such that $0\leq|\phi_1|\leq|\phi_2|\leq\ldots\leq|f|$ and $\phi_n\to f$ (Theorem 2.10b). Then $\int |\phi_n-f|<\epsilon$ for $n$ sufficiently large by the DCT, since $|\phi_n-f|\leq 2|f|$.
I'm paraphrasing Folland. Theorem 2.26 states that there is an integrable simple function $\phi$ such that $\int |f - \phi|\,d\mu < \epsilon$. I don't see how $\phi$ (i.e. $\phi_n$ in the proof for sufficiently large $n$) is measurable for instance. Theorem 2.10b just says there is a sequence of simple functions that converge to $f$, but not that they are measurable.
We apply the DCT to the sequence $|\phi_n-f|$, so this sequence must be in $L^1$.
Didn't we establish before that every non-negative function $f$ there exists $\phi_n\geq 0$ such that $\phi_n$ is increasing and $\phi_n\to f$ pointwise?
And didn't it follow directly from definition of $\phi_n$ that if $f$ is measurable then so is $\phi_n$ for each $n$?
And didn't we prove that it follows that there is a sequence of simple functions $\phi_n$ such that $|\phi_n|\leq |f|$, the absolute values are increasing and $\phi_n$ converges to $f$ pointwise?
And that if $f$ is measurable then $\phi_n$ is measurable?
I don't need to see it from Folland, we did this before. You have $\phi_n = (\phi^0_n-\phi^1_n)+i(\phi^3_n-\phi^4_n)$ where $\phi^0_n$ approximates $\text{Re}(f)_+$, $\phi^1_n$ approximates $\text{Re}(f)_-$, and so on
If $f$ is measurable then all of those four "components" are measurable
and if those four functions $\phi_n^k$ are measurable then $\phi_n$ is measurable too
@Jakobian ok, so it boils down to the components being measurable
Here's Folland's construction of the components. For $n=0,1,\ldots$ and $0\leq k\leq 2^{2n}-1$, let $$E_n^k=f^{-1}((k2^{-n},(k+1)2^{-n}])\quad\text{and}\quad F_n=f^{-1}((2^n,\infty]),$$ and define $$\phi_n=\sum_{k=0}^{2^{2n}-1}k2^{-n}\chi_{E_n^k}+2^n\chi_{F_n}.$$
he may not state that the ingredients used in each argument are measurable, because (in that instance at least) the ingredients are obviously measurable from his definition of "measurable" for a real valued function
jakobian, we may be dealing with an unreliable narrator
psie you might consider folland's section 2.1 as a prerequisite to everything that follows, and not something whose results would be cited to by name each time they are used
