# Tagged Questions

An important and fundamental axiom in set theory sometimes called Zermelo's axiom of choice. It was formulated by Zermelo in 1904 and states that, given any set of mutually disjoint nonempty sets, there exists at least one set that contains exactly one element in common with each of the nonempty ...

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### Partial Universes and the Axioms of $ZF$ Set Theory Without Choice

In his Senior Thesis, Samuel Coskey answered the question of which axioms of $ZFC$ hold at each stage of the cumulative hierarchy. Here is the list of his results: Axioms that always hold: ...
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### Without Choice: Are there filters of cardinality continuum?

Is it provable, in ZF (without Choice), that every filter can be extended to one of cardinality continuum? The extended filter is not requested to be an ultrafilter.
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### Choice and the Baire property in non-separable complete metric spaces

It's known to be consistent with ZF+DC that every subset of $\mathbb{R}$ has the Baire property (BP). (E.g. Shelah's model). If so, then every subset of every complete separable metric space has ...
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### What are the minimal requirements for the definable hyperreal field plus transfer?

It is interesting that to prove the transfer principle for the definable hyperreal field, one requires no more choice than for proving, for instance, the countable additivity of the Lebesgue measure. ...
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### Constructively, are all fibrations cloven?

A "cloven fibration" is a fibration for which we have an explicit choice of cartesian liftings; this is often phrased as, "We can pick a lifting without using the axiom of choice". Firstly, I'm a bit ...
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### $2M = M$ and its subset

I have some question concerning arithmetic of cardinal in ZF. Write $X = Y$ if there is a bijection between them. Let $M$ be a set such that $2M = M$. Can I show, in ZF, that any infinite subset $X$ ...
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### Well-ordering of power set of $\omega$

Assuming initially the background set theory ZF, what is the exact status of the existence of a well-ordering on the power set of $\omega$? How much needs to be added to guarantee this?
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### Transfer with minimal choice

Let FUF postulate the existence of a Free UltraFilter on $\mathbb{N}$ and ACC the axiom of countable choice. Consider the superstructure on $\mathbb{R}$ and its inclusion in the bounded ultrapower. ...
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### Is there any infinite set which is Dedekind finite and weakly Dedekind finite?

Is there any infinite set which is Dedekind finite and weakly Dedekind finite? If $X$ is a weakly Dedekind finite amorphous set, can we show that $\mathcal P(X)$, the power set of $X$, is also weakly ...
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### The axiom $I_0$ in the absence of $AC$

It is well-known that if $AC$ holds and if $j: L(V_{\lambda+1}) \to L(V_{\lambda+1})$ is a non-trivial elementary embedding with $crit(j) < \lambda,$ then $\lambda$ has countable cofinality (and in ...
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### Does Con(ZF + Reinhardt) really imply Con(ZFC + I0)?

The question is: if I assert in ZF that there exists a Reinhardt cardinal, do I really get a theory of higher consistency strength than when I assert in ZFC that there exists an I0 cardinal (the ...
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### Does “$|{\cal P}_2(X)| = |X|$ for $X$ infinite” imply ${\sf (AC)}$?

This comes from a comment made by user bof in this thread. Let $X$ be a set, define ${\cal P}_2(X) = \big\{\{a, b\}: a\neq b\in X\big\}$. Consider the statement ${\sf (S)}$ If $X$ is an ...
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### How much choice does a linear or well-order on cardinals imply?

It is well-known that if the natural (partial) order on the class of cardinal numbers is a linear order, then it is in fact a well-order and the axiom of choice holds. I was, however, interested in ...
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### Is the universality of the surreal number line a weak global choice principle?

I'd like to consider the principle asserting that the surreal number line is universal for all class linear orders, or in other words, that every linear order (including proper-class-sized) linear ...
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### What sort of large cardinal can continuum be?

I have stumbled across a related question asking which large cardinal properties can hold for $\aleph_1$. This question is probably also related, asking in what ways $\aleph_0$ is a "large" cardinal. ...
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### Is tree property inconsistent with Berkeley cardinals in the absence of Axiom of Choice?

