06:11
A paper where Arthur Fischer is one of the authors (Arthur Fischer, Martin Goldstern, Jakob Kellner, Saharon Shelah: Creature forcing and five cardinal characteristics in Cichon's diagram) was mentioned in this recently updated answer:
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Bumping an old question, because the situation has changed dramatically in the last year or so:
The most well-known cardinal characteristics of the continuum are those appearing in CichoĊ's Diagram, which also presents all $ZFC$-provable relations between pairs of these characteristics: besides ...
> I can't resist bumping this to add one fascinating result (from this morning!) of Goldstern, Kellner, and Shelah: that all the cardinal characteristics in Cichon's diagram can be separated simultaneously! Now granted, this isn't a consistency result relative to ZFC - at present, it takes four strongly compact cardinals, which is no small assumption - but still, this is absolutely amazing.
8 hours later…
14:04
in Mathematics, 12 mins ago, by Secret
First I wan to clarify: Is it necessary and sufficient that for $2^S$ to have a constructible linear order, $S$ must be well ordered?
14:21
in Mathematics, 7 mins ago, by Martin Sleziak
It seems that AC (and so also well-ordering principle) is strictly stronger than: "Every set can be linearly ordered."
in Mathematics, 7 mins ago, by Martin Sleziak
The latter is also called ordering principle. MO: Are all sets totally ordered ?
in Mathematics, 6 mins ago, by Martin Sleziak
So you can't prove in ZF: "$\mathcal P(X)$ is linearly orderable" $\implies$ "$X$ is well-orderable".
14:46
in Mathematics, 1 min ago, by Secret
Hmm... if we cannot proof $\mathcal{P}(X)$ is linearly orderable $\implies$ $X$ is well orderable, then by contrapositive, we cannot prove $X$ is not well orderable \implies $\mathcal{P}(X)$ is not linearly orderable...
in Mathematics, 29 mins ago, by Secret
The fact that $2^{\Bbb{Q}}$ is linearly ordered suggest the proposition is sufficient but not necessary, but then what is the ordering on $2^{\Bbb{Q}}$ explicitly?
in Mathematics, 1 hour ago, by Secret
is the usual order on the reals a lexicographical order or something different, because in our exercise yesterday on trying to define a lexicographical order on $\{0,1\}^{\Bbb{R}^+}$ that fails and Leaky said it is because the reals don't have a well order hence the powerset cannot have a linear order?
in Mathematics, 1 hour ago, by Tobias Kildetoft
@Secret The binary expansion has the naturals as "base", rather than the rationals
in Mathematics, 1 hour ago, by Leaky Nun
@Secret but the natural numbers do have a well-order, so the powerset can have a linear order
in Mathematics, 1 hour ago, by Leaky Nun
the usual order can be somehow regarded as a lexicographical order, just be reminded that 0.999... = 1
in Mathematics, 23 hours ago, by Secret
But I am trying to construct explicitly $\aleph_{\geq 2}$dense linear order when GCH holds. That MSE link said it is possible. The above discussion then suggest $\{0,1\}^{\Bbb{R}^+}$ under the lexicographical order can be a good candidate
in Mathematics, 22 hours ago, by Leaky Nun
2 hours ago, by Leaky Nun
@Secret $\chi_A$ or $\chi_B$ where $A=\{1,4,5,8,9,12,\cdots\}$ and $B=\{2,3,6,7,10,11,\cdots\}$?
in Mathematics, 22 hours ago, by Secret
@LeakyNun $\chi_A > \chi_B$ if $\exists$ smallest (infimum or minimum) pairs of $x \in A, y \in B$ s.t. $x > y$
in Mathematics, 22 hours ago, by Secret
Right, so the issue is that once any subset is allowed to have a sup/inf that does not coincide with a min/max, and when the topology is not discrete, then in general we cannot assign $>$ or $<$ on uncountably many of them. If these elements are considered equal under the ordering then it becomes a partial order
in Mathematics, 22 hours ago, by Secret
Hmm, I still cannot quite see in general why $2^{\Bbb{R}}$ breaks it while $\Bbb{R}$ don't:
In $\Bbb{R}$ a binary expansion is something of the form:
...bbbbbbbb.bbbbbbb...
