@LeakyNun so the issue is that I only have up to 5 datapoints, and one can be an enormous outlier. I thought that taking the root or the log before the sum might be a good idea to increase robustness
Can anyone clarify my understanding of saying what it means for something not to be a spanning tree?
Spanning tree: a spanning subgraph that is tree, so if I say that something is not a spanning tree, it is still a spanning subgraph BUT is not a tree (i.e. a spanning subgraph that is either not connected or contains a cycle)?
Okay thanks! I am unsure if this is the usual definition for it, I'm new to graph theory and I realized several books define some things quite differently
That supremum involves countable unions of countable sets, which lacking AC, cannot be shown to be countable. Unless the fact that ordinals are well ordered somehow avoided that problem
Let's think about the countable case like this: think of the
binary tree $2^{\lt\omega}$, which has size $\omega$, but has
$2^\omega$ many branches. Each branch describes a cut in the
natural lexical order on the nodes, and so we have a countable
linear order with $2^\omega$ many cuts.
So consid...
@Secret trying to understand the first paragraph lol
The first paragraph is basically the same as taking a bunch of nested sets created by dedekind cuts on the rationals. Each cut partition the set into two subsets, thus each node has two branches, making a total of $2^{\aleph_0}$ many branches
If we assume $AC$ we can construct an uncountable well-ordered chain of subsets of $\mathbb{N}$ by well-ordering the reals and then using the same Dedekind's cut construction as in this question.
edit: as pointed out in the comments and Asaf's answer the construction above doesn't work
What if ...
Recently in chat, we investigated the explicit construction of $\omega_1$. Regardless on whether we use ZFC or hartogs number, we seemed to hit a roadblock.
Using either ZFC or hartogs number, we managed to came up a set of countable ordinals $S$ that corresponds to well orderings on $\omega$. H...
I have the distinct memory of having often heard and read that intuitionism was inter alia geared to avoid Cantor's uncountable sets, and it may be that this was Brouwer's plan. But are there accounts which demonstrate that early intuitionism (i.e. before the advent of modern intuitionistic set t...
Let us for example give an example of diagonal proof of uncountability of the set of real numbers $\mathbb{R}$.
Would intuitionists accept this, or deny this? If they deny this argument, why would they?
yeah, but we cannot well order the reals without C
the above MSEs also said we cannot use powersets of well ordered sets to make an uncountable well ordering in the form of uncountable chains
The hartog's construction seemed to have issue with circularity as pointed out by user21820 and elaborated in my MSE on predicative uncountable well ordering
namely the circularity is that how are two computable ordinals $\alpha, \beta$ are given with the relation $\alpha < \beta$ without first showing that to be true, what guarentees it?
If $\alpha,\beta$ can be written in ordinal notations, then we can easily figure out which one injects into another, but the nature of $\omega_1^{CK}$ means we must ran out of ordinal notations before we even reach $\omega_1^{CK}$
How is $\alpha$,$\beta$ given? If $\beta$ is a successor of $\alpha$, then I can see we can show that $\alpha < \beta$ by injection. But if $\beta$ is a limit or an ordinal that is not a successor of $\alpha$, how can we show that $\alpha <\beta$ without having to assume that our given $\alpha$ injects into $\beta$? Or is the Hartogs number construction guarantee that for all $\alpha,\beta$, $\alpha < \beta$ and thus established that $S$ is a well ordering even though we cannot construct $\alpha,\beta$? — Secret4 hours ago
But in order to construct $\omega_1$ from hartog's number, we need to show the set of all countable well orderings is well ordered, but we cannot seemed to do so for any two computable ordinals that has no ordinal notation without assuming it is true
@AlessandroCodenotti The problem is we cannot show whether any two computable ordinals $\alpha,\beta$ (that lacked ordinal notations) there's an injection of one into another without first assuming that $\alpha < \beta$ or $\beta < \alpha$, but that is precisely what is needed to be proved for all ordinals in the set of all countable well orderings
"the inability to write down computable ordinals $x,y$ between $\omega_1^{CK}$ and the large Veblen ordinal means that given arbitrary $x,y$ we cannot justify that an injection exists from one into another." I don't understand this - can you explain more clearly what the problem is here? Given two ordinals $\alpha,\beta$ we define an injection from one to the other via transfinite recursion along them - and I believe this is predicative (the crucial point being that we're given $\alpha$ and $\beta$ in this situation, we don't have to build them predicatively). — Noah Schweber14 hours ago
Noah Schweber said that $\alpha, \beta$ were given, but how does doing power sets on the naturals will guarenteee we have $\alpha < \beta$ for all $\alpha,\beta$ in the set of all countable well orderings?
We pick $S=\Bbb{N}$ and construct $\mathcal{P}(\Bbb{N})$ by the axiom of power set. This set contains some subset of the naturals
Now, the set of all countable well ordering of the naturals are all the ways that the subsets of the naturals can be well ordered, and they live inside $\mathcal{P}^2(\Bbb{N})$
@ LeakyNun @ AlessandroCodenotti Actually, I am thinking, since ordinals are transitive under $\in$, if we have the set of all countable well orderings S (an element in $\mathcal{P}^2(\Bbb{N})$), then the union of them all must be an ordinal and thus it will be automatically well ordered because of transitivity, no?
If that's the case, then we can bypass showing that $\alpha < \beta$ for any $\alpha,\beta \in S$
@LeakyNun, I reasoned like this: We have $a$ distinguishable objects and $b$ distinguishable boxes $(a<b)$ and we have to distribute $a$ objects into $b$ boxes such that no box may contain more than 1 object. Therefore we have to calculate the number of ways in which we can fill up $b$ places when we have $a$ objects at our disposal and this is equal to $aPb$. But $a<b$ so this is not valid. What is wrong with this reasoning?
So even if we end up with something like e.g. $\{\omega,\omega^{\omega},\omega 2,...\}$ after the union because of how the countable ordinals are all over the place in the collection, it is automatically well ordered when we view it under $\in$
Sure, we still cannot deal with that $\alpha < \beta$ case because there must exists $\alpha,\beta \in S$ that lacked ordinal notation due to the nature of $\omega_1^{CK}$ and thus we cannot figure how to construct an (explicit) injection between them, but if the previous messages holds, then we can at least showed that $\omega_1$ exists
But suppose the argument does not work and there's still some excluded middle somewhere we don't see), then I am fine defining the aleph function (hence existence of well ordered uncountable sets) as an extra axiom in ZF to construct well ordered sets $\omega_{\alpha}$
Recently I've been thinking about constructing ZFC in graph theory, not because it's more intuitive, but because it's less intuitive, so people won't apply blind intuition.
"The probability of having no car is 0.3, the probability of having one is 0.6, and the probability of having two is 0.1. There are a thousand households utilizing a common parking lot with said probabilities. How many parking spots are needed in order for the parking lot to fit all cars with a 90% certainty?"
I've been staring at this one for a while. Does anyone have a hint of some sort?
If we assume $AC$ we can construct an uncountable well-ordered chain of subsets of $\mathbb{N}$ by well-ordering the reals and then using the same Dedekind's cut construction as in this question.
edit: as pointed out in the comments and Asaf's answer the construction above doesn't work
What if ...
Let us for example give an example of diagonal proof of uncountability of the set of real numbers $\mathbb{R}$.
Would intuitionists accept this, or deny this? If they deny this argument, why would they?
