Tagged Questions
5
votes
3answers
68 views
How to show that $(\Bbb Q, <)$ contains an order isomorphic copy of $\epsilon_0$ which is the least ordinal satisfies $\omega^\epsilon = \epsilon$?
We define $\epsilon_0 = \sup\{\omega, \omega^\omega,\omega^{\omega^\omega},\omega^{\omega^{\omega^\omega}},\ldots\}$. How to find a subset of $ \Bbb Q$, $(A, <_{\Bbb Q})$ that is isomorphic to ...
3
votes
2answers
78 views
Why does every countable limit ordinal have cofinality $\omega$?
According to Wikipedia, if $\alpha$ is a countable limit ordinal, then $\mathrm{cf}(\alpha)=\omega$. It is intuitively clear to me that it should be so. Certainly the cofinality of such an ordinal ...
6
votes
1answer
70 views
Why should the union of such a family of sets have cardinality $\aleph_1?$
Let $\mathcal A$ be a family of infinite countable sets which is linearly ordered by inclusion and such that $\bigcup\mathcal A$ is uncountable. I have to prove that $|\bigcup\mathcal A|=\aleph_1.$
I ...
2
votes
1answer
45 views
Two questions of set theory: necessary and sufficient conditions for a subset of a poset to be centered, posets with noncentered linked subsets.
$(1)\hspace{4pt}$ Let $\left\langle X,\leq\right\rangle$ be a poset, and let $Y\subseteq X$ with $Y\ne\emptyset$. I’m trying to prove that $Y$ is centered—i.e., $Y$ is a filterbase on ...
3
votes
1answer
76 views
Existence of a minimal set of linear order that is equivalent to the given partial order
Let $\{(P, \leq_{\alpha})\}_{\alpha \in A}$ be a set of linear order on set $P$, $(P,R)$ be a partial order on the same set $P$. We say that the former is equivalent to the later, iff: $$\forall a, b ...
6
votes
3answers
124 views
Ordinal exponentiation - $2^{\omega}=\omega$
This is my understanding of ordinal arithmetic - two ordinals are the same as one another if there is an order-preserving bijection between them. So for instance
$$1+\omega = \omega$$
because if
...
3
votes
1answer
83 views
Is every discrete topological space orderable?
I apologize for asking a question in topological terms when it's not really about topology, but here goes:
If $S$ is a set, is it always possible to totally order $S$ so that every non-minimal ...
6
votes
3answers
109 views
Incomparable subsets of real numbers with usual order relation
Consider linear order $(R,<)$, would there be two sub-orders $(X,<), (Y,<)$ of it such that $X,Y$ are uncountable, which are incomparable in the following sense:
There does not exist order ...
0
votes
1answer
117 views
Every finite partially ordered set has a maximum length chain.
Every finite partially ordered set, $(A, \leq)$, has a maximum length chain.
A chain is a sequence of distinct elements $a_1 \leq a_2 \leq .......\leq a_n$ with relation $"\leq"$ where $a_i \in A$ ...
5
votes
2answers
135 views
If every nonempty subset of a set $S$ has a least and greatest element, is $S$ finite?
Some sets are well-ordered; all of their nonempty subsets have least elements. You can also have sets where all of their nonempty subsets have greatest elements.
Some sets have both of these ...
1
vote
3answers
79 views
Question regarding well-ordering theorem
The well-ordering theorem states that every set can be well-ordered. That is, there is a total order on the set in which every non-empty subset has a least element. I was wondering whether it's also ...
4
votes
2answers
59 views
Showing there is only one isomorphism between well ordered sets using transfinite induction
I need to show specifically using transfinite induction that given two well-ordered sets $\left(A,<_{1}\right)$ and $\left(B,<_{2}\right)$ there is only one isomorphism between them. To do ...
2
votes
1answer
63 views
Question about Hausdorff Maximal principle and antichain
I have this prob and still haven't figured about
Let (A, $\leq$) be a partially ordered set. A subset B of A is called an antichain if and only if for any two distinct elements x and y in B, ...
1
vote
2answers
133 views
Well-Ordered Sets and Functions
I need some assistance with the following problems. I've come up with some ideas but may need some clarification.
1) Let $S$ be a well-ordered set.
a) Need to show $S$ has a first element.
...
2
votes
2answers
46 views
Reaching elements with finite applications of the successor relation in well-ordered sets
Given an well-ordered class $(A,\leq)$ we want to proof that every element $a \in A$ can be reached by applying the successor relation on a the smallest element of $A$ or on an limes element of $A$ ...
5
votes
1answer
105 views
Exercise on partially ordered sets from Kunen's *Set Theory*
This problem is Exercise (F4) from Chapter VII of Kunen's text. Apparently, it is a result of Tarski. It goes as follows:
Let $ \mathbb{P} $ be a partial order. Define $ \text{c.c.}(\mathbb{P}) $ ...