On one hand due to Kunen's inconsistency theorem it is known that within $\sf ZF$, large cardinal axioms beyond Reinhardt cardinal are inconsistent with $\sf AC$. Also some recent results of Bagaria, ...
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### Axiom of choice as zero dimensionality

In the paper Quantifiers and Sheaves by Lawvere, at the bottom of the second page, the author writes: "... the condition that every epi splits, which geometrically we would call 0-dimensionality ...
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### Maximal chains and antichains of statements weaker than AC

First, I would like to say that I asked this question (a more general one actually) on math.stackexchange.com. Consider the set $\varPhi$ of statements in the language of ZF that are weaker than AC (...
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### Does Łoś's theorem imply choice given a free ultrafilter?

In the paper "Łoś's theorem and the boolean prime ideal theorem imply axiom of choice" Howard has shown that Łoś's theorem and the boolean prime ideal theorem imply axiom of choice. At the end of the ...
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### Darboux property of non-atomic sigma-additive nonnegative measures equivalent to the AC?

A result commonly, and probably erroneously, attributed to W. Sierpiński is that every non-atomic, countably additive, nonnegative measure $\mu: \Sigma \to \bf R$, where $\Sigma$ is a sigma-algebra on ...
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### Axiom of choice and a set in the plane that intersects every line in two points

In this question Subset of the plane that intersects every line exactly twice someone ask for a reference of a paper where they proof the result : ''There exist a subset of the plane that intersects ...
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### Finding non convex functions satisfying a weak form of convexity, without the axiom of choice

If a real-valued function $f$ over reals satisfies $$(1) \; \; \; f({x+y\over2})\le {f(x)+f(y)\over2},$$and it is continuous, then it is not hard to see that $f$ is indeed convex. On the other hand,...
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### How close to being well-orderable does this make my powerset?

Let's work in a set theory without assuming AC (for instance, but not necessarily, ZF). Fix a set $k$ satisfying $k\times k \simeq k$, and consider its powerset $X = 2^k$. I have a technical condition ...
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### Replacing Axiom of Choice with Axiom of Countable Choice

Many people find ACC more intuitive than AC ("Pick something from the first set, then something from the second set, then...) and it also doesn't lead to "controversial consequences" (See for eg: ...
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### The role of the rigid relation principle ($RR$) in the Kunen inconsistency

Consider the rigid relation ($RR$) principle, i.e. "every set admits a rigid binary relation", that is,"that for every set $A$ there is a binary relation $R$ on $A$ such that the structure $(A,R)$ is ...
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### Notions of infinity in $\mathsf{ZF}$ without choice

Consider the following statements about a given set $X$ in in $\mathsf{ZF}$: (1) There is $x_0\in X$ such that there is a surjective map $\varphi: X\setminus\{x_0\}\to X$. (2) There is an injective ...
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### Independence of the countable axiom of choice

How does one proove that the Countable axiom of choice is not provable in ZF?Is there any brief proof?Does the Independence of the countable axiom of choice implies the independence of the axiom of ...
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### Quasi-disjoint subsets of an infinite set and $\neg \mathsf{AC}$

Is it consistent with $\mathsf{ZF}$ (without $\mathsf{AC}$) that there is an infinite set $X$ and a subset $S\subseteq\mathcal P(X)$ of the same cardinality as $\mathcal P(X)$ with the property that ...
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### Noetherianess of a finite module over a noethrian ring without Axiom of Choice

All rings are assumed to be commutative with 1. We say a module over a ring is strictly noetherian if every non-empty set of submodules has a maximal member. We say a ring is strictly noetherian if it ...
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### Matching power series to infinity

As pointed out by Makoto, on this question about power series rings and the axiom of choice, an idea I had needed the axiom of dependent choice to work. However, the construction raises another ...
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### Are two forms of the Dual Schroeder-Bernstein property equivalent?

We know the Shroeder-Bernstein (SB) theorem can be proved in ZF, while the Dual Schroeder-Bernstein (DSB) can be proved in ZF+AC but not in ZF. Define as ISB the property that whenever there are both ...
Wikipedia states that, in $ZF$, the Axiom of Choice ($AC$) implies the Hahn-Banach theorem, but that the Hahn-Banach theorem does not imply $AC$. It also states that in $ZF$, the Hahn-Banach theorem ...