Set theory - 2017-08-16

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Martin Sleziak

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:

A: Consistency Results Separating Three Cardinal Characteristics Simultaneously

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 Cichoń's Diagram, which also presents all $ZFC$-provable relations between pairs of these characteristics: besides ...

The today's addition to the answer was:

> 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…

Martin Sleziak

14:04

This seems kind of interesting:

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?

Martin Sleziak

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".

Secret

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, 30 mins ago, by Secret

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, 28 mins ago, by Leaky Nun

@Secret lexicographical.

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 rationals are still linearly ordered

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

and yesterday...

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, 23 hours ago, by Leaky Nun

@Secret no, you can't define an order on $2^{\Bbb R}$

in Mathematics, 22 hours ago, by Leaky Nun

55 mins ago, by Leaky Nun

@Secret $C=\dfrac1{2\Bbb N+1}$ or $D=\dfrac1{2\Bbb N+2}$?

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 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:

(counterexample being the C and D set example)

in Mathematics, 21 hours ago, by Leaky Nun

@Secret there is no "first" for there is no well-order.

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.

and just now:

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

So thissounds all convoluted. Let me clarify this again:

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

So this give rise to a few questions:

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?

2. How exactly is the lexicographical order of $2^{\Bbb{Q}}$ defined, is it arrange fractions from smallest to largest,or it is done by cosidering the numerator and denominator separately?

typo: from smallest to largest like in the usual ordering of the reals

Leaky Nun

15:13

@Secret no, $\Bbb Q$ isn't well-ordered so the lexicographical order on $2^\Bbb Q$ doesn't make sense

Secret

Ok, if I understood correctly, both $\Bbb{R}$ and $2^\Bbb{Q}$ have a linear ordering, but they don't biject in an order preserving manner into each other?

Leaky Nun

@Secret that isn't right

Secret

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?

(I think it will be clearer to write down explicitly the linear ordering of $2^{\Bbb{Q}}$ cause I think that's where I most confused at)

Leaky Nun

@Secret you cannot comment about order preserving maps if you haven't talked about the ordering

@Secret where your thoughts go wrong is when you think that a linear ordering is inherent of a set

Secret

15:29

Ok, using the above logic on $2^{\Bbb{Q}}$, I think $2^{\Bbb{R}^+}$ has no lexicographical ordering makes sense now, but is it possible for $2^{\Bbb{R}^+}$ to have an explicit dense linear order under ZF, or it is not provable since dense linear order is strickly weaker than the axiom of choice?

Leaky Nun

@Secret DPO is independent of ZF

So it would be not provable, I think.

Secret

I see

Reposting this back to set theory room:

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?

Martin Sleziak

15:58

in Mathematics, 11 mins ago, by Leaky Nun

in Mathematics, 8 mins ago, by Leaky Nun

@Semiclassical oops, I meant $|\kappa|$, not $|\Bbb N|$

@LeakyNun Are we allowed AC, or do you want to prove this in ZF?

Martin Sleziak

16:10

@LeakyNun I should have remembered this question: Does $\aleph_0\cdot\kappa=\kappa$ for every $\kappa\ge\aleph_0$ hold in ZF?

Asaf says it's not true in ZF.

Leaky Nun

@MartinSleziak thanks