2
votes
1answer
111 views
The relation between order isomorphism and homeomorphism
Let $X$ be a set and let $<_1,<_2$ be order relations on $X$.
Let $T_1,T_2$ be the topologies induced on $X$ respectively.
If $(X,T_1)$ is homeomorphic to $(X,T_2)$, does that imply that ...
2
votes
1answer
128 views
Is the set $S$ of functions $S = \{ f : \mathbb R \to \mathbb R \}$ from the reals to the reals a totally ordered set?
Is the set $S = \{ f : \mathbb R \to \mathbb R \}$ of functions $f$ from the reals to the reals a totally ordered set ?
I think the question is clear. Note that there is a bijection to $g(x)$ gives ...
4
votes
1answer
58 views
Generalization of ordinals to well-founded sets?
In set theory, the ordinal numbers are objects that represent the order types of well-ordered sets. Is there a similar class of objects that represent the order types of well-founded sets? If so, ...
9
votes
1answer
136 views
Atoms necessary for the existence of a generic filter?
I'm reading Some applications of the method of forcing by Todorchevich and Farah and one of the exercises seems to be wrong to me.
Definitions
We fix some partially ordered set $(P,\preceq)$.
A ...
5
votes
2answers
167 views
Infinite linear (non-well) orderings
I can rather easily imagine the infinite linear (non-well) orderings $(\mathbb{Z},\leq)$, $(\mathbb{Q},\leq)$, $(\mathbb{R},\leq)$, each one of its own order type.
Are there "essentially" other ...
1
vote
1answer
153 views
Totally ordered set with greater cardinality than the continuum
Does there exist a totally ordered set $S$ with cardinality greater than that of the real numbers? Sequences are continuous functions with domain $\mathbb{N}$ and paths are continuous functions with ...
8
votes
2answers
271 views
An order type $\tau$ equal to its power $\tau^n, n>2$
In this question we are concerned only with linear (aka total) order types. By a cardinality of an order type we understand a cardinality of an instance of this type, which obviously does not depend ...
0
votes
1answer
97 views
$<$ on a preorder is a strict partial order
Definition: Suppose $X$ is a preorder. Define $x < y$ as $x \le y$ and $y \not\le x$ for each $x, y \in X$.
Question: Show that this gives a strict partial order on $X$.
10
votes
1answer
300 views
Linearly ordered sets “somewhat similar” to $\mathbb{Q}$
$\pmb{\eta}$ - order type of $\mathbb{Q}$.
$\pmb{\eta} + \pmb{1}$ - order type of $\mathbb{Q}\cap(0, 1]$.
$\pmb{\omega_1}$ - order type of the first uncountable ordinal.
Let's say that a linear ...
14
votes
2answers
277 views
Is $(\pmb{1} + \pmb{\eta})\cdot\pmb{\omega_1} = \pmb{1} + \pmb{\eta}\cdot\pmb{\omega_1}$?
$\pmb{\eta}$ - order type of $\mathbb{Q}$.
$\pmb{1}$ - order type of a singleton set.
$\pmb{\omega_0}$ - order type of $\mathbb{N}$.
$\pmb{\omega_1}$ - order type of the first uncountable ordinal.
...
8
votes
1answer
135 views
Is $\pmb{\eta}\cdot\pmb{\omega_1} = (\pmb{\eta} + \pmb{1})\cdot\pmb{\omega_1}$?
$\pmb{\eta}$ - order type of $\mathbb{Q}$.
$\pmb{1}$ - order type of a singleton set.
$\pmb{\omega_0}$ - order type of $\mathbb{N}$.
$\pmb{\omega_1}$ - order type of the first uncountable ordinal.
...
4
votes
2answers
129 views
Is every order isomorphism of sets induced by a bijection?
Let $f$ is an order isomorphism $\mathscr{P}A \rightarrow \mathscr{P}B$ (where $A$ and $B$ are some sets, the order is set-inclusion $\subseteq$).
Is it true that it always exist a bijection $F: A ...
2
votes
1answer
131 views
Preorders, chains, cartesian products, and lexicographical order
Definitions:
A preorder on a set $X$ is a binary relation $\leq$ on $X$ which is reflexive and transitive.
A preordered set $(X, \leq)$ is a set equipped with a preorder.... Where confusion cannot ...
4
votes
2answers
111 views
$<$ in a preorder
The author of the book I am studying defines $<$ for a poset as
If $x, y \in X$, where $X$ is a poset, then we shall write $x < y$ to mean that $x \le y$ and $x \ne y$.
From this, I can ...
3
votes
1answer
218 views
Does every set have a well ordering with greatest element?
Does every non-empty set have a well ordering with greatest element?
It is well known that every set has a well ordering. But can we also assume that this well ordering has greatest element?
[Edited ...
2
votes
3answers
94 views
Equal elements vs isomorphic elements in a preoder
While introducing preorders, Roy L. Crole in Cateories for Types states
If $x \leq y$ and $y \leq x$ then we shall write $x \cong y$ and say that $x$ and $y$ are isomorphic elements.