So moving to the left, it can become unbounded, while moving to the right the importance of the digits tends to zero
There are $\aleph_1$ many possible diverging sequences and $\aleph_1$ may possible sequences that converges to zero, yet for all reals we have no problem telling which is bigger than which
For $2^{\Bbb{R}}$, a "binary expansion" should look something like this:
In $\Bbb{R}$ a binary expansion is something of the form:
...bbbbbbbb.bbbbbbb...
So moving to the left, it can become unbounded, while moving to the right the importance of the digits tends to zero
There are $\aleph_1$ many possible diverging sequences and $\aleph_1$ may possible sequences that converges to zero, yet for all reals we have no problem telling which is bigger than which
For $2^{\Bbb{R}}$, a "binary expansion" should look something like this:
in Mathematics, 21 hours ago, by Leaky Nun
@Secret because you don't just need an order. you need a well-order. and you can't construct one for $\Bbb R$.
in Mathematics, 21 hours ago, by Secret
Do defining lexicographic order need well ordering, or it is the binary representation of some element in the set is the one that need to be well ordered?
in Mathematics, 21 hours ago, by Tobias Kildetoft
@Secret Lexicographic order has nothing to do with well-order. It works for the product of any two ordered sets
in Mathematics, 21 hours ago, by Secret
If I tried to define the lexicographic ordering by finding the first pair of elemnts that differ and then x < y, then I ran into the following counterexample:
in Mathematics, 21 hours ago, by Tobias Kildetoft
@LeakyNun Hmm, where do we need the indexing set to be well-ordered?
in Mathematics, 21 hours ago, by Leaky Nun
@TobiasKildetoft when we are finding the smallest element that differ
in Mathematics, 21 hours ago, by Tobias Kildetoft
@LeakyNun Ahh, in order for this to give a total ordering, right. I think you can define it without this, but it will not necessarily be total.
in Mathematics, 5 mins ago, by Alessandro Codenotti
@Secret the second thing, $X$ not well orderable implies $\mathcal{P}(X)$ is not linearly orderable is false, "every set can be linearly ordered" is stricly weaker than "every set can be well ordered" (which is AC) over ZF
The primary thing that lead to the above discussion is I am trying to define a dense linear ordering that is analogous to the usual ordering of the reals, on the set $2^{\Bbb{R}^+}$ which can be thought of as a $\aleph_2$ base number system
This means I am trying to get the dense linear ordering by using the lexicographical ordering of the reals since in $2^{\Bbb{R}^+}$, the (positive) reals basically forms the positions of this number system
But as Leaky and I found out, this is not possible because sets $C$ and $D$ is infinitely decreasing, hence screw up the lexicographical order
Leaky then said that in order to get a linear order on $2^S$, the indexing set $S$ has to be well ordered.
Later on, Martin and Alessandro found that we cannot prove $\mathcal{P}(X)$ is linearly orderable $\implies$ $X$ is well orderable as linear order is strictly weaker than well order
1. How does the lexicographical ordering on $2^{\Bbb{Q}}$ avoid the example illustrated in $C$ and $D$, since in $2^{\Bbb{Q}}$, we can get the sets $A,B,C,D$ as above and hence having both a infinitely increasing and a infinitely decreasing sequence hence cannot assign a consistent lexicographical order?
but we know that $\Bbb{R}$ has a linear ordering in the form of a lexicographical ordering for any given base expansion, and you just said that lexicographical orderr on $2^{\Bbb{Q}}$ does not make sense since $\Bbb{Q}$ is not well ordered. So how could there exists an order preseriving map between these two sets?
15:29
in Mathematics, 2 mins ago, by Secret
> Is there an explicit, dense linear order, non-well ordered on a set $S$ with $|S| > \aleph_1$ under GCH without AC (i.e. using ZF). If so, must they all be isomorphic to some subset of the surreals?
15:58
in Mathematics, 11 mins ago, by Leaky Nun
How to prove that $|\kappa \times \Bbb N| = |\Bbb N|$ where $|\kappa| > |\Bbb N|$?
16:10
@LeakyNun I should have remembered this question: Does $\aleph_0\cdot\kappa=\kappa$ for every $\kappa\ge\aleph_0$ hold in ZF?
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