# Questions tagged [transfinite-induction]

Transfinite induction is an extension of mathematical induction to well-ordered sets, for example to sets of ordinal numbers or cardinal numbers.

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### Question on the hierarchy of named Ordinals

I was wondering how many named transfinite ordinals there are, and what the notations and names are, arranged as a hierarchy which ignores the different levels on a single level of Ordinals (ω+1, 2ω, ...
• 179
1 vote
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### What contradiction arises from asuming strong induction on a set with no minimum?

Proposition: Let X be a totally ordered non-empty set such that whenever a subset A⊆X satisfies ∀x [(∀y<x ⟹ y∈A)⟹x∈A]; x,y∈X then A=X. Then X is well-ordered. This is the proposition that I'm ...
• 137
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### A quick question about Set Theory and Uncountable Infinity [duplicate]

I noticed a factor about Set Theory that has somewhat confused me. Assuming CH, $2^{\aleph_0} = \aleph_1$. Taking into account Combinatorics, you can treat a number as a set of numbers; 0, 1, 2, 3, 4, ...
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### Confusing about how the lemma can be stated

Let's consider the following: Lemma Let $g\in F(\beta)$ and $f\in \Bbb R^\Bbb R$ such that $f\restriction M= g\restriction M$ for some $M\subset \mathbb R$. Then $f\in F(\beta).$ Of course, no one ...
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### Prove by induction on $\alpha < 2^{\aleph_0}$ that $|\mathcal G_\alpha| < 2^{\aleph_0}$ and $\mathcal G_\alpha$ is not an ultrafilter over $\omega$

The following question is exercise 7.4 from Ernest Schimmerling's A Course on Set Theory which involves a construction from Tarski's theorem that extends filters to ultrafilters. I couldn't find much ...
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1 vote
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### If we can prove transfinite induction, why can't we prove induction?

Let $(J, <)$ be a well ordered set. Let $S_\alpha \equiv \{ j \in J: j < \alpha \}$ ( section by $\alpha$ of $J$). Transfinite induction says that for any subset $J_0$, if the following property ...
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### Does there exist a Bernstein set such that a finite sum still Bernstein?

Recall that $B\subset\Bbb R$ is a Bernstein set if $B\cap P\neq\emptyset \neq P\setminus B$ for every perfect set $P\subset\Bbb R.$ It can be constructed by an easy transfinite induction. Moreover, ...
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### About Construction a one-to-one function from $(a,b)$ onto $[a,b]$

This question asked to construct one-to-one function from $(a,b)$ onto $[a,b]$. I know there is a function but it seems the question to define this function explicitly. How this can be done? Edit I ...
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### Recursion theorem for ordinals proof

I'm trying to understand the proof of the recursion principle of ordinals, the theorem is: The proof of this theorem uses this other theorem: The proof is pretty long (I'm sorry) so I will try to ...
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### Couldn't we state the principle of transfinite induction without the $0 \in A$ condition?

Principle of transfinite induction for ordinals: Let $A$ be a subset of the ordinal $\alpha$ such that $0 \in A$ for all $\beta \in \alpha$ if $\beta \subseteq A$ then $\beta \in A$ Then $A=\alpha$ ...
1 vote
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### Prove the limit step for $\Gamma (\alpha, \alpha)\le \omega^\alpha$

I'm working through Jechs Set Theory right now and encountered an exercise (3.5) to prove that the order type of $\Gamma(\alpha, \alpha):=\{(\xi, \eta): (\xi, \eta)<(\alpha, \alpha)\}$ is less than ...
1 vote
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### Partition $\mathfrak c$-dense set to $\mathfrak c$- many dense set.

A subset of $A\subset\mathbb R$ is called $\mathfrak c$-dense if $|A\cap I|=\mathfrak c$ for any open interval $I\subset\mathbb R.$ Then, there is a partition for $A$ to continuum many dense set. ...
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### $\mathbb R$ as continuum many of pairwise disjoint of Bernstein sets

The set $B\subset\mathbb R$ is called Bernstein set if neither $B$ nor $\mathbb R\setminus B$ contains any perfect sets. Theorem: $\mathbb R$ can be written as continuum many of pairwise disjoint ...
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### family of pairwise disjoint sets in the complement of meager. set

Let $M$ be a meager subset of $\mathbb R$. I want to construct the following family in $\mathbb R\setminus M$ $$F:= \{A_{r}^\xi\colon r\in\mathbb R \ \&\ \xi<c\}$$ Such that all elemnets of $F$...
• 2,153
1 vote
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### Extension linear independent set to Hamel basis

If we have a linear independent set, then it is well known that by using Zorn's lemma it can be extended to Hamel basis. My question I have a linear independent set, call it $B_{0}$, I want to ...
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### Proof with transfinite induction

I'm trying to prove the following statement: Suppose that for every $r\in\mathbb{R}$ we are given a finite set $A_r\subseteq\mathbb{R}$ and that for any finite set $D\subseteq\mathbb{R}$, there ...
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### split perfect set into countable many pairwise disjoint perfect sets

We know that each perfect set can be written as a continuum many pairwise disjoint many perfect set. This will rely on the well know theorem which says: Let $X$ be a nonempty perfect polish space. ...
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### How does Reinhardt's extension of the set-theoretic universe beyond $V_\Omega$ work?

In this answer it is stated that in William Reinhardt's paper 'Remarks on reflection principles, large cardinals, and elementary embeddings' (1974) Reinhardt suggests extending the set-theoretic ...
• 307
1 vote
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### Transfinite induction

Is it true that while using transfinite induction we dont need to prove the zero case? because, if we want to prove some property $\psi$ , we assume that for any $x\in A$ if for any $y\leq x$ it ...
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Over classical logic, the induction and well-ordering schemas are equivalent. These schemas state the following, given any linear ordering $(W,<)$ and property $Q$ on $W$: Induction: \$∀k{∈}W\ ( \ ∀...