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Prove that an infinite well ordered set X has equal cardinality to the set X∪{a}, where 'a' does not belong to X.

Found this question in a book of analysis as a corollary. Before the question is introduced (as an exercise), the book introduced Theorem of Recursion on Wosets and Comparability Theorem. For ...
Arbbor1's user avatar
3 votes
1 answer
73 views

Is it consistent with ZC that a well-order of type $\omega_\omega$ does not exist?

Working in Zermelo's set theory (with choice for simplicity) - the construction in Hartogs' theorem shows that starting with a set $X$, there is a set $X'$ in at most $\mathcal{P}^4(X)$ (where $\...
Chad K's user avatar
  • 4,963
8 votes
1 answer
109 views

What does the cardinality alone of a totally ordered set say about the ordinals that can be mapped strictly monotonically to it?

For any cardinal $\kappa$ and any totally ordered set $(S,\le)$ such that $|S| > 2^\kappa$, does $S$ necessarily have at least one subset $T$ such that either $\le$ or its opposite order $\ge$ well-...
Transfinite Pyramid Scheme's user avatar
13 votes
1 answer
670 views

Why do we need "canonical" well-orders?

(I asked this question on MO, https://mathoverflow.net/questions/443117/why-do-we-need-canonical-well-orders) Von-Neumann ordinals can be thought of "canonical" well-orders, Indeed every ...
Vivaan Daga's user avatar
  • 5,704
1 vote
1 answer
126 views

Cardinality of set of well-ordered sequences

We think of $A=\mathbb{R}^\mathbb{N}$ as the set of all functions $f:\mathbb{N}\to\mathbb{R}$. Consider the following subset of $A$: $$ B=\{f\in A\mid f(\mathbb{N}) \text{ is a well-ordered subset of $...
user173558's user avatar
6 votes
3 answers
188 views

How big is the smallest not well-orderable set in $\mathsf{ZF}$? [duplicate]

I know that $\mathsf{ZF}$ alone (i.e., without the Axiom of Choice) cannot prove (nor disprove) that $\Bbb R$ can be well-ordered. Then again, without the Continuum Hypothesis, we cannot know whether ...
Hagen von Eitzen's user avatar
0 votes
1 answer
106 views

Linear ordering and cardinal numbers - Problem 5.3 Jech's book

I am solving problems of the Jech's book (set theory). I need help to solve problem 5.3: Let $(P, <)$ be a linear ordering and let κ be a cardinal. If every initial segment of $P$ has cardinality $&...
Aram's user avatar
  • 89
0 votes
1 answer
42 views

Is the proper segment of a TOSET is initial segment? [closed]

I can't fimd its answer. I have learnt that for a well ordered set proper segment is the initial segment. But i am unable to find segment of a toset which is proper and not initial.
Abhya Gupta's user avatar
3 votes
2 answers
361 views

Infinite subset of natural numbers

I want to show that if the set $A$ is an infinite subset of natural numbers $N$, then $\text{card}(A)=\text{card}(N)$. To do so, it suffices to find an injective function $f:N\rightarrow A$. One ...
S_Alex's user avatar
  • 991
-1 votes
1 answer
87 views

Regular Cardinal of a Power set and its Chains

My attempt(almost sure that it is not quite working): It is known that the greatest cardinal is P (ω) which is not a well-orderable set. Another fact is that P (ω) is a regular cardinal which by the ...
Tiki's user avatar
  • 123
0 votes
1 answer
611 views

How is the Continuum Hypothesis equivalent to the existence of a well-ordering on $\Bbb R$ whose bounded initial segments are countable?

There exists an well-ordering $(<)$ on $\Bbb R$ such that the set $\{x \in \Bbb R\mid x < y \}$ is countable for every $y \in \Bbb R.$ How to prove that the above statement is equivalent to ...
math maniac.'s user avatar
  • 2,013
3 votes
1 answer
89 views

Long sequences in singular cardinals can't be cofinal

My attempts to understand a solution to the Kunen exercise (two well orderings on a cardinal $\kappa$ necessarily agree on a set of size $\kappa$) have foundered on the following claim. If someone ...
Robert Shore's user avatar
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0 votes
1 answer
116 views

Cardinal of all well-orders of $\mathbb{N}$ [duplicate]

Lets consider the set $$ A = \{R\subset\mathbb{N}^2:R \text{ is a well-order of } \mathbb{N}\} $$ Now, it's clear that $\aleph_1\leq|A|\leq2^{\aleph_0}$(Since all the well-orders of $\mathbb{N}$ are ...
miraunpajaro's user avatar
  • 1,560
0 votes
1 answer
543 views

Indexing of uncountable sets and uncountable collections of sets, uncountable intersections containing a point

Definitions Let $\mathcal{A}$ be an uncountable collection of sets so that if $I_{\mathcal{A}}$ is the index set of elements of $\mathcal{A}$ then $|I_{\mathcal{A}}|\not=\aleph_{0}$ (I mean this to ...
dandar's user avatar
  • 1,010
1 vote
1 answer
113 views

Is there a maximum ordinal associated to a totally ordered set?

Let $R$ be a totally ordered set. We can construct a well-ordered cofinal set $S \subset R$ by transfinite induction and the axiom of choice. (Choose an element from $R$ as the base case; then if we'...
Twiffy's user avatar
  • 929

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