I'm trying ...
0
votes
1answer
60 views
From an equality to a comparison
Let $X$ be a set. Let $0$ be an element of $X$.
For a function $P$ defined on tuples of $n$ elements of the set $X$ we know (for every tuples $f$ and $g$ each having $n$ elements)
$$\forall i \in n : ...
1
vote
1answer
122 views
Are lexicographic and multiset the only extensions of sets that preserve well-foundedness?
It is well known that for a given set $S$ with well-founded order relation $R$, the lexicographic order that extends $R$ on tuples of $S$ is also well-founded.
Also, the multiset order on the ...
3
votes
1answer
196 views
Question of an isomorphism of $\epsilon_ 0$ and a subset of the rationals.
I don't know if this question is appropriated for this site. Anyway, I'm searching for an isomorphism of order $f:K \longrightarrow \epsilon_o $, such that $(K, \leq)$ is a subset(proper or not) of ...
3
votes
1answer
163 views
Cardinality of the set of ultrafilters on an infinite Boolean algebra
Let $\mathfrak B$ be a Boolean algebra with an infinite power $\kappa$. My question is how many ultrafilters does it have? $\kappa$ or $2^\kappa$? Or even smaller?
-2
votes
1answer
160 views
Concatenating an infinite number of arguments
(I asked this question at MathOverflow but it was closed as "too localized". I re-state the question at math.stackexchange.com and hope my repeated questions will be not considered spamming, as a ...
0
votes
2answers
255 views
Example of total order with some properties that is not well ordered
Is there an example of a total order with properties
there is a least element and
every element has a (unique) successor
not is not also a well ordering?
2
votes
1answer
160 views
Is a chain-complete lattice a complete lattice without the axiom of choice?
This question is inspired by this question. Consider the following result:
Let $(L,\leq)$ be a chain-complete lattice. Then $(L,\leq)$ is a complete lattice.
Can this result be proven without ...
5
votes
1answer
331 views
Quickest way to understand Kruskal's Tree Theorem
I came across the Kruskal Tree Theorem the other day and thought it looked pretty interesting (especially the stronger finite form due to Friedman). I'm currently a first year mathematics ...
3
votes
1answer
110 views
number of linear orders
It is well known that for every infinite cardinal $\kappa$ the number of non-isomorphic total orders of cardinality $\kappa$ is $2^\kappa$. Who first proved this, and in what context? Was it proved ...
2
votes
2answers
334 views
A question regarding the Continuum Hypothesis (Revised)
Given that the order type of the reals $(\mathbb R,<)$ has no definable points, can it be proven that $|\mathbb R|= \aleph_1$ by virtue of the fact that every uncountable proper subset $S$ of ...
2
votes
1answer
109 views
How to show the ascending chain condition
Let $(A,\le)$ be a poset. Suppose that for any $a < b \in A$ and for any chain $Q$ of $A$ whose maximum and minimum are $a$ and $b$ respectively, $Q$ is finite. Let $C$ be the set of all chains of ...
4
votes
1answer
253 views
Totally ordering the power set of a well ordered set.
Let's say I take a set $S$, where $S$ can be well ordered. From what I understand, one can use that well ordering to totally order $\mathscr{P}(S)$.
How does a body actually use the well ordering of ...
3
votes
1answer
176 views
The “canonical” representative of an order type
An ordinal (in von Neumann sense) can be thought as the "canonical" representative of the order type (i.e. of the class of all order-isomorphic sets) of a well-ordered set. This is very convenient, ...
10
votes
2answers
182 views
Given an infinite poset of a certain cardinality, does it contains always a chain or antichain of the same cardinality?
It is known that given an infinite poset, it always contains an infinite chain or antichain;
moreover, there is a constructive proof that we can find a continuous chain in $P(\mathbb{N})$;
so, in ...
10
votes
2answers
266 views
Any good decomposition theorems for total orders?
I'm learning about set theory and partial orders, well orders, total orders, etc. I'm curious to know of any good decomposition theorems that apply to all (or very general types of) total orders.
...
2
votes
1answer
151 views
Rudin–Keisler equivalence
Wikipedia says that if $a\le b$ and $b\le a$ in Rudin–Keisler order for ultrafilters $a$ and $b$, then $a$ and $b$ are Rudin–Keisler equivalent. How to prove this?
2
votes
2answers
244 views
Showing that well-ordered subsets of $P(\omega)$ are countable
I have the following problem:
Show that no uncountable subset of $P(\omega)$ is well-ordered by the inclusion relation.
I think they want me to do by embedding it in a separable complete dense ...
2
votes
2answers
196 views
Separative Quotients, and the Induced Order
Let $P$ be some partial order.
We say that $x$ and $y$ are compatible if $\exists r\in P (r\le x \wedge r\le y)$, we denote this by $x \perp y$. Otherwise, we say that $x$ and $y$ are incompatible.
...
