I'm trying to work my way through Baumgartner's proof of Hindman's theorem, as published in $\textit{Journal of Combinatorial Theory}$, specifically, a line in the proof of Lemma 2.
Definitions
-
$\mathcal{F}$ is the family of all finite non-empty subsets of $\mathbb{N}$.
-
A $\textbf{disjoint union}\ \mathcal{D}$ is an infinite subset of $\mathcal{F}$, all of whose elements are pairwise disjoint.
-
FU$(\mathcal{D})$ is the family of all finite unions of elements of $\mathcal{D}$, excluding the empty union.
-
Finally, $\mathcal{X}$ is $\textbf{large for}$ $\mathcal{D}$ if for every disjoint collection $\mathcal{D}'\subseteq\text{FU}(\mathcal{D})$, $\text{FU}(\mathcal{D}')\cap\cal{X}\neq\emptyset$.
Now, Lemma 2 states that suppose $\cal{X}$ is large for $\cal{D}$, then there is a finite set $\mathcal{E}\subseteq\text{FU}(\cal{D})$ such that for all $X\in\text{FU}(\cal{D})$, if $X\cap\bigcup_{E\in\cal{E}} E=\emptyset$, then there is a $D\in\text{FU}(\cal{E})$ such that $X\cup D\in\cal{X}$.
Baumgartner constructs, for a contradiction, a sequence $X_0,X_1,\cdots$ of pairwise disjoint elements of FU$(\cal{D})$ such that for all $n\geq 1$ and all $D\in\text{FU}(\{X_i\ :\ i<n\})$, $X_n\cup D\not\in\cal{X}$. He then defines $Y_n=X_{2n}\cup X_{2n+1}$ and claims that $\mathcal{D}'=\{Y_n\ :\ n\geq 0\}$ is a disjoint collection $\subseteq\text{FU}(\cal{D})$ and that FU$(\cal{D}')\cap X=\emptyset$, contradicting $\cal{X}$ large for $\cal{D}$.
My question is this: why bother with the $Y_n$? If we defined $\mathcal{D}''=\{X_n\ :\ n\geq 0\}$, then it is a disjoint collection by construction and FU$(\mathcal{D}'')\cap\cal{X}=\emptyset$, also by construction. What am I missing?
Best Answer
The negation of Lemma 2 guarantees that for every $x_n$ that you add to the sequence, the following holds: $x_n \bigcup{d} \notin{\cal{X}}$ for all $d\in{FU(\lbrace{x_0, x_1, ..., x_{n-1}}\rbrace)}$. Nevertheless, it might be the case that some (or all) of the $x_i$ are members of $\cal{X}$. This problem is avoided by taking the union of pairs of elements from the sequence since we know that $x_i\bigcup{x_j}\notin{\cal{X}}$ for all different $i,j\geq{0}$.