All Questions
Tagged with ordinals proof-explanation
38
questions
-2
votes
1
answer
57
views
Prove that the order type of $\alpha\cdot\beta$ is the antilexicographic order in $\alpha\times\beta$. [closed]
This question is related to this one, but not a duplicate, since I am struggling with injectivity and monotonicity, rather than proving that $\{\alpha\cdot\eta + \xi:\eta<\beta\textrm{ and }\xi<\...
0
votes
1
answer
68
views
Proving $\alpha\subset\beta\implies\alpha\in\beta$ for ordinals $\alpha$ and $\beta$
From Jech's Set Theory:
Lemma 2.11.
(iii) If $α\ne β$ are ordinals and $α ⊂ β$, then $α ∈ β$.
Proof: If $α ⊂ β$, let $γ$ be the least element of the set $β − α$. Since $α$ is transitive, it follows ...
1
vote
1
answer
194
views
How to define ordinal addition
From Jech's Set Theory:
We shall now define addition, multiplication and exponentiation of ordinal numbers, using Transfinite Recursion.
Definition 2.18 (Addition). For all ordinal numbers $\alpha$
$...
2
votes
0
answers
67
views
Step in Jech's proof that any well-ordered set is isomorphic to an ordinal
From Jech's Set Theory:
Proof. The uniqueness follows from Lemma 2.7. Given a well-ordered set $W$,
we find an isomorphic ordinal as follows: Define $F(x) = \alpha$ if $\alpha$ is isomorphic
to the ...
0
votes
1
answer
84
views
Trouble understanding Proof Wiki's proof that every well-ordered set is order isomorphic to an ordinal.
I'm struggling to understand a couple of steps in this proof showing that every well-ordered set is order isomorphic to an ordinal. Calling the following steps
first and second respectively, my ...
0
votes
1
answer
85
views
Question about a proof that addition from the left preserves the order of ordinals
I found the following proof that ordinal addition from the left preserves strict inequalities (see b)).
I am having trouble understanding the limit stage. The proof uses the inequality
$\gamma + \...
1
vote
1
answer
137
views
Question about a proof that any well-ordered set is isomorphic to a unique ordinal
I am studying a proof that every well-ordered set is isomorphic to a unique ordinal. However, I don't understand why $A = pred(\omega)$ (see yellow).
One direction is clear:
Let $x \in pred(\omega)$, ...
2
votes
0
answers
155
views
Existence of supremum of a set of ordinals
I'm trying to understand why the statement in yellow below is true.
The definition of an ordinal in the lecture notes is as follows:
I am also adding Theorem 29, Lemma 33 and Lemma 34 for reference:
...
1
vote
1
answer
82
views
Explanation - $\alpha = \sup{(C \cap \alpha)}$
I need help understanding a section of the proof of lemma 7.9 of the book "A course in Set Theory" by Ernest Schimmerling, more precisely the step where the author says
[...] Easily, we see ...
2
votes
1
answer
226
views
Why is cf(α) a cardinal for any limit ordinal α? [duplicate]
In Jech, one of the lemmas state that for every limit ordinal α, cf(α) is a regular cardinal. Every source I tried to search on the Internet claimed that it was obvious to see that cf(α) should be an ...
1
vote
1
answer
161
views
Questions about the induction on cardinals
From Hereditary Cardinality and Rank :
For an infinite cardinal $\kappa$, $$\forall x,\ \textrm{hcard
}x<\kappa\rightarrow\textrm{rank }x<\kappa$$
We can show this by induction on $\kappa$. ...
3
votes
2
answers
106
views
The axiom of regularity in a fact about $V_\omega$
I'm working out the details in this proof (from here), and I have some questions about the second part ("For the other direction, ...").
First, why is it possible to assume WLOG that $A$ is ...
1
vote
1
answer
355
views
Is true that $\bigcup α=α$ if $\alpha$ is a limit ordinal?
Definition
Given a set $A$ the membership relation on $A$ is the relation defined by the identity
$$
\in_A:=\{a\in A\times A:a_1\in a_2\}
$$
Definition
A set $A$ is said transitive if ech its element $...
1
vote
1
answer
108
views
Prove: If $A$ and $B$ are closed subsets of $[0,\Omega]$ then at least $A$ or $B$ is bounded
As usual, I am self studying topology and my knowledge of ordinals is meagre. Have
done some research on it.
Theorem 5.1 Any countable subset of $[0,\Omega)$ is bounded above.
(This exercise requires ...
0
votes
2
answers
209
views
Prove, using the definition of a closed set, that $S = [0, \Omega)$ is not a closed subset of $X = [0, \Omega]$ with order topology
I am using the book A First Course in Topology by Robert Conover. I can assume everything up to this point and info on ordinals.
Prove using the definition of a closed set that S = [0,$\Omega$) is not ...