# Tag Info

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### A rare property of Hausdorff spaces

Yes, there is such a space. Let $X=2^{\omega_1}$ be the space of binary sequences of length $\omega_1$, in the order topology generated by the lexical order. So $X$ consists of the branches through ...

### A rare property of Hausdorff spaces

How about this example: $[0,1]^A$ with the product topology, where $A$ is uncountable. Then every nonempty $G_\delta$ set is uncountable. Since your sets $f^{-1}(x)$ are $G_\delta$ sets, this has ...
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### Closed balls vs closure of open balls

The following theorem (or its corollary) implies negative answer to the original question. Theorem. For any point $x$ of a metric space $(X,d)$ the set $R_x:=\{r>0:cl(B(x,r))\ne \bar B(x,r)\}$ ...
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### How (non-)computable is set theory?

The question is extremely interesting, and I have looked into this kind of thing with various colleagues (including Russell Miller and Kameryn Williams), although our investigation has not yet ...

### Examples of statements with a high quantifier complexity

My favorite example is the 5-quantifier definition of an almost-periodic function: $f$ is almost-periodic iff \forall \epsilon>0\, \exists t>0\ \forall a\ \exists s \in [a,a+t]\ \forall x\, |f(...
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### Is there a Borel subset of $\mathbb{R}^{2}$, with finite vertical cross-sections, whose projection onto the first component is non-Borel?

No, no such set exists. This is a special case of the Lusin–Novikov theorem; see e.g. Kechris, Classical Descriptive Set Theory, Theorem 18.10. In general, let $X,Y$ be standard Borel spaces, ...
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### Two strengthenings of "strong measure zero"

Strategically strong measure zero is equivalent to countable. To prove the nontrivial direction, suppose $X$ is strategically strong measure zero and $s$ is a winning strategy for player II. Consider ...
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### Application of Fraïssé construction in set theory

A Fraïssé construction lies at the heart of the proof of my embedding theorems. Theorem. (J. D. Hamkins, Every countable model of set theory embeds into its own constructible universe, JML 13(2), ...
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### Partitions of the real line into Borel subsets

The answer to both problems is no! If the Cohen forcing is used to add lots of reals to a countable transitive model of GCH, then in the resulting extension, any partition of the real line into Borel ...
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### Woodin on Posner-Robinson for the hyperjump and sharp

MR2449474 (2009j:03067) Woodin, W. Hugh. A tt version of the Posner-Robinson theorem. Computational prospects of infinity. Part II. Presented talks, 355–392, Lect. Notes Ser. Inst. Math. Sci. Natl....
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### Can an ultrapower be undone by forcing?

For set-forcing, the answer is no, see the following article Joel David Hamkins, Greg Kirmayer, and Norman Lewis Perlmutter, Generalizations of the Kunen inconsistency, Ann. Pure Appl. Logic 163 (...
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Accepted

### Does Turing determinacy imply full determinacy?

This is open. In $L(\mathbb R)$ the answer is yes. Hugh has several proofs of this, and it remains one of the few unpublished results in the area. The latest version of the statement (that I know of) ...

### Application of Fraïssé construction in set theory

If you are willing to allow uncountable generalizations of Fraisse's/Hrushovski construction then the conditions you need essentially are those that make up an Abstract Elementary Class. Abstract ...
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### Descriptive Complexity of Knot Equivalence

A perfect timing for this question, since I just uploaded a paper on this topic to arXive (see below). Let us specify the definitions. A knot is a homeomorphic image of the circle in $\mathbb{R}^3$. ...

### Can Sacks forcing add a Cohen generic real over $L$?

Regarding Question 2: Zapletal's solution to the "Half-Cohen problem" gives an example of a 2-stage iteration $P*\dot Q$ such that $P$ does not add Cohen reals, and $P$ forces that $\dot Q$ does not ...
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### Ordinal-definable witnesses to the perfect set property?

Let $A$ be the set of those reals that are not OD. Then $A$ is OD (since I've just defined it), and it's uncountable (since its complement, being a well-orderable set of reals, must be countable ...
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### What is the descriptive complexity of a set added by Cohen forcing?

Since Cohen forcing is weakly homogeneous, all hereditarily ordinal definable sets in the Cohen extension are already in the ground model. That applies in particular to any ordinal definable real, ...
This is known as Rudin-Shelah problem. Note that, by Stone duality, this is equivalent to determine the self-homeomorphism group of the Stone-Cech boundary of $N$. Notably, consider the group induced ...
### Uncountable disjoint closed coverings of $[0,1]$
This is question has a long and interesting history, which is discussed in Arnie Miller's paper cited below. The first construction of a model of ZFC + $\aleph_1 < 2^{\aleph_0}$ where $[0,1]$ can ...