New answers tagged infinite-combinatorics
5
votes
Shifting an irrational binary sequence
I'll prove a general theorem. Unlike in my comments, I won't use any specific theorems.
Let $A$ be a finite alphabet and $X \subset A^{\mathbb{N}}$ a subshift of finite type, or SFT, meaning $X$ is ...
- 6,652
12
votes
Accepted
Shifting an irrational binary sequence
No, there's no irrational $s$ with this property. Here's a concrete hands-on argument. (This may be equivalent to Ville Salo's comment, but I'm not familiar with that terminology.)
The sequence $d = s\...
- 6,486
2
votes
Accepted
Image and pre-image integer choice function
Yes. Denote $X_k=(k, f(k))\in \newcommand{\Nplus} {\mathbb{N}^+}\Nplus\times \Nplus$. We want to construct the point set $A:=\{X_1,X_2,\ldots\}\subset \Nplus\times \Nplus
$ so that:
$A$ contains ...
- 109k
2
votes
Is there an uncountable extension of the Ramsey set $[\omega]^2$?
If $N$ is an infinite proper subset of $\omega$ then the family $\mathcal A=[\omega]^2\cup\{A\subseteq\omega:A\not\subseteq N\}$ is a Ramsey family and $|\mathcal A|=2^{\aleph_0}$.
On the other hand, ...
- 13.4k
4
votes
Accepted
Is there an uncountable extension of the Ramsey set $[\omega]^2$?
Yes. Just take $\mathcal{A}$ to be $[\omega]^2$ together with the powerset of the even integers.
- 32.1k
13
votes
Accepted
Is there a sparse almost disjoint family over $\omega$ of cardinality $2^{\aleph_0}$?
Yes. Let $\mathcal F$ be an almost disjoint family of subsets of $\{n^2:n\in\omega\}$ of cardinality $2^{\aleph_0}$.
- 13.4k
8
votes
Is there a sparse almost disjoint family over $\omega$ of cardinality $2^{\aleph_0}$?
Yes. Consider the function $f:2^{<\omega}\rightarrow 2^{<\omega}$ defined as follows:
$f(\langle\rangle)=\langle\rangle$.
Having defined $f$ on all strings of length $\le n$, we define $f$ on ...
- 21.1k
0
votes
Ramsey-theoretic properties of Erdős cardinals
This answer will answer question 1 and almost answer question 2, giving a positive answer to question 2 when $\beta$ is a limit ordinal.
The linked paper cites theorem 6.1 of Schmerl's "On $\...
- 2,031
9
votes
Accepted
Maximal Ramsey families
No. Consider the family $\mathcal{R}$ of all sets of size 1. It is certainly a Ramsey family. Consider a Ramsey extension $\mathcal{R}_1\supset \mathcal{R}$. Take a map $f$ which maps all sets from $\...
- 109k
4
votes
Accepted
Non-Ramsey function $f:[\omega]^{<\omega}\to\{0,1\}$
For $P \subset \omega$ finite, let $f(P) = \#P \bmod{2}$, i.e., $f(P) = 0$ if $P$ has even cardinal, and $1$ otherwise.
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