# proof of delta system lemma

Since there are only $\aleph_{0}$ possible cardinalities for any element of $S$, there must be some $n$ such that there are an uncountable number of elements of $S$ with cardinality $n$. Let $S^{*}=\{a\in S\mid|a|=n\}$ for this $n$. By induction, the lemma holds:

If $n=1$ then there each element of $S^{*}$ is distinct, and has no intersection with the others, so $X=\emptyset$ and $S^{\prime}=S^{*}$.

Suppose $n>1$. If there is some $x$ which is in an uncountable number of elements of $S^{*}$ then take $S^{**}=\{a\setminus\{x\}\mid x\in a\in S^{*}\}$. Obviously this is uncountable and every element has $n-1$ elements, so by the induction hypothesis there is some $S^{\prime}\subseteq S^{**}$ of uncountable cardinality such that the intersection of any two elements is $X$. Obviously $\{a\cup\{x\}\mid a\in S^{\prime}\}$ satisfies the lemma, since the intersection of any two elements is $X\cup\{x\}$.

On the other hand, if there is no such $x$ then we can construct a sequence $\langle a_{i}\rangle_{i<\omega_{1}}$ such that each $a_{i}\in S^{*}$ and for any $i\neq j$, $a_{i}\cap a_{j}=\emptyset$ by induction. Take any element for $a_{0}$, and given $\langle a_{i}\rangle_{i<\alpha}$, since $\alpha$ is countable, $A=\bigcup_{i<\alpha}a_{i}$ is countable. Obviously each element of $A$ is in only a countable number of elements of $S^{*}$, so there are an uncountable number of elements of $S^{*}$ which are candidates for $a_{\alpha}$. Then this sequence satisfies the lemma, since the intersection of any two elements is $\emptyset$.

Title proof of delta system lemma ProofOfDeltaSystemLemma 2013-03-22 12:55:03 2013-03-22 12:55:03 Henry (455) Henry (455) 5 Henry (455) Proof msc 03E99