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forcing, large cardinals, descriptive set theory, infinite combinatorics, cardinal characteristics, forcing axioms, ultrapowers, measures, reflection, pcf theory, models of set theory, axioms of set theory, independence, axiom of choice, continuum hypothesis, determinacy, Borel equivalence relations, Boolean-valued models, embeddings, orders, relations, transfinite recursion, set theory as a foundation of mathematics, the philosophy of set theory.

6 votes

"Universal" infinite graph (with respect to minors)

For countably infinite graphs you can take the Rado Graph, which contains all countable graphs (even as induced subgraphs). For higher cardinals, see this paper of Shelah. As I mention in a comment, …
Tony Huynh's user avatar
  • 32.1k
8 votes

Games that never begin

As a supplement to Joel's answer, you may want to look at this nice paper of Bollobas, Leader, and Walters concerning continuous games. As a starting point they discuss the classical Lion and Man gam …
Tony Huynh's user avatar
  • 32.1k
7 votes

Uncountable family of infinite subsets with pairwise finite intersections

I think that this was answered by Andres Caicedo in a comment to an answer to this question. I quote: Given an infinite sequence of 1s and 2s, its initial segments are numbers (written in decim …
Tony Huynh's user avatar
  • 32.1k
5 votes
Accepted

"Strongly" almost disjoint subsets

No. For each element $x \in \kappa$, let $g(x)$ be the set of elements in $\frak{E}$ that contain $x$. By assumption, for all $x \in \kappa$, we have $|g(x)| \leq \kappa$. We may clearly assume $\e …
Tony Huynh's user avatar
  • 32.1k
1 vote

Terminology for relation on sets

There is a related concept dealing with separations as opposed to sets. A separation of a set $U$ is an ordered pair $(A,B)$ such that $A$ and $B$ are disjoint subsets of $U$ whose union is $U$. We …
Tony Huynh's user avatar
  • 32.1k
13 votes
2 answers
1k views

Ultrafilters vs Well-orderings

This question was actually asked by John Stillwell in a comment to an answer to this question. I thought I would advertise it as a separate question since no one has yet answered and I am also curiou …
Tony Huynh's user avatar
  • 32.1k
15 votes

A function that is defined everywhere but has unknown values

What about the (diagonal) Ramsey Numbers, $k \to R(k,k)$? These are fairly natural to define but notoriously hard to compute. Indeed, only the first two Ramsey numbers $R(3,3)=6$ and $R(4,4)=18$ are …
Tony Huynh's user avatar
  • 32.1k
3 votes
Accepted

Coloring hypergraphs with no singleton intersections

To answer Jon Noel's question in the comments, there is no such example for finite hypergraphs. Claim. Let $H=(V,E)$ be a finite hypergraph such that $|e| > 1$ for all $e \in E$ and for all distinc …
Tony Huynh's user avatar
  • 32.1k
5 votes
Accepted

Clutters with no maximum-size matchings

Yes, this is possible. For each prime $p$ and $c \in \{0,1, \dots, p-1\}$ let $A_{c,p}=\{c+kp \mid k \in \mathbb{Z}\}$. Clearly, the set of all $A_{c,p}$ is a clutter $\mathcal C$ with ground set $\ …
Tony Huynh's user avatar
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10 votes
Accepted

Hadwiger-Nelson problem for $\ell^\infty$

No. The set of all $\{0,1\}$-sequences is also a clique in $G$. Thus, $\chi(G) \geq 2^{\aleph_0}$. On the other hand, the set of all bounded real sequences has size $2^{\aleph_0}$, so $\chi(G)=2^{\ …
Tony Huynh's user avatar
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12 votes
5 answers
6k views

Subset of the plane that intersects every line exactly twice

In a comment to this question, Tim Gowers remarked that using the axiom of choice, one can show that there exists a subset of the plane that intersects every line exactly twice (although it has yet to …
Tony Huynh's user avatar
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96 votes
16 answers
34k views

Most 'unintuitive' application of the Axiom of Choice?

It is well-known that the axiom of choice is equivalent to many other assumptions, such as the well-ordering principle, Tychonoff's theorem, and the fact that every vector space has a basis. Even tho …